ELECTRIC FORCE AND ELECTRIC CHARGE
6
1. Introduction
6
2. Charge Quantization and Charge Conservation
8
3. Conductors and Insulators
8
THE ELECTRIC FIELD
10
1. Introduction
10
2. The Superposition of Electric Forces Example Problem 1: Electric Field of a Charged Rod
10 11
3. The Electric Field Example Problem 2: Electric Field of Point Charge Q. Example Problem 3: Electric Field of Charge Sheet. Example Problem 4: Electric Field of Multiple Charge Sheets
13 13 13 15
4. Field Lines
16
5. Electric Dipole in an Electric Field
17
GAUSS' LAW
19
1. Introduction
19
2. Gauss' Law Example Problem 1: Electric Field of a Point Charge. Example Problem 2: Electric Field of a Charge Sheet
19 20 20
3. Conductors in Electric Fields
22
THE ELECTROSTATIC POTENTIAL
24
1. Introduction
24
2. Calculating the Electrostatic Potential Example Problem 1: Electrostatic Potential of a Charged Rod Example Problem 2: Distance of Closest Approach
25 26 27
3. The Electrostatic Field as a Conservative Field
28
4. The Gradient of the Electrostatic Potential Example Problem 3: Electric Field derived from Electrostatic Potential Example Problem 4: Potential and Field of a Charged Annulus
28 30 31
5. The Potential and Field of a Dipole
32
1
ELECTRIC ENERGY OF A SYSTEM OF POINT CHARGES
34
1 Introduction Example Problem: Model of a Carbon Nucleus
34 35
2 Energy of a System of Conductors Example Problem: Fission of Uranium
35 38
CAPACITORS AND DIELECTRICS.
42
1. Introduction
42
2. The parallel-plate capacitor Example Problem: The Geiger Counter
42 43
3. Capacitors in Combination Example: Multi-plate Capacitor
44 46
Example Problem: Capacitors in Series/Parallel
47
4. Dielectrics Example Problem: The Parallel Plate Capacitor
49 50
5. Gauss Law in Dielectrics Example Problem: The Spherical Capacitor.
51 52
6 Energy in Capacitors Example Problem: Capacitors in Parallel. Example Problem: Energy Stored in Capacitors.
54 54 55
CURRENTS AND OHMS LAW
57
1. Electric Current Example: Resistance of a Wire
57 59
2. The resistivity of materials Example: Connecting an AC Example: HV Lines
60 60 61
3. Resistance in combination Example: Superconducting Cables Example: Resistors in Circuits
61 63 64
DC CIRCUITS
67
1. Electromotive Force
67
27.2. Single-loop currents
68
3. Multi-loop circuits. Example Problem: Multi-loop Circuit
70 73 2
4. Energy in circuits Example Problem: High-Voltage Transmission Line Example Problem: Charging a Battery Example Problem: Draining a Battery
74 75 76 77
5. The RC circuit
78
THE MAGNETIC FORCE AND FIELD.
81
1. The magnetic force Example Problem: Magnetic Field of a Neutron Star
81 82
2. The Biot-Savart Law Example Problem: Helmholtz Coils Example Problem: Magnetic Field due to a Long Wire
83 84 86
3. The magnetic dipole Example Problem: Spinning Charged Disk
89 89
AMPERES LAW
91
1. Introduction Example Problem: Field due to six parellel wires
91 92
2. The solenoid Example Problem: Superposition of magnetic fields Example Problem: Coaxial cable
93 94 94
3. Motion of charges in electric and magnetic fields
96
4. Crossed electric and magnetic fields
98
5. Forces on a wire Example Problem: Magnetic balance
100 101
6. Torque on a current loop
102
105
ELECTROMAGNETIC INDUCTION 1. Motional emf Example Problem: Metal Rod in Magnetic Field Example Problem: Induced EMF in a Solenoid
105 106 108
2. The Induced Electric-Field
109
3 Inductance Example Problem: Mutual Induction
110 111
4. Magnetic Energy Example Problem: The Toroid
111 113
5. The RL circuit
114 3
Example Problem: Joule Heat in RLCircuit
115
117
MAGNETIC MATERIALS 1. Magnetic Moments Example Problem: Two-electron Interactions
117 118
2. Paramagnetism Example Problem: Filled Solenoid
119 120
3. Ferromagnetism Example Problem: Number of Aligned Electrons
121 121
4. Diamagnetism
122
AC CIRCUITS
125
1. Alternating Current
125
2. AC Resistor Circuits
125
3. AC Capacitor Circuits
126
4. AC Inductive Circuit Example Problem: AC Circuit
127 128
5. LC Circuits
129
6. The Phasor Diagram Example Problem: LCC Circuit Example Problem: RC Circuit
136 137 139
7. The Transformer
140
THE DISPLACEMENT CURRENT AND MAXWELLS EQUATIONS 143 1. THE DISPLACEMENT CURRENT
143
Example Problem: Parallel-Plate Capacitor
144
2. Maxwells Equations Example Problem: Conservation of Charge
147 147
3. Cavity Oscillations
148
4. The Electric Field of an Accelerated Charge Example Problem: Radio Antenna
150 153
5. The magnetic field of an accelerated charge
154
4
LIGHT AND RADIO WAVES
155
1. Electromagnetic waves Example Problem: Radio Receiver
155 158
2. Plane Harmonic Waves Example Problem: Circularly Polarized Waves Example Problem: Polarization of Electromagnetic Waves
159 160 161
3. The Generation of Electromagnetic Waves
163
4. Energy of a Wave Example Problem: Energy of a Laser Beam Example Problem: Energy and Current Flow
164 166 166
5. Momentum of a Wave Example Problem: Solar Sails
168 169
6. The Doppler Shift of Light
170
5
ELECTRIC FORCE AND ELECTRIC CHARGE 1. Introduction Ordinary matter consists of atoms. Each atom consists of a nucleus, consisting of protons and neutrons, surrounded by a number of electrons. The masses of the electrons, protons and neutrons are listed in Table 1. Most of the mass of the atom is due to the mass of the nucleus.
particle
mass (kg)
electron 9.11 x 10-31 proton 1.673 x 10-27 neutron 1.675 x 10-27 Table 1. Masses of the building blocks of atoms. The diameter of the nucleus is between 10E-15 and 10E-14 m. The electrons are contained in a roughly spherical region with a diameter of about 2 x 10E-10 m. In Physics 121 it was shown that an object can only carry out circular motion if a radial force (directed towards the center of the circle) is present. Measurements of the velocity of the orbital electrons in an atom have shown that the attractive force between the electrons and the nucleus is significantly stronger than the gravitational force between these two objects. The attractive force between the electrons and the nucleus is called the electric force. Experiments have shown that the electric force between two objects is proportional to the inverse square of the distance between the two objects. The electric force between two electrons is the same as the electric force between two protons when they are placed as the same distance. This implies that the electric force does not depend on the mass of the particle. Instead, it depends on a new quantity: the electric charge. The unit of electric charge q is the Coulomb (C). The electric charge can be negative, zero, or positive. Per definition, the electric charge on a glass rod rubbed with silk is positive. The electric charge of electrons, protons and neutrons are listed in Table 2. Detailed measurements have shown that the magnitude of the charge of the proton is exactly equal to the magnitude of the charge of the electron. Since atoms are neutral, the number of electrons must be equal to the number of protons. The precise magnitude of the electric force that a charged particle exerts on another is given by Coulomb's law: " The magnitude of the electric force that a particle exerts on another particle is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The direction of the force is along the line joining the particles. "
6
particle charge (C) electron - 1.6 x 10-19 proton 1.6 x 10-19 neutron 0 Table 2. Electric charges of the building blocks of atoms The electric force Fc can be written as
(1) where q1 and q 2 are the charges of particle 1 and particle 2, respectively r is the distance between particle 1 and particle 2 (see Figure 1) [epsilon]0 is he permittivity constant: [epsilon]0 = 8.85 x 10-12 C2/(N . m2) This formula applies to elementary particles and small charged objects as long as their sizes are much less than the distance between them.
Figure 1. Electric force between two charged objects. An important difference between the electric force and the gravitational force is that the gravitational force is always attractive, while the electric force can be repulsive (Fc > 0), zero, or attractive (Fc < 0), depending on the charges of the particles. Table 3 lists the gravitational and the Coulomb force between electrons, protons and neutrons when they are separated by 1 x 10-10 m. This table shows clearly that the electric force dominates the motion of electrons in atoms. However, on a macroscopic scale, the gravitational force dominates. Since most macroscopic objects are neutral, they have an equal number of protons and electrons. The attractive force between the electrons in one body and the protons in the other body is exactly canceled by the repulsive force between the electrons in the two bodies. Our discussion of the electric force will initially concentrate on those cases in which the charges are at rest or are moving very slowly. The electric force exerted under these circumstances is called the electrostatic force. If the charges are moving with a uniform velocity, they will experience both the electrostatic force and a magnetic force. The combined electrostatic and magnetic force is called the electromagnetic force.
7
particle-particle
Fg (N)
Fc (N)
-51
electron - electron -5.5 x 10
2.3 x 10-8
electron - proton -1.0 x 10-47 - 2.3 x 10-8 electron - neutron -1.0 x 10-47 -44
proton - proton - 1.9 x 10
proton - neutron - 1.9 x 10-44
0 2.3 x 10-8 0
-44
neutron - neutron - 1.9 x 10 0 Table 3. The gravitational (F g) and Coulomb (F c) between the building blocks of atoms.
2. Charge Quantization and Charge Conservation An important experiment in which the charge of small oil droplets was determined was carried out by Millikan. Millikan discovered that the charge on the oil droplets was always a multiple of the charge of the electron (e, the fundamental charge). For example, he observed droplets with a charge equal to +/- e, +/- 2 e, +/- 3 e, etc., but never droplets with a charge equal to +/- 1.45 e, +/- 2.28 e, etc. The experiments strongly suggested that charge is quantized. Another important property of charge is that charge a conserved quantity. No reaction has ever been found that creates or destroys charge. For example, the annihilation of an electron and an anti electron (positron) produces two photons:
(2) This reaction does not violate conservation of charge. The initial charge is equal to
(3) Note that the charge of an antiparticle is opposite that of the particle. The final charge is equal to zero since photons are uncharged. The following reaction however violates conservation of electric charge
(4) This reaction has never been observed.
3. Conductors and Insulators A conductor is a material that permits the motion of electric charge through its volume. Examples of conductors are copper, aluminum and iron. An electric charge placed on the end of a conductor will spread out over the entire conductor until an equilibrium distribution is
8
established. In contrast, electric charge placed on an insulator stays in place: an insulator (like glass, rubber and Mylar) does not permit the motion of electric charge.
Figure 2. Induction of Charge on Metal Sphere. The properties of a conductor are a result of the presence of free electrons in the material. These electrons are free to move through the entire volume of the conductor. Because of the free electrons, the charge distribution of a conductor can be changed by the presence of external charges. For example, the metal sphere shown in Figure 2 is initially uncharged. This implies that the free electrons (and positive ions) are distributed uniformly over its surface. If a rod with a positive charge is placed in the vicinity of the sphere, it will produce an attractive force on the free electrons. As a consequence of this attractive force the free electrons will be redistributed, and the top of the conductor will get a negative charge (excess of electrons). Since the number of free electrons on the sphere is unchanged, the bottom of the sphere will have a deficit of free electrons (and will have a positive charge). The positive ions are bound to the lattice of the material, and their distribution is not affected by the presence of the charged rod. If we connect the bottom of the sphere to ground (a source or drain of electrons) electrons will be attracted by the positive charge. The number of electrons on the sphere will increase, and the sphere will have a net negative charge. If we break the connection to the ground before removing the charged rod, we are left with a negative charge on the sphere. If we first remove the charged rod, the excess of electrons will drain to the ground, and the sphere will become uncharged.
9
THE ELECTRIC FIELD 1. Introduction The presence of an electric charge produces a force on all other charges present. The electric force produces action-at-a-distance; the charged objects can influence each other without touching. Suppose two charges, q 1 and q 2, are initially at rest. Coulomb's law allows us to calculate the force exerted by charge q 2 on charge q 1 (see Figure 1). At a certain moment charge q 2 is moved closer to charge q 1. As a result we expect an increase of the force exerted by q 2 on q 1. However, this change can not occur instantaneous (no signal can propagate faster than the speed of light). The charges exert a force on one another by means of disturbances that they generate in the space surrounding them. These disturbances are called electric fields. Each electrically charged object generates an electric field which permeates the space around it, and exerts pushes or pulls whenever it comes in contact with other charged objects. The electric field E generated by a set of charges can be measured by putting a point charge q at a given position. The test charge will feel an electric force F. The electric field at the location of the point charge is defined as the force F divided by the charge q:
Figure 1. Electric force between two electric charges. (1) The definition of the electric field shows that the electric field is a vector field: the electric field at each point has a magnitude and a direction. The direction of the electric field is the direction in which a positive charge placed at that position will move. In this chapter the calculation of the electric field generated by various charge distributions will be discussed.
2. The Superposition of Electric Forces From the definition of the electric field it is clear that in order to calculate the field strength generated by a charge distribution we must be able to calculate the total electric force exerted on a test charge by this charge distribution.
10
Figure 2. Superposition of electric forces. Suppose a charge q is placed in the vicinity of three other charges, q 1, q2, and q 3, as is shown in Figure 2. Coulomb's law can be used to calculate the electric force between q and q 1, between q and q 2, and between q and q 3. Experiments have shown that the total force exerted by q 1, q2 and q 3 on q is the vector sum of the individual forces:
(2)
Example Problem 1: Electric Field of a Charged Rod A total amount of charge Q is uniformly distributed along a thin, straight, plastic rod of length L (see Figure 3). a) Find the electric force acting on a point charge q located at point P, at a distance d from one end of the rod (see Figure 3). b) Find the electric force acting on a point charge q located at point P', at a distance y from the midpoint of the rod (see Figure 3).
Figure 3. Example Problem 1. a) Figure 4 shows the force dF acting on point charge q, located at point P, as a result of the Coulomb interaction between charge q and a small segment of the rod. The force is directed along the x-axis and has a magnitude given by 11
(3)
Figure 4. Relevant dimensions for Example Problem 1. The total force acting on charge q can be found by summing over all segments of the rod:
(4) b) Figure 5 shows the force acting on charge q, located at P', due to two charged segments of the rod. The net force dF exerted on q by the two segments of the rod is directed along the yaxis (vertical axis), and has a magnitude equal to
(5) Note: the x-component of dFl cancels the x-component of dFr, and the net force acting on q is therefore equal to the sum of the y-components of dFl and dFr. The magnitude of dFl and dFr can be obtained from Coulomb's law:
(6)
Figure 5. Relevant dimension for Example Problem 1 Substituting eq. (6) into eq. (5) we obtain 12
(7) The net force acting on charge q can be obtained by summing over all segments of the rod.
(8)
3. The Electric Field Equation (1) shows that the electric field generated by a charge distribution is simply the force per unit positive charge. The procedure to measure the electric field, outlined in the introduction, assumes that all charges that generate the electric field remain fixed at their position while the test charge is introduced. To avoid disturbances to these charges, it is usually convenient to use a very small test charge.
Example Problem 2: Electric Field of Point Charge Q. A test charge placed a distance r from point charge Q will experience an electric force Fc given by Coulomb's law:
(9) The electric field generated by the point charge Q can be calculated by substituting eq.(9) into eq.(1)
(10)
Example Problem 3: Electric Field of Charge Sheet. Suppose a very large sheet has a uniform charge density of [sigma] Coulomb per square meter. The charge sheet can be regarded as made up of a collection of many concentric rings, centered around the z-axis (which coincides with the location of the point of interest). The total electric field at this point can be obtained by vector addition of the electric field generated by all small segments of the sheet. Figure 6 shows the relevant dimension used to 13
calculate the electric field generated by a ring with radius r and width dr. The strength of the electric field generated by each ring is directed along the z-axis and has a strength equal to
(11) where dQ is the charge of the ring and z is the z-coordinate of the point of interest. The charge dQ can be expressed in terms of r, dr, and [sigma]
(12)
Figure 6. Electric field above large charge sheet. The angle [theta] depends on the radius of the ring and the z-coordinate of the point of interest
(13) Substituting eq.(12) and eq.(13) into eq.(11) one obtains
(14) The total electric field can be found by summing the contributions of all rings that make up the charge sheet
14
(15)
Figure 7. Field generated by 2 large parallel charged plates. Equation (15) shows that a uniform electric field is produced by an infinitely large charged sheet. However, in many practical applications in which a uniform electric field is required, two parallel charge sheets are used. The electric field between the two charged plates (with charge density [sigma] and - [sigma]) can be obtained by vector addition of the fields generated by the individual plates (see Figure 7):
(16) The electric fields above and below the plates have opposite directions (see Figure 7), and cancel. Therefore, two charged plates generate a homogeneous electric field confined to the region between the plates, and no electric field outside this region (note: this in contrast to a single charged sheet which produces an electric field everywhere).
Example Problem 4: Electric Field of Multiple Charge Sheets Two large sheets of paper intersect each other at right angles. Each sheet carries a uniform distribution of positive charge of [sigma] C/m2. Find the magnitude of the electric field in each of the four quadrants.
15
Figure 8. Example Problem 2. This problem can be solved easily by applying the superposition principle of electric fields generated by each sheet individually (see Figure 8). The strength of the electric field produced by each plate is given by eq.(15). The direction of the electric field is perpendicular to the plate, and pointing away from it. The strength of the total electric field in each of the quadrants is given by
(17) and its direction in each of the four quadrants is indicated in Figure 8.
4. Field Lines The electric field can be represented graphically by field lines. These lines are drawn in such a way that, at a given point, the tangent of the line has the direction of the electric field at that point. The density of lines is proportional to the magnitude of the electric field. Each field line starts on a positive point charge and ends on a negative point charge. Since the density of field lines is proportional to the strength of the electric field, the number of lines emerging from a positive charge must also be proportional to the charge. An example of field lines generated by a charge distributions is shown in Figure 9.
16
Figure 9. Electric field produced by two point charges q = + 4
5. Electric Dipole in an Electric Field The net force acting on a neutral object placed in a uniform electric field is zero. However, the electric field can produce a net torque if the positive and negative charges are concentrated at different locations on the object. An example is shown in Figure 10. The figure shows a charge Q located on one end of a rod of length L and a charge - Q located on the opposite end of the rod. The forces acting on the two charges are given by
(18)
Figure 10. Electric Dipole in an Electric Field. Clearly, the net force acting on the system is equal to zero. The torque of the two forces with respect to the center of the rod is given by
(19) As a result of this torque the rod will rotate around its center. If [theta] = 0deg. (rod aligned with the field) the torque will be zero. The distribution of the charge in a body can be characterized by a parameter called the dipole moment p. The dipole moment of the rod shown in Figure 10 is defined as 17
(20) In general, the dipole moment is a vector which is directed from the negative charge towards the positive charge. Using the definition of the dipole moment from eq.(20) the torque of an object in an electric field is given by
(21)
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GAUSS' LAW 1. Introduction The electric field of a given charge distribution can in principle be calculated using Coulomb's law. The examples discussed in the previous Chapter show however, that the actual calculations can become quit complicated.
2. Gauss' Law An alternative method to calculate the electric field of a given charge distribution relies on a theorem called Gauss' law. Gauss' law states that " If the volume within an arbitrary closed mathematical surface holds a net electric charge Q, then the electric flux [Phi] though its surface is Q/[epsilon]0 " Gauss' law can be written in the following form:
(1)
Figure 1. Electric flux through surface area A. The electric flux [Phi] through a surface is defined as the product of the area A and the magnitude of the normal component of the electric field E: (2) where [theta] is the angle between the electric field and the normal of the surface (see Figure 1). To apply Gauss' law one has to obtain the flux through a closed surface. This flux can be obtained by integrating eq.(2) over all the area of the surface. The convention used to define the flux as positive or negative is that the angle [theta] is measured with respect to the perpendicular erected on the outside of the closed surface: field lines leaving the volume
19
make a positive contribution, and field lines entering the volume make a negative contribution.
Example Problem 1: Electric Field of a Point Charge. The field generated by a point charge q is spherical symmetric, and its magnitude will depend only on the distance r from the point charge. The direction of the field is along the direction (see Figure 2). Consider a spherical surface centered around the point charge q (see Figure 2). The direction of the electric field at any point on its surface is perpendicular to the surface and its magnitude is constant. This implies that the electric flux [Phi] through this surface is given by
(3)
Figure 2. Electric field generated by point charge q. Using Gauss's law we obtain the following expression
(4) or
(5) which is Coulomb's law.
Example Problem 2: Electric Field of a Charge Sheet Charge is uniformly distributed over the volume of a large slab of plastic of thickness d. The charge density is [rho] C/m3. The mid-plane of the slab is the y-z plane (see Figure 3). What is the electric filed at a distance x from the mid-plane ?
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Figure 3. Problem 16. As a result of the symmetry of the slab, the direction of the electric field will be along the xaxis (at every point). To calculate the electric field at any given point, we need to consider two separate case: - d/2 < x < d/2 and x > d/2 or x < -d/2. Consider surface 1 shown in Figure 3. The flux through this surface is equal to the flux through the planes at x = x 1 and x = - x1. Symmetry arguments show that
(6) The flux [Phi]1 through surface 1 is therefore given by
(7) The amount of charge enclosed by surface 1 is given by
(8) Applying Gauss' law to eq.(7) and eq.(8) we obtain
(9) or
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(10) Note: this formula is only correct for - d/2 < x 1 < d/2. The flux [Phi]2 through surface 2 is given by
(11) The charge enclosed by surface 2 is given by
(12) This equation shows that the enclosed charge does not depend on x 2. Applying Gauss's law one obtains
(13) or
(14)
3. Conductors in Electric Fields A large number of electrons in a conductor are free to move. The so called free electrons are the cause of the different behavior of conductors and insulators in an external electric filed. The free electrons in a conductor will move under the influence of the external electric field (in a direction opposite to the direction of the electric field). The movement of the free electrons will produce an excess of electrons (negative charge) on one side of the conductor, leaving a deficit of electrons (positive charge) on the other side. This charge distribution will also produce an electric field and the actual electric field inside the conductor can be found by superposition of the external electric field and the induced electric field, produced by the induced charge distribution. When static equilibrium is reached, the net electric field inside the conductor is exactly zero. This implies that the charge density inside the conductor is zero. If the electric field inside the conductor would not be exactly zero the free electrons would continue to move and the charge distribution would not be in static equilibrium. The electric field on the surface of the conductor is perpendicular to its surface. If this would not be the case, the free electrons would move along the surface, and the charge distribution would not be in equilibrium. The redistribution of the free electrons in the conductor under the influence of an external electric field, and the cancellation of the external electric field inside the conductor is being used to shield sensitive instruments from external electric fields.
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The strength of the electric field on the surface of a conductor can be found by applying Gauss' law (see Figure 4). The electric flux through the surface shown in Figure 4 is given by
(15) where A is the area of the top of the surface shown in Figure 4. The flux through the bottom of the surface shown in Figure 4 is zero since the electric field inside a conductor is equal to zero. Note that eq.(15) is only valid close to the conductor where the electric field is perpendicular to the surface. The charge enclosed by the surface shown in Figure 4 is equal to
Figure 4. Electric field of conductor. (16) where [sigma] is the surface charge density of the conductor. Eq.(16) is correct if the charge density [sigma] does not vary significantly over the area A (this condition can always be met by reducing the size of the surface being considered). Applying Gauss' law we obtain
(17) Thus, the electric field at the surface of the conductor is given by
(18)
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THE ELECTROSTATIC POTENTIAL 1. Introduction The electrostatic force is a conservative force. This means that the work it does on a particle depends only on the initial and final position of the particle, and not on the path followed. With each conservative force, a potential energy can be associated. The introduction of the potential energy is useful since it allows us to apply conservation of mechanical energy which simplifies the solution of a large number of problems. The potential energy U associated with a conservative force F is defined in the following manner
(1) where U(P0) is the potential energy at the reference position P0 (usually U(P0) = 0) and the path integral is along any convenient path connecting P0 and P1. Since the force F is conservative, the integral in eq.(1) will not depend on the path chosen. If the work W is positive (force and displacement pointing in the same direction) the potential energy at P1 will be smaller than the potential energy at P0. If energy is conserved, a decrease in the potential energy will result in an increase of the kinetic energy. If the work W is negative (force and displacement pointing in opposite directions) the potential energy at P1 will be larger than the potential energy at P0. If energy is conserved, an increase in the potential energy will result in an decrease of the kinetic energy. If In electrostatic problems the reference point P0 is usually chosen to correspond to an infinite distance, and the potential energy at this reference point is taken to be equal to zero. Equation (1) can then be rewritten as:
(2) To describe the potential energy associated with a charge distribution the concept of the electrostatic potential V is introduced. The electrostatic potential V at a given position is defined as the potential energy of a test particle divided by the charge q of this object:
(3) 24
In the last step of eq.(3) we have assumed that the reference point P0 is taken at infinity, and that the electrostatic potential at that point is equal to 0. Since the force per unit charge is the electric field (see Chapter 23), eq. (3) can be rewritten as
(4) The unit of electrostatic potential is the volt (V), and 1 V = 1 J/C = 1 Nm/C. Equation (4) shows that as the unit of the electric field we can also use V/m. A common used unit for the energy of a particle is the electron-volt (eV) which is defined as the change in kinetic energy of an electron that travels over a potential difference of 1 V. The electron-volt can be related to the Joule via eq.(3). Equation (3) shows that the change in energy of an electron when it crosses over a 1 V potential difference is equal to 1.6 . 10-19 J and we thus conclude that 1 eV = 1.6 . 10-19 J.
2. Calculating the Electrostatic Potential A charge q is moved from P0 to P1 in the vicinity of charge q' (see Figure 1). The electrostatic potential at P1 can be determined using eq. (4) and evaluating the integral along the path shown in Figure 1. Along the circular part of the path the electric field and the displacement are perpendicular, and the change in the electrostatic potential will be zero. Equation (4) can therefore be rewritten as
(5) If the charge q' is positive, the potential increases with a decreasing distance r. The electric field points away from a positive charge, and we conclude that the electric field points from regions with a high electrostatic potential towards regions with a low electrostatic potential.
Figure 1. Path followed by charge q between P 0 and P 1. From the definition of the electrostatic potential in terms of the potential energy (eq.(3)) it is clear that the potential energy of a charge q under the influence of the electric field generated by charge q' is given by
25
(6)
Example Problem 1: Electrostatic Potential of a Charged Rod A total charge Q is distributed uniformly along a straight rod of length L. Find the potential at point P at a distance h from the midpoint of the rod (see Figure 2). The potential at P due to a small segment of the rod, with length dx and charge dQ, located at the position indicated in Figure 3 is given by
(7) The charge dQ of the segment is related to the total charge Q and length L
(8) Combining equations (7) and (8) we obtain the following expression for dV:
(9)
Figure 2. Problem 21.
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Figure 3. Solution of Problem 21. The total potential at P can be obtained by summing over all small segments. This is equivalent to integrating eq.(9) between x = - L/2 and x = L/2.
(10)
Example Problem 2: Distance of Closest Approach An alpha particle with a kinetic energy of 1.7 x 10-12 J is shot directly towards a platinum nucleus from a very large distance. What will be the distance of closest approach ? The electric charge of the alpha particle is 2e and that of the platinum nucleus is 78e. Treat the alpha particle and the nucleus as spherical charge distributions and disregard the motion of the nucleus. The initial mechanical energy is equal to the kinetic energy of the alpha particle
(11) Due to the electric repulsion between the alpha particle and the platinum nucleus, the alpha particle will slow down. At the distance of closest approach the velocity of the alpha particle is zero, and thus its kinetic energy is equal to zero. The total mechanical energy at this point is equal to the potential energy of the system
(12) where q 1 is the charge of the alpha particle, q 2 is the charge of the platinum nucleus, and d is the distance of closest approach. Applying conservation of mechanical energy we obtain
(13) The distance of closest approach can be obtained from eq.(13)
(14)
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3. The Electrostatic Field as a Conservative Field The electric field is a conservative field since the electric force is a conservative force. This implies that the path integral
(15) between point P0 and point P1 is independent of the path between these two points. In this case the path integral for any closed path will be zero:
(16) Equation (16) can be used to prove an interesting theorem: " within a closed, empty cavity inside a homogeneous conductor, the electric field is exactly zero ".
Figure 4. Cross section of cavity inside spherical conductor. Figure 4 shows the cross section of a possible cavity inside a spherical conductor. Suppose there is a field inside the conductor and one of the field lines is shown in Figure 4. Consider the path integral of eq.(16) along the path indicated in Figure 4. In Chapter 24 it was shown that the electric field within a conductor is zero. Thus the contribution of the path inside the conductor to the path integral is zero. Since the remaining part of the path is chosen along the field line, the direction of the field is parallel to the direction of the path, and therefore the path integral will be non-zero. This obviously violates eq.(16) and we must conclude that the field inside the cavity is equal to zero (in this case the path integral is of course equal to zero).
4. The Gradient of the Electrostatic Potential The electrostatic potential V is related to the electrostatic field E. If the electric field E is known, the electrostatic potential V can be obtained using eq.(4), and vice-versa. In this 28
section we will discuss how the electric field E can be obtained if the electrostatic potential is known.
Figure 5. Calculation of the electric field E. Consider the two points shown in Figure 5. These two nearly identical positions are separated by an infinitesimal distance dL. The change in the electrostatic potential between P1 and P2 is given by (17) where the angle [theta] is the angle between the direction of the electric field and the direction of the displacement (see Figure 5). Equation (17) can be rewritten as
(18) where EL indicates the component of the electric field along the L-axis. If the direction of the displacement is chosen to coincide with the x-axis, eq.(18) becomes
(19) For the displacements along the y-axis and z-axis we obtain
(20) (21) The total electric field E can be obtained from the electrostatic potential V by combining equations (19), (20), and (21):
(22) Equation (22) is usually written in the following form
29
(23) where --V is the gradient of the potential V. In many electrostatic problems the electric field of a certain charge distribution must be evaluated. The calculation of the electric field can be carried out using two different methods: 1. the electric field can be calculated by applying Coulomb's law and vector addition of the contributions from all charges of the charge distribution. 2. the total electrostatic potential V can be obtained from the algebraic sum of the potential due to all charges that make up the charge distribution, and subsequently using eq.(23) to calculate the electric field E. In many cases method 2 is simpler since the calculation of the electrostatic potential involves an algebraic sum, while method 1 relies on the vector sum.
Example Problem 3: Electric Field derived from Electrostatic Potential In some region of space, the electrostatic potential is the following function of x, y, and z:
(24) where the potential is measured in volts and the distances in meters. Find the electric field at the points x = 2 m, y = 2 m. The x, y and z components of the electric field E can be obtained from the gradient of the potential V (eq.(23)):
(25) (26) (27) Evaluating equations (25), (26), and (27) at x = 2 m and y = 2 m gives
(28) (29) (30) 30
Thus
(31)
Example Problem 4: Potential and Field of a Charged Annulus An annulus (a disk with a hole) made of paper has an outer radius R and an inner radius R/2 (see Figure 6). An amount Q of electric charge is uniformly distributed over the paper. a) Find the potential as a function of the distance on the axis of the annulus. b) Find the electric field on the axis of the annulus. We define the x-axis to coincide with the axis of the annulus (see Figure 7). The first step in the calculation of the total electrostatic potential at point P due to the annulus is to calculate the electrostatic potential at P due to a small segment of the annulus. Consider a ring with radius r and width dr as shown in Figure 7. The electrostatic potential dV at P generated by this ring is given by
(32) where dQ is the charge on the ring. The charge density [rho] of the annulus is equal to
(33)
Figure 6. Problem 36. Using eq. (33) the charge dQ of the ring can be calculates
(34) Substituting eq.(34) into eq.(32) we obtain 31
(35) The total electrostatic potential can be obtained by integrating eq.(35) over the whole annulus:
(36)
Figure 7. Calculation of electrostatic potential in Problem 36. Due to the symmetry of the problem, the electric field will be directed along the x-axis. The field strength can be obtained by applying eq.(23) to eq.(36):
(37) Since the electrostatic field and the electrostatic potential are related we can replace the field lines by so called equipotential surfaces. Equipotential surfaces are defined as surfaces on which each point has the same electrostatic potential. The component of the electric field parallel to this surface must be zero since the change in the potential between all points on this surface is equal to zero. This implies that the direction of the electric field is perpendicular to the equipotential surfaces.
5. The Potential and Field of a Dipole Figure 8 shows an electric dipole located along the z-axis. It consists of two charges + Q and Q, separated by a distance L. The electrostatic potential at point P can be found by summing the potentials generated by each of the two charges:
32
(38)
Figure 8. The electric dipole. If the point P is far away from the dipole (r >> L) we can make the approximation that r1 and r2 are parallel. In this case (39) and
(40) The electrostatic potential at P can now be rewritten as
(41) where p is the dipole moment of the charge distribution. The electric field of the dipole can be obtained from eq.(41) by taking the gradient (see eq.(23)).
33
ELECTRIC ENERGY OF A SYSTEM OF POINT CHARGES 1 Introduction
Figure 1. System of three charges. The electric potential energy U of a system of two point charges was discussed in our previous Chapter and is equal to
(1) where q 1 and q 2 are the electric charges of the two objects, and r is their separation distance. The electric potential energy of a system of three point charges (see Figure 1) can be calculated in a similar manner
(2) where q 1, q2, and q 3 are the electric charges of the three objects, and r12, r13, and r23 are their separation distances (see Figure 1). The potential energy in eq.(2) is the energy required to assemble the system of charges from an initial situation in which all charges are infinitely far apart. Equation (2) can be written in terms of the electrostatic potentials V:
(3) where Vother(1) is the electric potential at the position of charge 1 produced by all other charges
(4)
34
and similarly for Vother(2) and Vother(3).
Example Problem: Model of a Carbon Nucleus According to the alpha-particle model of the nucleus some nuclei consist of a regular geometric arrangement of alpha particles. For instance, the nucleus of 12C consists of three alpha particles on an equilateral triangle (see Figure 2). Assuming that the distance between pairs of alpha particles is 3 x 10-15 m, what is the electric energy of this arrangement of alpha particles ? Treat the alpha particles as pointlike.
Figure 2. Alpha-particle model of 12C. The electric potential at the location of each alpha particle is equal to
(5) where d = 3.0 x 10-15 m. The electric energy of this configuration can be calculated by combining eq.(5) and eq.(3):
(6)
2 Energy of a System of Conductors
Figure 3. The capacitor. 35
The electrostatic energy of a system of conductors can be calculated using eq.(3). For example, a capacitor consists of two large parallel metallic plates with area A. Suppose that charges +Q and -Q are placed on the two plates (see Figure 3). Suppose the electrostatic potential of plate 1 is V1 and the potential of plate 2 is V2. The electrostatic energy of the capacitor is then equal to (7) The electric field E between the plates is a function of the charge density [sigma]
(8) The potential difference V1 - V2 between the plates can be obtained by a path integration of the electric field
(9) Combining eq.(9) and eq.(7) we can calculate the electrostatic energy of the system:
(10) This equation shows that electrostatic energy can be stored in a capacitor. Equation (10) can be rewritten as
(11) where Volume is the volume between the capacitor plates. The quantity [epsilon]0 . E2/2 is called the energy density (potential energy per unit volume).
Figure 4. Field lines at the edge of a capacitor. 36
In the calculation of the energy density carried out for the capacitor we assumed that the electric field was homogeneous in the region between the plates. In a real capacitor the field at the edge is not homogeneous, and the calculation will have to be modified. Figure 4 shows a couple of field lines at the edge of a capacitor. Consider the two small sections of the capacitor plates with charges dQ and -dQ, respectively, shown in Figure 4. The contribution of these two sections to the total electrostatic energy of the capacitor is given by (12) where V1 and V2 are the electrostatic potential of the top and bottom plate, respectively. The potential difference, V1 - V2, is related to the electric field between the plates
(13) The electric field E(l) can be related to the charges on the small segments of the capacitor plates via Gauss' law. Consider a volume with its sides parallel to the field lines (see Figure 5). The electric flux through its surface is equal to
(14) where E(l) is the strength of the electric field at a distance l from the bottom capacitor plate (see Figure 5) and dS(l) is the area of the top of the integration volume. The flux is negative since the field lines are entering the integration volume. The flux through the sides of the integration volume is zero since the sides are chosen to be parallel to the field lines. The flux through the bottom of the integration volume is also zero, since the electric field in any conductor is zero. Gauss' law requires that the flux through the surface of any volume is equal to the charge enclosed by that volume divided by [epsilon]0:
(15)
Figure 5. Integration volume discussed in the text. Combining eq.(14) and eq.(15) we obtain
(16)
37
Equations (12), (13) and (16) can be combined to give
(17) This calculation can be generalized to objects of arbitrary shapes, and the electrostatic energy of any system can be expressed as the volume integral of the energy density u which is defined as
(18) Thus
(19) where the volume integration extends over all regions where there is an electric field.
Example Problem: Fission of Uranium In symmetric fission, the nucleus of uranium (238U) splits into two nuclei of palladium (119Pd). The uranium nucleus is spherical with a radius of 7.4 x 10-15 m. Assume that the two palladium nuclei adopt a spherical shape immediately after fission; at this instant, the configuration is as shown in Figure 6. The size of the nuclei in Figure 6 can be calculated from the size of the uranium nucleus because nuclear material maintains a constant density.
Figure 6. Two palladium nuclei right after fission of 238U. a) Calculate the electric energy of the uranium nucleus before fission b) Calculate the total electric energy of the palladium nuclei in the configuration shown in Figure 6, immediately after fission. Take into account the mutual electric potential energy of the two nuclei and also the individual electric energy of the two palladium nuclei by themselves. c) Calculate the total electric energy a long time after fission when the two palladium nuclei have moved apart by a very large distance. d) Ultimately, how much electric energy is released into other forms of energy in the complete fission process ? e) If 1 kg of uranium undergoes fission, how much electric energy is released ? 38
a) The electric energy of the uranium nucleus before fission can be calculated using the known electric field distribution generated by a uniformly charged sphere of radius R:
(20) For the uranium nucleus q = 92e and R = 7.4 x 10-15 m. Substituting these values into eq.(20) we obtain
(21) b) Suppose the radius of a palladium nucleus is R Pd. The total volume of nuclear matter of the system shown in Figure 6 is equal to
(22) Since the density of nuclear matter is constant, the volume in eq.(22) must be equal to the volume of the original uranium nucleus
(23) Combining eq.(23) and (22) we obtain the following equation for the radius of the palladium nucleus:
(24) The electrostatic energy of each palladium nucleus is equal to
(25) where we have used the radius calculated in eq.(24) and a charge q Pd = 46e. Besides the internal energy of the palladium nuclei, the electric energy of the configuration must also be included in the calculation of the total electric potential energy of the nuclear system
(26)
39
where q Pd is the charge of the palladium nucleus (q Pd = 26e) and R int is the distance between the centers of the two nuclei (R int = 2 RPd = 11.7 x 10-15 m). Substituting these values into eq.(26) we obtain
(27) The total electric energy of the system at fission is therefore
(28) c) Due to the electric repulsion between the positively charge palladium nuclei, they will separate and move to infinity. At this point, the electric energy of the system is just the sum of the electric energies of the two palladium nuclei:
(29) d) The total release of energy is equal to the difference in the electric energy of the system before fission (eq.(21)) and long after fission (eq.(29)):
(30) e) Equation (30) gives the energy released when 1 uranium nucleus fissions. The number of uranium nuclei in 1 kg of uranium is equal to
(31) The total release of energy is equal to
(32) To get a feeling for the amount of energy released when uranium fissions, we can compare the energy in eq.(32) with the energy released by falling water. Suppose 1 kg of water falls 100 m. The energy released is equal to the change in the potential energy of the water:
(33) The mass of water needed to generate an amount of energy equal to that released in the fission of 1 kg uranium is
40
(34)
41
CAPACITORS AND DIELECTRICS. 1. Introduction A capacitor is an arrangement of conductors that is used to store electric charge. A very simple capacitor is an isolated metallic sphere. The potential of a sphere with radius R and charge Q is equal to
(1) Equation (1) shows that the potential of the sphere is proportional to the charge Q on the conductor. This is true in general for any configuration of conductors. This relationship can be written as
(2) where C is called the capacitance of the system of conductors. The unit of capacitance is the farad (F). The capacitance of the metallic sphere is equal to
(3)
2. The parallel-plate capacitor Another example of a capacitor is a system consisting of two parallel metallic plates. In Chapter 26 it was shown that the potential difference between two plates of area A, separation distance d, and with charges +Q and -Q, is given by
(4) Using the definition of the capacitance (eq.(2)), the capacitance of this system can be calculated:
(5) Equation (2) shows that the charge on a capacitor is proportional to the capacitance C and to the potential V. To increase the amount of charge stored on a capacitor while keeping the 42
potential (voltage) fixed, the capacitance of the capacitor will need to be increased. Since the capacitance of the parallel plate capacitor is proportional to the plate area A and inversely proportional to the distance d between the plates, this can be achieved by increasing the surface area A and/or decreasing the separation distance d. These large capacitors are usually made of two parallel sheets of aluminized foil, a few inches wide and several meters long. The sheets are placed very close together, but kept from touching by a thin sheet of plastic sandwiched between them. The entire sandwich is covered with another sheet of plastic and rolled up like a roll of toilet paper.
Example Problem: The Geiger Counter The tube of a Geiger counter consists of a thin straight wire surrounded by a coaxial conducting shell. The diameter of the wire is 0.0025 cm and that of the shell is 2.5 cm. The length of the tube is 10 cm. What is the capacitance of a Geiger-counter tube ?
Figure 1. Schematic of a Geiger counter. The problem will be solved under the assumption that the electric field generated is that of an infinitely long line of charge. A schematic side view of the tube is shown in Figure 1. The radius of the wire is rw, the radius of the cylinder is rc, the length of the counter is L, and the charge on the wire is +Q. The electric field in the region between the wire and the cylinder can be calculated using Gauss' law. The electric field in this region will have a radial direction and its magnitude will depend only on the radial distance r. Consider the cylinder with length L and radius r shown in Figure 1. The electric flux [Phi] through the surface of this cylinder is equal to
(6) According to Gauss' law, the flux [Phi] is equal to the enclosed charge divided by [epsilon]0. Therefore
(7) The electric field E(r) can be obtained using eq.(7):
43
(8) The potential difference between the wire and the cylinder can be obtained by integrating the electric field E(r):
(9) Using eq.(2) the capacitance of the Geiger tube can be calculated:
(10) Substituting the values for rw, rc, and L into eq.(10) we obtain
(11)
3. Capacitors in Combination The symbol of a capacitor is shown in Figure 2. Capacitors can be connected together; they can be connected in series or in parallel. Figure 3 shows two capacitors, with capacitance C 1 and C 2, connected in parallel. The potential difference across both capacitors must be equal and therefore
(12)
Figure 2. Symbol of a Capacitor.
44
Figure 3. Two capacitors connected in parallel. Using eq.(12) the total charge on both capacitors can be calculated
(13) Equation (13) shows that the total charge on the capacitor system shown in Figure 3 is proportional to the potential difference across the system. The two capacitors in Figure 3 can be treated as one capacitor with a capacitance C where C is related to C 1 and C 2 in the following manner
(14) Figure 4 shows two capacitors, with capacitance C 1 and C 2, connected in series. Suppose the potential difference across C 1 is [Delta]V1 and the potential difference across C 2 is [Delta]V2. A charge Q on the top plate will induce a charge -Q on the bottom plate of C 1. Since electric charge is conserved, the charge on the top plate of C2 must be equal to Q. Thus the charge on the bottom plate of C 2 is equal to -Q. The voltage difference across C 1 is given by
(15) and the voltage difference across C 2 is equal to
(16)
Figure 4. Two capacitors connected in series. 45
The total voltage difference across the two capacitors is given by
(17) Equation (17) again shows that the voltage across the two capacitors, connected in series, is proportional to the charge Q. The system acts like a single capacitor C whose capacitance can be obtained from the following formula
(18)
Example: Multi-plate Capacitor A multi-plate capacitor, such as used in radios, consists of four parallel plates arranged one above the other as shown in Figure 5. The area of each plate is A, and the distance between adjacent plates is d. What is the capacitance of this arrangement ?
Figure 5. A Multi-plate Capacitor. The multiple capacitor shown in Figure 5 is equivalent to three identical capacitors connected in parallel (see Figure 6). The capacitance of each of the three capacitors is equal and given by
(19) The total capacitance of the multi-plate capacitor can be calculated using eq.(14):
(20)
46
Figure 6. Schematic of Multi-plate Capacitor shown in Figure 5.
Example Problem: Capacitors in Series/Parallel Three capacitors, of capacitance C 1 = 2.0 uF, C 2 = 5.0 uF, and C 3 = 7.0 uF, are initially charged to 36 V by connecting each, for a few instants, to a 36-V battery. The battery is then removed and the charged capacitors are connected in a closed series circuit, with the positive and negative terminals joined as shown in Figure 7. What will be the final charge on each capacitor ? What will be the voltage across the points PP' ?
Figure 7. Capacitors in Series/Parallel. The initial charges on each of the three capacitors, q 1, q2, and q 3, are equal to
(21) After the three capacitors are connected, the charge will redistribute itself. The charges on the three capacitors after the system settles down are equal to Q1, Q2, and Q3. Since charge is a conserved quantity, there is a relation between q 1, q2, and q 3, and Q1, Q2, and Q3:
(22) The voltage between P and P' can be expressed in terms of C 3 and Q3, or in terms of C 1, C2, Q1, and Q2:
47
(23) and
(24) Using eq.(22) the following expressions for Q1 and Q2 can be obtained:
(25) (26) Substituting eq.(25) and eq.(26) into eq.(24) we obtain
(27) Combining eq.(27) and eq.(23), Q3 can be expressed in terms of known variables:
(28) Substituting the known values of the capacitance and initial charges we obtain
(29) The voltage across P and P' can be found by combining eq.(29) and eq.(23):
(30) The charges on capacitor 1 and capacitor 2 are equal to
(31) (32)
48
4. Dielectrics If the space between the plates of a capacitor is filled with an insulator, the capacitance of the capacitor will chance compared to the situation in which there is vacuum between the plates. The change in the capacitance is caused by a change in the electric field between the plates. The electric field between the capacitor plates will induce dipole moments in the material between the plates. These induced dipole moments will reduce the electric field in the region between the plates. A material in which the induced dipole moment is linearly proportional to the applied electric field is called a linear dielectric. In this type of materials the total electric field between the capacitor plates E is related to the electric field Efree that would exist if no dielectric was present:
(33) where [kappa] is called the dielectric constant. Since the final electric field E can never exceed the free electric field Efree, the dielectric constant [kappa] must be larger than 1. The potential difference across a capacitor is proportional to the electric field between the plates. Since the presence of a dielectric reduces the strength of the electric field, it will also reduce the potential difference between the capacitor plates (if the total charge on the plates is kept constant):
(34) The capacitance C of a system with a dielectric is inversely proportional to the potential difference between the plates, and is related to the capacitance C free of a capacitor with no dielectric in the following manner
(35) Since [kappa] is larger than 1, the capacitance of a capacitor can be significantly increased by filling the space between the capacitor plates with a dielectric with a large [kappa]. The electric field between the two capacitor plates is the vector sum of the fields generated by the charges on the capacitor and the field generated by the surface charges on the surface of the dielectric. The electric field generated by the charges on the capacitor plates (charge density of [sigma]free) is given by
(36) Assuming a charge density on the surface of the dielectric equal to [sigma]bound, the field generated by these bound charges is equal to 49
(37) The electric field between the plates is equal to Efree/[kappa] and thus
(38) Substituting eq.(36) and eq.(37) into eq.(38) gives
(39) or
(40)
Example Problem: The Parallel Plate Capacitor A parallel plate capacitor of plate area A and separation distance d contains a slab of dielectric of thickness d/2 (see Figure 8) and dielectric constant [kappa]. The potential difference between the plates is [Delta]V. a) In terms of the given quantities, find the electric field in the empty region of space between the plates. b) Find the electric field inside the dielectric. c) Find the density of bound charges on the surface of the dielectric.
Figure 8. The Parallel-Plate Capacitor. a) Suppose the electric field in the capacitor without the dielectric is equal to E0. The electric field in the dielectric, Ed, is related to the free electric field via the dielectric constant [kappa]:
50
(41) The potential difference between the plates can be obtained by integrating the electric field between the plates:
(42) The electric field in the empty region is thus equal to
(43) b) The electric field in the dielectric can be found by combining eq.(41) and (43):
(44) c) The free charge density [sigma]free is equal to
(45) The bound charge density is related to the free charge density via the following relation
(46) Combining eq.(45) and eq.(46) we obtain
(47)
5. Gauss Law in Dielectrics The electric field in an "empty" capacitor can be obtained using Gauss' law. Consider an ideal capacitor (with no fringing fields) and the integration volume shown in Figure 9. The area of 51
each capacitor plate is A and the charges on the plates are +/-Q. The charge enclosed by the integration volume shown in Figure 9 is equal to +Q. Gauss' law states that the electric flux [Phi] through the surface of the integration volume is related to the enclosed charge:
(48) If a dielectric is inserted between the plates, the electric field between the plates will change (even though the charge on the plates is kept constant). Obviously, Gauss' law, as stated in eq.(48), does not hold in this case. The electric field E between the capacitor plates is related to the dielectric-free field Efree:
(49) where [kappa] is the dielectric constant of the material between the plates. Gauss' law can now be rewritten as
(50) Gauss' law in vacuum is a special case of eq.(50) with [kappa] = 1.
Figure 9. Ideal Capacitor.
Example Problem: The Spherical Capacitor. A metallic sphere of radius R is surrounded by a concentric dielectric shell of inner radius R, and outer radius 3R/2. This is surrounded by a concentric, thin, metallic shell of radius 2R (see Figure 10). The dielectric constant of the shell is [kappa]. What is the capacitance of this contraption ? Suppose the charge on the inner sphere is Qfree. The electric field inside the dielectric can be determined by applying Gauss' law for a dielectric (eq.(50)) and using as the integration volume a sphere of radius r (where R < r < 3R/2)
52
(51) The electric field in this region is therefore given by
(52)
Figure 10. Problem 25. The electric field in the region between 3R/2 and 2R can be obtained in a similar manner, and is equal to
(53) Using the electric field from eq.(52) and eq.(53) we can determine the potential difference [Delta]V between the inner and outer sphere:
(54) The capacitance of the system can be obtained from eq.(54) using the definition of the capacitance in terms of the charge Q and the potential difference [Delta]V:
(55)
53
6 Energy in Capacitors The electric potential energy of a capacitor containing no dielectric and with charge +/-Q on its plates is given by
(56) where V1 and V2 are the potentials of the two plates. The electric potential energy can also be expressed in terms of the capacitance C of the capacitor
(57) This formula is also correct for a capacitor with a dielectric; the properties of the dielectric enters into this formula via the capacitance C.
Example Problem: Capacitors in Parallel. Ten identical 5 uF capacitors are connected in parallel to a 240-V battery. The charged capacitors are then disconnected from the battery and reconnected in series, the positive terminal of each capacitor being connected to the negative terminal of the next. What is the potential difference between the negative terminal of the first capacitor and the positive terminal of the last capacitor ? If these terminals are connected via an external circuit, how much charge will flow around this circuit as the series arrangement discharges ? How much energy is released in the discharge ? Compare this charge and this energy with the charge and energy stored in the original, parallel arrangement, and explain any discrepancies. The charge on each capacitor, after being connected to the 240-V battery, is equal to
(58) The potential difference across each capacitor will remain equal to 240 V after the capacitors are connected in series. The total potential difference across the ten capacitors is thus equal to
(59) If the two end terminals of the capacitor network are connected, a charge of 1.2 mC will flow from the positive terminal to the negative terminal (see Figure 11).
54
Figure 11. Problem 40. The electric energy stored in the capacitor network before discharge is equal to (60) The energy stored in each capacitor, after being charged to 240 V, is equal to
(61) Clearly no energy is lost in the process of changing the capacitor configuration from parallel to serial.
Example Problem: Energy Stored in Capacitors. Three capacitors are connected as shown in Figure 12. Their capacitances are C 1 = 2.0 uF, C 2 = 6.0 uF, and C 3 = 8.0 uF. If a voltage of 200 V is applied to the two free terminals, what will be the charge on each capacitor ? What will be the electric energy of each ?
Figure 12. problem 39. Suppose the voltage across capacitor C 1 is V1, and the voltage across capacitor (C 2 + C3) is V2. If the charge on capacitor C 1 is equal to Q1, then the charge on the parallel capacitor is also equal to Q1. The potential difference across this system is equal to
55
(62) The charge on capacitor 1 is thus determined by the potential difference [Delta]V
(63) The voltage V23 across the capacitor (C 2 + C3) is related to the charge Q1
(64) The charge on capacitor C 2 is equal to
(65) The charge on capacitor C 3 is equal to
(66) The electric potential energy stored in each capacitor is equal to
(67) For the three capacitors in this problem the electric potential energy is equal to
(68)
(69)
(70)
56
CURRENTS AND OHMS LAW 1. Electric Current
Figure 1. Electric field in a wire. When a wire is connected to the terminals of a battery, an electric field is generated inside the wire (see Figure 1). The free electrons in the wire will move in a direction opposite to that of the field lines. The electric charge will try to redistribute itself in such a way that the net electric field in the wire is equal to zero. However, the positive terminal of a battery acts as a sink for electrons and the negative terminal acts as a source of electrons, and a continuous flow of electrons will be created. This continuous flow of electrons is called an electric current. The symbol of current is I and its SI unit is the Ampere (A). The current is defined as (1) where dq is the amount of charge that flows past some given point on the wire during a time period dt. A current of 1 A is equal to 1 C/s. The current density j is defined as
(2) where I is the current flowing through the conductor, and A is the cross-sectional area of the conductor. Even though the electrons feel an electric field inside the conductor, they will not accelerate. The electrons will experience significant friction as a result of collisions with the positive ions in the conductor. On average, the electrons will move with a constant speed from the negative terminal of the battery to the positive terminal. Their average velocity, also called the drift velocity vd, is proportional to the electric field E
(3) For a given density of electrons in the conductor, an increase of the drift velocity of each of the electrons will increase the number of electrons passing by a given point on the conductor 57
per unit of time. This is illustrated in Figure 2. During a time interval dt the electrons will cover, on average, a distance equal to dx where
(4)
Figure 2. Motion of average electron in conductor. All electrons within a distance dx from the point P will therefore pass this point during the time interval dt. Suppose the density of electrons in the conductor is n electrons/m3. The number of electrons dN that will pass P during the time interval dt is then equal to (5) Since each electron carries a charge e, the total charge dQ that will pass point P in a time interval dt is equal to
(6) The current through the conductor is therefore equal to
(7) Equation (7) shows that the current in the conductor is proportional to the cross-sectional area of the conductor and proportional to the drift velocity. Since the drift velocity is proportional to the electric field E the following relation holds for the current in the conductor:
(8) The electric field in the conductor is determined by its length L and the potential difference [Delta]V between its two ends (E = [Delta]V/L). Equation (8) can therefore be rewritten as
(9) Equation (9) can be rewritten as
58
(10) The constant of proportionality [rho] is called the resistivity of the material. The resistivity [rho] depends on the characteristics of the conductor ([rho] is small for a good conductor, and [rho] is very large for an insulator). The resistance R of a conductor is defined as
(11) The SI unit of resistance is the ohm ([Omega]). Using the resistance R we can rewrite eq.(10)
(12) Equation (12) is called Ohm's Law. Equation (12) shows that the current through a conductor is proportional to the potential difference between the ends of the conductor and inversely proportional to its resistance. Equation (12) also shows that 1 [Omega] equals 1 V/A.
Example: Resistance of a Wire An aluminum wire has a resistance of 0.10 [Omega]. If you draw this wire through a die, making it thinner and twice as long, what will be its new resistance ? The initial resistance Ri of the aluminum wire with length L and cross-sectional area A is equal to
(13) The initial volume of the wire is L . A. After passing the wire through the die, it s length has changed to L' and its cross-sectional area is equal A'. Its final volume is therefore equal to L' A'. Since the density of the aluminum does not change, the volume of the wire does not change, and therefore the initial and final dimensions of the wire are related:
(14) or
(15) The problem states that the length of the wire is doubled (L' = 2 L). The final cross-sectional area A' is therefore related to the initial cross-sectional area A in the following manner:
(16) 59
The final resistance R f of the wire is given by
(17) The resistance of the wire has increased by a factor of four and is now 0.40 [Omega].
2. The resistivity of materials The resistivity [rho] has as units ohm-meter ([Omega] . m). The resistivity of most conductors is between 10-8 [Omega] . m and 10-7 [Omega] . m. The resistivity of a conductor depends not only on the type of the material but also on its temperature. The resistivity of an insulator varies between 1011 [Omega] . m and 1017 [Omega] . m. In all materials the resistivity decreases with decreasing temperature. In some materials, such as lead, zinc, tin and niobium, the resistivity vanishes as the temperature approaches absolute zero. At these low temperatures, these materials exhibit superconductivity.
Example: Connecting an AC The air conditioner in a home draws a current of 12 A. Suppose that the pair of wires connecting the air conditioner to the fuse box are No. 10 copper wires with a diameter of 0.259 cm and a length of 25 m each. a) What is the potential drop along each wire ? Suppose that the voltage delivered to the home is exactly 110 V at the fuse box. What is the voltage delivered to the air conditioner ? b) Some older homes are wired with No. 12 copper wire with a diameter of 0.205 cm. Repeat the calculation of part (a) for this wire.
Figure 3. Wiring diagram of air conditioner in problem 17. a) The resistivity of copper is 1.7 x 10-8 [Omega] . m (see Table 1). The resistance R Cu of each copper wire is equal to
60
(18) where L is the length of the wire and d is its diameter. A current I is flowing through the wires and I = 12 A. The voltage drop [Delta]V across each wire is equal to
(19) Figure 3 shows schematically a wiring diagram of the air conditioner circuit. The voltage across the air-conditioner unit is equal to 110 - 2 . [Delta]V, where [Delta]V is given by eq.(19). The length of each copper cable is 25 m, and its diameter is equal to 0.259 cm. The voltage drop across each wire is thus equal to
(20) The voltage across the AC unit is therefore equal to 108.1 V. b) A No. 12 wire has a diameter equal to 0.205 cm. The voltage drop across this wire is equal to
(21) and the voltage across the AC unit is equal to 106.9 V.
Example: HV Lines A high voltage transmission line has an aluminum cable of diameter 3.0 cm, 200 km long. What is the resistance of this cable ? The resistivity of aluminum is 2.8 x 10-8 [Omega] m. the length of the cable is 200 km or 2 x 105 m. The diameter of the cable is 3 cm and its cross-sectional area is equal to [pi] (d/2)2 or 7.1 x 10-4 m2. Substituting these values into eq.(11) the resistance of the cable can be determined
(22)
3. Resistance in combination A device that is specifically designed to have a high resistance is called a resistor. The symbol of a resistor in a circuit diagram is a zigzag line (see Figure 4). 61
Figure 4. Symbol of a resistor. Figure 5 shows two resistors with resistance R 1 and R 2 connected in series. Suppose the current flowing through the circuit is equal to I. The voltage drop [Delta]V1 across resistor R 1 is equal to (23) and the voltage drop [Delta]V2 across resistor R 2 is equal to
(24) The potential difference [Delta]V across the series circuit is equal to
(25) Equation (25) shows that two resistors connected in series act like one resistor with a resistance equal to the sum of the resistance of resistor 1 and the resistance of resistor 2
(26)
Figure 5. Two resistors connected in series. Figure 6 shows two resistors connected in parallel. In this circuit, the current through each resistor will be different, but the voltage drop [Delta]V across each resistor will be the same. Using Ohm's law the current I1 flowing through resistor R 1 can be calculated
(27) and the current I2 flowing through resistor R 2 is equal to
(28) The total current flowing through the circuit is equal to the sum of the currents through each resistor
(29) 62
The resistor network shown in Figure 6 is therefore equivalent to a single resistor R where R can be obtained from the following relation
(30) Equation (30) shows that the resistance of a parallel combination of resistors is always less than the resistance of each of the individual resistors.
Figure 6. Two resistor connected in parallel.
Example: Superconducting Cables Commercially manufactured superconducting cables consist of filaments of superconducting wire embedded in a matrix of copper. As long as the filaments are superconducting, all the current flows in them, and no current flows in the copper. But if the superconductivity suddenly fails because of a temperature increase, the current can spill into the copper; this prevents damage to the filaments of the superconductor. Calculate the resistance per meter of length of a copper matrix. The copper matrix has a diameter of 0.7 mm, and each of the 2100 filaments has a diameter of 0.01 mm. Consider 1 meter of cable. The cross-sectional area of each filament is [pi] . (d/2)2 = 7.9 x 1011 m2. The cross-sectional area of 2100 filaments is equal to 1.65 x 10-7 m2. The diameter of the copper matrix is equal to 0.7 mm, and its cross-sectional area is equal to 1.54 x 10-6 m2. The area of the copper itself is thus equal to 1.37 x 10-6 m2. The resistance of the copper matrix per unit length is equal to
(31) Suppose the resistivity of the filament at room temperature is the same as the resistivity of copper. The resistance of each superconducting filament is equal to
(32) The wire can be treated as a parallel circuit of one resistor representing the resistance of the copper matrix and 2100 resistors representing the 2100 strands of superconducting wire. The fraction of the current flowing through the copper matrix can be determined easily. Suppose
63
that the potential difference across the conductor is equal to [Delta]V. The current ICu flowing through the copper matrix is equal to
(33) The current Ifil flowing through the 2100 filaments is equal to
(34) The fraction F of the total current flowing through the copper matrix is equal to
(35) Two special cases will need to be considered. 1. The temperature is below the critical temperature. At or below this temperature the resistance of the filaments vanishes (R fil = 0 [Omega]). Equation (35) shows that in this case no current will flow through the copper matrix. 2. If the temperature of the wire is above the critical temperature, the current flow will change drastically. In this case, the fraction of the current flowing through the copper is equal to
(36) The copper matrix will carry 90% of the total current.
Example: Resistors in Circuits What is the resistance of the combination of four resistors shown in Figure 7. Each of the resistors has a value of R.
Figure 7. Problem 42. 64
To find the net resistance of the circuit shown in Figure 7, we start calculating the net resistance R 34 of the parallel circuit of resistors R 3 and R 4:
(37) or
(38) The circuit shown in Figure 7 is therefore equivalent with the circuit shown in Figure 8. Resistors R 2 and R 34 form a series network and can be replaced by a single resistor with a resistance R 234 where
(39)
Figure 8. Problem 42.
Figure 9. Problem 42. The circuit shown in Figure 8 can now be replaced by an equivalent circuit shown in Figure 9. The resistance R tot of this circuit can be obtained from the following relation
(40) or
(41) In the special case considered, R 1 = R2 = R3 = R4 = R. Thus
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(42) (43)
(44) For R = 3 [Omega] the total resistance is equal to 1.8 [Omega].
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DC CIRCUITS 1. Electromotive Force
Figure 1. Electron in electronic circuit.
To keep a current flowing in an electronics circuit we need a source of electric potential. Consider the circuit shown in Figure 1. The electrons in the conductor move from the end with the low (negative) potential towards the end with the high (positive) potential. The velocity of the electron is limited by the friction it experiences while traveling through the conductor, and on average will not change. However, the electron will lose potential energy, and its total mechanical energy is therefore reduced when it arrives at the end of the circuit (positive terminal of the source). In order to sustain the continuous flow of electrons around the circuit, the source must force the electrons from the terminal with the positive potential to the terminal with the negative potential. During this process, the source does work on the electrons, and increases their total mechanical energy. The strength of the source is measured in terms of the electromotive force (emf). The emf of a source is defined as the amount of electric energy delivered by the source per Coulomb of positive charge as charge passes through the source from the low-potential terminal to the high-potential terminal. A steady current will flow through the circuit in Figure 1 if the increase of potential energy of the electrons in the source is equal to the change in the potential energy of the electrons along their path through the external circuit. The unit of emf is the Volt (V) and usually the emf is simply called the voltage of the source. The symbol for emf is [epsilon]. The most important sources of emf are: 1. Batteries: Batteries convert chemical energy into electric energy. In a lead-acid battery the following reactions take place when the battery delivers a current:
(1) and
(2)
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These reactions deposit electrons on the negative electrode and absorb electrons from the positive electrodes. The reactions in eq.(1) and eq.(2) will continue until the sulfuric acid is depleted. At this point the battery is discharged. The battery can be recharged by forcing a current through the battery in the reverse direction. The reverse of the reactions listed in eq.(1) and eq.(2) will then occur. 2. Electric Generators: Electric generators convert mechanical energy into electric energy. The principle of operation of electric generators will be discussed later. 3. Fuel cells: A mixture of chemicals are combined in the fuel cell. The chemical reactions that occur are
(3) and
(4) The net result of these reactions is the conversion of hydrogen and oxygen into water which is removed from the fuel cell in the form of water vapor. A fuel cell therefore burns fuel. The efficiency of the best fuel cells is about 45%. 4. Solar Cell: A solar cell converts the energy of the sunlight directly into electric energy. The emf of a silicon solar cell is 0.6 V. However, the amount of current that can be extracted is rather small.
27.2. Single-loop currents A source with a time-independent emf is represented by the symbol shown in Figure 2. The long line in Figure 2 represents the positive terminal of the source, while the short line represents the negative terminal.
Figure 2. Schematic symbol of source of emf.
An example of a simple circuit in which a resistor is connected between the terminals of the emf source is shown in Figure 3. The current through this circuit can be determined by applying Kirchhoff's rule, which states: 68
Around a closed loop in a circuit, the sum of all the emfs and all the potential changes across resistors and other circuit elements must equal zero. In this sum, the emf of a source is reckoned as positive if the current flows through the source in the forward direction (from the negative terminal to the positive terminal) and negative if it flows in the backward direction. (Note: the direction of the current indicates the direction of the positive charge carriers).
Figure 3. Simple single-loop circuit.
The current flowing in the circuit shown in Figure 3 flows through the source from the negative terminal to the positive terminal. The emf of the source is therefore positive. The potential drop [Delta]V across the resistor can be determined using Ohm's law, and is equal to I R. Applying Kirchhoff's rule one obtains the following equation: (5) The current in the circuit is therefore equal to
(6)
Figure 4. Real source consisting of internal resistance Ri and ideal emf [epsilon].
In a real circuit we will have to take the internal resistance R i of the source into account. The internal resistance R i of the source can be regarded as connected in series to an ideal emf (see Figure 4). If a current I is flowing though the source, the emf across the external terminals of the source is equal to [epsilon] - I . Ri. The external emf therefore depends on the current delivered by the source. 69
3. Multi-loop circuits. The procedure used to calculate the current flowing through a complicated circuit (with several resistors and emfs) is called the loop method. This procedure can be summarized as follows (see Figure 5):
Figure 5. Multi-loop circuit 1. Regard the circuit as a collection of several closed current loops. The loops may overlap, but each loop must have at least one portion that does not overlap with other loops. In Figure 5 there are clearly 3 current loops. 2. Label the current in the loops I1, I2, I3, ....., and arbitrarily assign a direction to each of the currents. A useful policy can be to define the direction of the current in a loop as going from the positive terminal of the emf to the negative terminal of the emf (if an emf is present in the loop). This policy defines the direction of the current in loop 1, loop 2 and in loop 3 (see Figure 6). 3. Apply Kirchhoff's rule to each loop.
Figure 6. Current loops in multi-loop circuit.
For loop 1 Kirchhoff's law states that (7) 70
For loop 2 we find
(8) Finally, for loop 3 we find
(9) Using eq.(8) we can rewrite eq.(9) in the following manner
(10) or
(11) The current I2 can be obtained by substituting eq.(11) into eq.(8):
(12) The current I1 can be obtained from eq.(7):
(13) An alternative method to obtain the currents in a multi-loop circuit is the branch method which is based on Kirchhoff's first rule: The sum of all currents entering a branch point of a circuit (where three or more wires merge) must be equal to the sum of the currents leaving the branch point. The branch method involves the following steps (and is illustrated by discussing its application to the circuit in Figure 7):
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Figure 7. Branch method applied to multi-loop circuit. 1. Regard the given circuit as a collection of branches which begin and end at the points where wires merge. The circuit in Figure 5 has five branches. 2. Label the currents in each branch I1, I2, I3, ....., and arbitrarily assign a direction to each of these currents. The currents in the five branches of circuit 5 are indicated in Figure 7. 3. Apply Kirchhoff's second law to each loop. 4. Apply Kirchhoff's first law to each branch point. For loop 1 of the circuit shown in Figure 7 Kirchhoff's law dictates that
(14) For loop 2 we obtain
(15) For loop 3 we obtain
(16) Apply Kirchhoff's first law to the three branch points of the circuit shown in Figure 7:
(17) (18) (19) This procedure produces six equations with 5 unknown and the system is over-defined.
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The current I1 can be obtained from eq.(14):
(20) Equation (15) can be used to determine I4:
(21) The current I3 can be obtained from eq.(), using the solution for I4 just derived (eq.(21)):
(22) The current I2 can be obtained from eq.(17):
(23) The current I5 can be obtained from eq.(19)
(24) Note: since the loop method involves fewer unknown and fewer equations, it is usually quicker to use than the branch method.
Example Problem: Multi-loop Circuit Consider the circuit shown in Figure 8. Given that [epsilon]1 = 6.0 V, [epsilon]2 = 10.0 V, and R1 = 2.0 [Omega], what must be the value of the resistance R 2 if the current through this resistance is to be 2.0 A ?
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Figure 8. Multi-loop Circuit. Consider the two loops shown in Figure 8. We will assume that the current in each loop flows in a counter clockwise direction (as shown in Figure 8). The voltage drop across resistor R 1 is equal to R 1 . (I1 - I2). The sum of all emfs and all potential drops across the resistors in loop 1 is
(25) The sum of all emfs and all potential drops across the resistors in loop 2 is equal to
(26) Equation (26) can be rewritten as
(27) where we have used eq.(25). The current I2 is thus equal to
(28) The problems states that [epsilon]1 = 6.0 V, [epsilon]2 = 10.0 V and I2 = 2.0 A. These values combined with eq.(28) can be used to determine R 2:
(29)
4. Energy in circuits To keep a current flowing in a circuit, work must be done on the circulating charges. If a charge dq passes through a battery with emf [epsilon], the work done dW will be equal to
(30) The rate at which the emf source does work is given by
(31) 74
The rate of work is called the power P, and the unit of power is the Watt (W, 1 W = 1 VA). The moving charges dissipate some of their energy when passing through resistors. Suppose the potential drop across a resistor is [Delta]V. For an electron moving through the resistor the loss of potential energy is equal to e . [Delta]V. The energy lost is converted into heat, and the rate at which energy is dissipated is equal to
(32) or
(33) the conversion of electric energy into thermal energy is called Joule heating.
Example Problem: High-Voltage Transmission Line A high-voltage transmission line that connects a city to a power plant consists of a pair of copper cables, each with a resistance of 4 [Omega]. The current flows to the city along one cable, and back along the other. a) The transmission line delivers to the city 1.7 x 105 kW of power at 2.3 x 105 V. What is the current in the transmission line ? How much power is lost as Joule heat in the transmission line ? b) If the transmission line delivers the same 1.7 x 105 kW of power at 110 V, how much power would be lost in Joule heat ? Is it more efficient to transmit power at high voltage or at low voltage ? a) The power delivered to the city, Pdelivered, is equal to 1.7 x 105 kW. The voltage delivered, [Delta]Vdelivered, is equal to 2.3 x 105 V. The current through the cables can determined by applying eq.(32):
(34) This is also the current flowing through the transmission cables. The electric energy dissipated in the cables is equal to
(35) The power generated by the power plant must therefore be equal to
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(36) Comparing the power generated with the power delivered, we conclude that 98% of the generated power is delivered to the city. b) If the voltage delivered to the city is 110 V, than current through the transmission cables must be equal to
(37) This current is roughly 2000 times the current flowing through the transmission cables if the power is delivered at high voltage. The power dissipated in the transmission cables is
(38) The power generated by the power plant must therefore be equal to
(39) Comparing the power generated with the power delivered, we conclude that only 0.0009% of the generated power is delivered to the city. Clearly, our conclusion should be that the transmission of electric energy at high voltage is much more efficient that the transmission at low voltage. The voltage of the transmission line is therefore reduced to 110 V as close as possible to the house of the customer.
Example Problem: Charging a Battery A 12-V battery of internal resistance R i = 0.20 [Omega] is being charged by an external source of emf delivering 6.0 A. a) What must be the minimum emf of the external source ? b) What is the rate at which heat is developed in the internal resistance of the battery ? a) The circuit describing this problem is shown in Figure 9. In Figure 9 [epsilon]1 is the emf of the battery being recharged, and [epsilon]2 is the emf of the recharger. Applying Kirchhoff's second rule to the single loop circuit we obtain the following relation between the charging current and the emfs:
(40) Equation (40) can be rewritten as 76
(41) The emf of the battery to be recharged is [epsilon]1 = 12.0 V and it has an internal resistance Ri = 0.20 [Omega]. If the recharger is to deliver a current I = 6.0 A then the emf of the recharger must be equal to
(42)
Figure 9. Problem 24. b) The power dissipation in the internal resistance of the battery can be calculated by using eq.(33) (43)
Example Problem: Draining a Battery Suppose that a 12-V battery has an internal resistance of 0.40 [Omega]. a) If this battery delivers a steady current of 1.0 A into an external circuit until it is completely discharged, what fraction of the initial stored energy is wasted in the internal resistance ? b) What if the battery delivers a current of 10.0 A ? Is it more efficient to use the battery at low current or at high current ? a) Suppose the battery as an emf [epsilon]b and delivers a current I. The internal resistance of the battery is equal to R i. The voltage drop across the internal resistance is equal to I R i. The external voltage of the battery if thus equal to ([epsilon]b - I Ri). The power delivered by the battery to the external circuit is therefore equal to
(44) The total power delivered by the emf is equal to
(45)
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The fraction of the total delivered power that is dissipated in the internal resistance of the battery is equal to
(46) The ratio in eq.(46) is proportional to the current I. Using the values of the parameters specified in the problem we can calculate the ratio:
(47)
5. The RC circuit The current through the circuits discussed so far have been time-independent, as long as the emf of the source is time-independent. The currents in these circuits can be determined by applying Kirchhoff's first and/or second rule. A simple circuit in which the current is time dependent is the RC circuit which consists of a resistor R connected in series with a capacitor C (see Figure 10). Applying Kirchhoff's second rule to the current loop I gives
(48) However, the current I through the resistor and the charge Q on the capacitor are related:
(49) Substituting eq.(49) into eq.(48) we obtain
(50)
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Figure 10. Simple RC circuit. Equation (50) is a simple differential equation which can be solved for Q. The first step is to rewrite eq.(50) in the following manner: (51) The second step is to integrate each side of eq.(51):
(52) where Q0 is the charge on the capacitor at time t = 0. After evaluating both integrals in eq.(52) we obtain
(53) Equation (53) can be rewritten as
(54) or
(55) Let us consider the case in which the capacitor is discharged at time t = 0 (that is Q0 = 0 C). Equation (55) can then be rewritten as
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(56) The charge on the capacitor will increase as function of time and the final charge Qf on the capacitor is equal to
(57) The time constant of the charging process of the capacitor is equal to RC. When the charged capacitor is connected in series across a resistor, the charge on the capacitor will decrease. The time constant for this process is also RC. Once we know the charge on the capacitor as function of time we can immediately find the current as function of time by applying eq.(48):
(58) The current at time t = 0 is equal to [epsilon]/R and it decreases to zero with increasing time.
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THE MAGNETIC FORCE AND FIELD. 1. The magnetic force Up to now we have only considered the electrostatic forces acting on charges at rest. When the charges are in motion, an extra force acts on them. This extra force is called the magnetic force. The magnetic force between two charges q 1 and q 2, moving with velocities v1 and v2, is equal to
(1) where u 0 is called the permeability constant which is equal to 4[pi] x 10-7 Ns2/C2, and r is the distance between the two charges (see Figure 1). The ratio R of the magnetic force and the electric force is equal to
(2)
Figure 1. Relevant vectors for definition of magnetic force. Substituting the numerical values of [epsilon]0 and u 0 into eq.(2), the ratio R can be rewritten as
(3) where c is the velocity of light in vacuum (c = 3 x 108 m/s). Clearly, the magnetic force is small compared with the electric force unless the speed of the particles is high (a significant fraction of the velocity of light). A magnetic field B can be associated with the magnetic force. The magnetic field at some point in the vicinity of a moving charge can be determined by placing a test charge at that point and moving it with some velocity v. The test charge will experience, besides the electric
81
force, a magnetic force Fmag. Per definition, the magnetic field B is related to the magnetic force Fmag via
(4) A measurement of the magnetic force acting on the test charge for various directions of v can be used to determine the magnetic field B. The magnetic force is always perpendicular to the velocity vector and the direction of the magnetic field. The unit of magnetic field strength is the Tesla (T). Comparing eq.(1) and eq.(4) we can determine the magnetic field generated by a point charge q 2 moving with a velocity v2:
(5) Similar to electric field lines we can graphically represent the magnetic field by field lines. The density of field lines indicates the strength of the magnetic field. The tangent of the field lines indicates the direction of the magnetic field. The magnetic field lines form closed loops, that is they do not begin or end anywhere in the way that the electric field lines begin and end on positive and negative charges. This immediately implies that the magnetic flux through an arbitrary closed surface is equal to zero:
(6) The principle of superposition is also valid for the magnetic field.
Example Problem: Magnetic Field of a Neutron Star At the surface of a pulsar, or neutron star, the magnetic field may be as strong as 108 T. Consider the electron in a hydrogen atom on the surface of the neutron star. The electron is at a distance of 0.53 x 10-10 m from the proton and has a speed of 2.2 x 106 m/s. Compare the electric force that the proton exerts on the electron with the magnetic force that the magnetic field of the neutron star exerts on the electron. Is it reasonable to expect that the hydrogen atom will be strongly deformed by the magnetic field ? The electron in a hydrogen atom is at a distance r equal to 0.53 x 10-10 m from the proton. The electric force acting on the electron is equal to
(7)
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The maximum magnetic force acting on the electron occurs when the direction of the electron is perpendicular to the direction of the magnetic field. The maximum magnetic force is equal to
(8) Comparing eq.(7) and eq.(8) we conclude that the magnetic field is significantly stronger hat the electric field, and we expect that the orbits of the electrons are strongly affected by the intense magnetic field.
2. The Biot-Savart Law The definition of the magnetic force showed that two moving charges experience a magnetic force. In other words, a moving charge produces a magnetic field which results in a magnetic force acting on all charges moving in this field. A current flowing through a wire is equivalent to a collection of electrons moving with a certain velocity along the direction of the wire. Each of the moving electrons produces a magnetic field that is given by eq.(5). Consider a small segment of the wire with a length dL (see Figure 2). At any given time, a charge dq will be located in this segment. The magnetic field, dB, generated by this charge at point P is equal to
(9) where v is the velocity of the charge carriers. The time dt that it takes for all original charge carriers to leave the segment dL is given by
(10) The current I in the wire can now be obtained easily
(11)
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Figure 2. Calculation of magnetic field produced by an electric current. This equation can be rewritten as (12) and substituted into eq.(9):
(13) Equation (13) is called the Biot-Savart Law.
Example Problem: Helmholtz Coils Helmholtz coils are often used to make reasonably uniform magnetic fields in laboratories. These coils consist of two thin circular rings of wire parallel to each other and on a common axis, the z-axis. The rings have a radius R and they are separated by a distance which is also R. These rings carry equal currents in the same direction. Find the magnetic field at any point on the z-axis.
Figure 3. Calculation of magnetic field produced by one ring. The first step to calculate the field of a pair of Helmholtz coils is to calculate the magnetic field produced by each ring. Suppose the ring is located in the x-y plane and we are interested in the field at point P, a distance z above the x-y plane (see Figure 3). The net magnetic field 84
of the ring at point P will be directed along the z-axis. The magnitude dB of the magnetic field produced by a small segment of the ring with length dL is equal to
(14) To obtain eq.(14) we have used the fact that for any point on the ring, the position vector r is perpendicular to the direction of dL. The z-component of the magnetic field dB is equal to
(15) The magnitude of the position vector r is related to R and z:
(16) The angle a is also related to R and z:
(17) Combining eqs.(15), (16) and (17) we obtain
(18) Integrating eq.(18) over the whole ring we obtain for the total field generated by the ring
(19) Figure 4 shows the magnetic field generated by one coil with a radius of 1 m, located at z = 0 m. To find the field generated by a pair of Helmholtz coils, we assume that the coils are centered at z = 0 and at z = R. The magnetic field generated by the coil located at z = 0 is given by eq.(19). The magnetic field generated by the coil located at z = R is given by
(20)
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The total field on the axis of a pair of Helmholtz coils is equal to the sum of the field generated by coil 1 and the field generated by coil 2:
(21) The total magnetic field generated by a pair of Helmholtz coils is shown in Figure 5 where also the contributions of the two coils are shown individually. We observe that the field is very homogeneous between the coils (0 < z < R).
Figure 4. Magnetic field generated by a coil with R = 1 m.
Figure 5. Magnetic field generated by a pair of Helmholtz coils.
Example Problem: Magnetic Field due to a Long Wire A very long wire is bent at a right angle near its midpoint. One branch of it lies along the positive x-axis and the other along the positive y-axis (see Figure 6). The wire carries a current I. What is the magnetic field at a point in the first quadrant of the x-y plane ?
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Figure 6. Example Problem "Magnetic Field due to a Long Wire"
Figure 7. Field generated by wire. The first step to solve this problem is to look at the magnetic field produced by this single wire (see Figure 7). The direction of the magnetic field generated by a small segment of the wire is pointing out of the paper. The magnitude of the field, dB, is equal to
(22) The position x of the segment under consideration is determined by the angle a:
(23) or
(24) From eq.(24) we can obtain a relation between dx and da:
(25) 87
Furthermore,
(26) Substituting eq.(25) and eq.(26) into eq(22) we obtain
(27) The total field can be obtained by integrating eq.(27) over the wire. The integration limits are
(28) and
(29) The result of the integration is
(30) The field of the vertical wire can be obtained in a similar fashion:
(31) The magnitude of the total field is thus equal to
(32)
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3. The magnetic dipole The magnetic field on the axis of a current loop was discussed when we studied the field generated by Helmholtz coils. At large distances from the current loop (z >> R) the field is approximately equal to
(33) which shows that the magnetic field strength decreases as 1/z3. This dependence of the magnetic field strength on distance is similar to the dependence observed for the electric field strength of an electric dipole:
(34) Equation (33) is often rewritten as
(35) where
(36) is called the magnetic dipole moment of the loop. In general, the dipole moment of a current loop is equal to
(37) Magnetic dipole moments exist for objects as small as electrons and as large as the earth.
Example Problem: Spinning Charged Disk An amount of charge Q is uniformly distributed over a disk of paper of radius R. The disk spins about its axis with angular velocity [omega]. Find the magnetic dipole moment of the disk. The first step to solve this problem is to determine the dipole moment of a ring of the disk, with radius r and with a width dr. The amount of charge dq on this ring is equal to
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(38) The angular velocity of the disk is [omega] and its period T is equal to
(39) During one period the charge dq will pass any given point on the ring. The current dI is thus equal to
(40) The magnetic dipole moment du of the ring is equal to
(41) The total dipole moment of the disk can be found by integrating eq.(41) between r = 0 and r = R:
(42)
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AMPERES LAW 1. Introduction The magnetic field at a distance r from a very long straight wire, carrying a steady current I, has a magnitude equal to
(1) and a direction perpendicular to r and I. The path integral along a circle centered around the wire (see Figure 1) is equal to
(2) Here we have used the fact that the magnetic field is tangential at any point on the circular integration path.
Figure 1. Magnetic field generated by current. Any arbitrary path can be thought of as a collection of radial segments (r changes and [theta] remains constant) and circular segments ([theta] changes and r remains constant). For the radial segments the magnetic field will be perpendicular to the displacement and the scaler product between the magnetic field and the displacement is zero. Consider now a small circular segment of a trajectory around the wire (see Figure 2). The path integral along this circular segment is equal to
(3)
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Figure 2. Path integral along a small circular path. Equation (3) shows that the contribution of this circular segment to the total path integral is independent of the distance r and only depends on the change in the angle [Delta][theta]. For a closed path, the total change in angle will be 2[pi], and eq.(3) can be rewritten as (4) This expression is Ampere's Law: " The integral of B around any closed mathematical path equals u 0 times the current intercepted by the area spanning the path "
Example Problem: Field due to six parellel wires Six parallel aluminum wires of small, but finite, radius lie in the same plane. The wires are separated by equal distances d, and they carry equal currents I in the same direction. Find the magnetic field at the center of the first wire. Assume that the currents in each wire is uniformly distributed over its cross section. A schematic layout of the problem is shown in Figure 3. The magnetic field generated by a single wire is equal to
(5) where r is the distance from the center of the wire. Equation (5) is correct for all points outside the wire, and can therefore be used to determine the magnetic field generated by wire 2, 3, 4, 5, and 6. The field at the center of wire 1, due to the current flowing in wire 1, can be determined using Ampere's law, and is equal to zero. The total magnetic field at the center of wire 1 can be found by vector addition of the contributions of each of the six wires. Since the direction of each of these contributions is the same, the total magnetic field at the center of wire 1 is equal to
(6) 92
Figure 3. Six parallel wires.
2. The solenoid A solenoid is a device used to generate a homogeneous magnetic field. It can be made of a thin conducting wire wound in a tight helical coil of many turns. The magnetic field inside a solenoid can be determined by summing the magnetic fields generated by N individual rings (where N is the number of turns of the solenoid). We will limit our discussion of the magnetic field generated by a solenoid to that generated by an ideal solenoid which is infinitely long, and has very tightly wound coils. The ideal solenoid has translational and rotational symmetry. However, since magnetic field lines must form closed loops, the magnetic field can not be directed along a radial direction (otherwise field lines would be created or destroyed on the central axis of the solenoid). Therefore we conclude that the field lines in a solenoid must be parallel to the solenoid axis. The magnitude of the magnetic field can be obtained by applying Ampere's law.
Figure 4. The ideal solenoid. Consider the integration path shown in Figure 4. The path integral of the magnetic field around this integration path is equal to (7) where L is the horizontal length of the integration path. The current enclosed by the integration path is equal to N . I0 where N is the number of turns enclosed by the integration path and I0 is the current in each turn of the solenoid. Using Ampere's law we conclude that
(8) or
(9) 93
where n is the number of turns of the solenoid per unit length. Equation (9) shows that the magnetic field B is independent of the position inside the solenoid. We conclude that the magnetic field inside an ideal solenoid is uniform.
Example Problem: Superposition of magnetic fields A long solenoid of n turns per unit length carries a current I, and a long straight wire lying along the axis of this solenoid carries a current I'. Find the net magnetic field within the solenoid, at a distance r from the axis. Describe the shape of the magnetic field lines. The magnetic field generated by the solenoid is uniform, directed parallel to the solenoid axis, and has a magnitude equal to
(10) The magnetic field if a long straight wire, carrying a current I' has a magnitude equal to
(11) and is directed perpendicular to the direction of r and I'. The direction of Bwire is therefore perpendicular to the direction of Bsol. The net magnetic field inside the solenoid is equal to the vector sum of Bwire and Bsol. Its magnitude is equal to
(12) The angle a between the direction of the magnetic field and the z-axis is given by
(13)
Example Problem: Coaxial cable A coaxial cable consists of a long cylindrical copper wire of radius r1 surrounded by a cylindrical shell of inner radius r2 and outer radius r3 (see Figure 5). The wire and the shell carry equal and opposite currents I uniformly distributed over their volumes. Find formulas for the magnetic field in each of the regions r < r1, r1 < r < r2, r2 < r < r3, and r > r3. The magnetic field lines are circles, centered on the symmetry axis of the coaxial cable. First consider an integration path with r < r1. The path integral of B along this path is equal to
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(14) The current enclosed by this integration path is equal to
(15) Applying Faraday's law we can relate the current enclosed to the path integral of B
(16) Therefore, the magnetic field is B is equal to
(17)
Figure 5. The coaxial cable. In the region between the wire and the shell, the enclosed current is equal to I and the path integral of the magnetic field is given by eq.(14). Ampere's law states then that (18) and the magnetic field is given by
(19) In the third region (r2 < r < r3) the path integral of the magnetic field along a circular path with radius r is given by eq.(14). The enclosed current is equal to
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(20) The magnetic field is therefore equal to
(21) The current enclosed by an integration path with a radius r > r3 is equal to zero (since the current in the wire and in the shell are flowing in opposite directions). The magnetic field in this region is therefore also equal to zero.
3. Motion of charges in electric and magnetic fields The magnetic force acting on particle with charge q moving with velocity v is equal to
(22) This force is always perpendicular to the direction of motion of the particle, and will therefore only change the direction of motion, and not the magnitude of the velocity. If the charged particle is moving in a uniform magnetic field, with strength B, that is perpendicular to the velocity v, then the magnitude of the magnetic force is given by
(23) and its direction is perpendicular to v. As a result of this force, the particle will carry out uniform circular motion. The radius of the circle is determined by the requirement that the strength magnetic force is equal to the centripetal force. Thus
(24) The radius r of the orbit is equal to
(25) where p is the momentum of the charged particle. The distance traveled by the particle in one revolution is equal to
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(26) The time T required to complete one revolution is equal to
(27) The frequency of this motion is equal to
(28) and is called the cyclotron frequency. Equation (28) shows that the cyclotron frequency is independent of the energy of the particle, and depends only on its mass m and charge q. The effect of a magnetic field on the motion of a charged particle can be used to determine some of its properties. One example is a measurement of the charge of the electron. An electron moving in a uniform magnetic field will described a circular motion with a radius given by eq.(25). Suppose the electron is accelerated by a potential V0. The final kinetic energy of the electron is given by
(29) The momentum p of the electron is determined by its kinetic energy
(30) The radius of curvature of the trajectory of the electron is thus equal to
(31) Equation (31) shows that a measurement of r can be used to determine the mass over charge ratio of the electron. Another application of the effect of a magnetic field on the motion of a charged particle is the cyclotron. A cyclotron consists of an evacuated cavity placed between the poles of a large electromagnet. The cavity is cut into two D-shaped pieces (called dees) with a gap between them. An oscillating high voltage is connected to the plates, generating an oscillating electric field in the region between the two dees. A charged particle, injected in the center of the cyclotron, will carry out a uniform circular motion for the first half of one turn. The frequency of the motion of the particle depends on its mass, its charge and the magnetic field strength. The frequency of the oscillator is chosen such that each time the particle crosses the gap between the dees, it will be accelerated by the electric field. As the energy of the ion 97
increases, its radius of curvature will increase until it reaches the edge of the cyclotron and is extracted. During its motion in the cyclotron, the ion will cross the gap between the dees many times, and it will be accelerated to high energies. Up to now we have assumed that the direction of the motion of the charged particle is perpendicular to the direction of the magnetic field. If this is the case, uniform circular motion will result. If the direction of motion of the ion is not perpendicular to the magnetic field, spiral motion will result. The velocity of the charged particle can be decomposed into two components: one parallel and one perpendicular to the magnetic field. The magnetic force acting on the particle will be determined by the component of its velocity perpendicular to the magnetic field. The projection of the motion of the particle on the x-y plane (assumed to be perpendicular to the magnetic field) will be circular. The magnetic field will not effect the component of the motion parallel to the field, and this component of the velocity will remain constant. The net result will be spiral motion.
4. Crossed electric and magnetic fields A charged particle moving in a region with an electric and magnetic field will experience a total force equal to
(32) This force is called the Lorentz force.
Figure 6. Charged particle moving in crossed E and B fields. Consider a special case in which the electric field is perpendicular to the magnetic field. The motion of a charged particle in such a region can be quit complicated. A charged particle with a positive charge q and velocity v is moving in this field (see Figure 6). The direction of the particle shown in Figure 6 is perpendicular to both the electric field and the magnetic field. The electric force acting on the particle is directed along the direction of the electric field and has a magnitude equal to (33) The magnetic force acting on the charge particle is directed perpendicular to both v and B and has a magnitude equal to
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(34) The net force acting on the particle is the sum of these two components and has a magnitude equal to
(35) If the charged particle has a velocity equal to
(36) then the net force will be equal to zero, and the motion of the particle will be uniform linear motion. A device with crossed electric and magnetic fields is called a velocity selector. If slit are placed in the appropriate positions, it will transport only those particles that have a velocity defined by the magnitudes of the electric and magnetic fields.
Figure 7. Current in a magnetic field. A technique used to determine the density and sign of charge carriers in a metal is based on the forces exerted by crossed E and B fields on the charge carriers. The diagram shown in Figure 7 shows a metallic strip carrying a current in the direction shown and placed in a uniform magnetic field with the direction of the magnetic field being perpendicular to the electric field (which generates the current I). Suppose the charge carriers in the material are electrons, than the electrons will move in a direction opposite to that of the current (see Figure 7). Since the magnetic field is perpendicular to the electric field, it is also perpendicular to the direction of motion of the electrons. As a result of the magnetic force, the electrons are deflected downwards, and an excess of negative charge will be created on the bottom of the strip. At the same time, a deficit of negative charge will be created at the top of the strip. This charge distribution will generate an electric field that is perpendicular to the external electric field and, under equilibrium conditions, the electric force produced by this field will cancel the magnetic force acting on the electrons. When this occurs, the internal electric field, Ein, is equal to the product of the electron velocity, vd, and the strength of the magnetic field, B. As a result of the internal electric field, a potential difference will be created between the top and bottom of the strip. If the metallic strip has a width w, then the potential difference [Delta]V will be equal to
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(37) This effect is called the Hall effect. The drift velocity of the electrons depend on the current I in the wire, its cross sectional area A and the density n of electrons (see Chapter 28):
(38) Combining eq.(38) and eq.(37) we obtain the following expression for [Delta]V
(39) A measurement of [Delta]V can therefore be used to determine n.
5. Forces on a wire A current I flowing through a wire is equivalent to a collection of charges moving with a certain velocity vd along the wire. The amount of charge dq present in a segment dL of the wire is equal to
(40) If the wire is placed in a magnetic field, a magnetic force will be exerted on each of the charge carriers, and as a result, a force will be exerted on the wire. Suppose the angle between the direction of the current and the direction of the field is equal to [theta] (see Figure 8). The magnetic force acting on the segment dL of the wire is equal to
(41) The total force exerted by the magnetic field on the wire can be found by integrating eq.(41) along the entire wire.
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Figure 8. Magnetic force on wire.
Example Problem: Magnetic balance A balance can be used to measure the strength of the magnetic field. Consider a loop of wire, carrying a precisely known current, shown in Figure 9 which is partially immersed in the magnetic field. The force that the magnetic field exerts on the loop can be measured with the balance, and this permits the calculation of the strength of the magnetic field. Suppose that the short side of the loop measured 10.0 cm, the current in the wire is 0.225 A, and the magnetic force is 5.35 x 10-2 N. What is the strength of the magnetic field ? Consider the three segments of the current loop shown in Figure 9 which are immersed in the magnetic field. The magnetic force acting on segment 1 and 3 have equal magnitude, but are directed in an opposite direction, and therefore cancel. The magnitude of the magnetic force acting on segment 2 can be calculated using eq.(41) and is equal to
(42) This force is measured using a balance and is equal to 5.35 x 10-2 N. The strength of the magnetic field is thus equal to
(43)
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Figure 9. Current loop in immersed in magnetic field.
6. Torque on a current loop If a current loop is immersed in a magnetic field, the net magnetic force will be equal to zero. However, the torque on this loop will in general not be equal to zero. Suppose a rectangular current loop is placed in a uniform magnetic field (see Figure 10). The angle between the normal of the current loop and the magnetic field is equal to [theta]. The magnetic forces acting on the top and the bottom sections of the current loop are equal to
(44) where L is the length of the top and bottom edge. The torque exerted on the current loop, with respect to its axis, is equal to
(45)
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Figure 10. Current loop placed in uniform magnetic field. Using the definition of the magnetic dipole moment u, discussed in Chapter 30, eq.(45) can be rewritten as (46) where
(47) Using vector notation, eq.(45) can be rewritten as
(48) where the direction of the magnetic moment is defined using the right-hand rule. The work that must be done against the magnetic field to rotate the current loop by an angle d[theta] is equal to - [tau] d[theta]. The change in potential energy of the current loop when it rotates between [theta]0 and [theta]1 is given by
(49) A common choice for the reference point is [theta]0 = 90deg. and U([theta]0) = 0 J. If this choice is made we can rewrite eq.(50) as
(50) In vector notation: 103
(51) The potential energy of the current loop has a minimum when u and B are parallel, and a maximum when u and B are anti-parallel.
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ELECTROMAGNETIC INDUCTION 1. Motional emf Figure 1 shows a rod, made of conducting material, being moved with a velocity v in a uniform magnetic field B. The magnetic force acting on a free electron in the rod will be directed upwards and has a magnitude equal to
(1)
Figure 1. Moving conductor in magnetic field. As a result of the magnetic force electrons will start to accumulate at the top of the rod. The charge distribution of the rod will therefore change, and the top of the rod will have an excess of electrons (negative charge) while the bottom of the rod will have a deficit of electrons (positive charge). This charge distribution will produce an electric field in the rod. The strength of this electric field will increase until the electrostatic force produced by this field is equal in magnitude to the magnetic force. As this point the upward flow of electrons will stop and (2) or
(3) The induced electric field will generate a potential difference [Delta]V between the ends of the rod, equal to
(4) where L is the length of the rod. If the ends of the rod are connected with a circuit providing a return path for the accumulated charge, the rod will be a source of emf. Since the emf is associated with the motion of the rod through the magnetic field it is called motional emf. Equation (4) shows that the magnitude of the emf is proportional to the velocity v. Looking at
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Figure 1 we observe that vL is the area swept across by the rod per second. The quantity BvL is the magnetic flux swept across by the rod per second. Thus
(5) Although this formula was derived for the special case shown in Figure 1, it is valid in general. It holds for rods and wires of arbitrary shape moving through arbitrary magnetic fields. Equation (5) relates the induced emf to the rate at which the enclosed magnetic flux changes. In the system shown in Figure 1 the enclosed flux changes due to the motion of the rod. The enclosed magnetic flux can also be changed if the strength of the enclosed magnetic field changes. In both cases the result will be an induced emf. The relation between the induced emf and the change in magnetic flux is known as Faraday's law of induction: " The induced emf along a moving or changing mathematical path in a constant or changing magnetic field equals the rate at which magnetic flux sweeps across the path. " If we consider a closed path, Faraday's law can be stated as follows: " The induced emf around a closed mathematical path in magnetic field is equal to the rate of change of the magnetic flux intercepted by the area within the path " or
(6) The minus sign in eq.(6) indicates how polarity of the induced emf is related to the sign of the flux and to the rate of change of flux. The sign of the flux is fixed by the right-hand rule: " Curl the fingers of your right hand in the direction in which we are reckoning the emf around the path; the magnetic flux is then positive if the magnetic field lines point in the direction of the thumb, and negative otherwise. "
Example Problem: Metal Rod in Magnetic Field A metal rod of length L and mass m is free to slide, without friction, on two parallel metal tracks. The tracks are connected at one end so that they and the rod form a closed circuit (see Figure 2). The rod has a resistance R, and the tracks have a negligible resistance. A uniform magnetic field is perpendicular to the plane of this circuit. The magnetic field is increasing at a constant rate dB/dt. Initially the magnetic field has a strength B0 and the rod is at rest at a distance x 0 from the connected end of the rails. Express the acceleration of the rod at this instant in terms of the given quantities. 106
Figure 2. Metal Rod in Magnetic Field. The magnetic flux [Phi] enclosed by the rod and the tracks at time t = 0 s is given by
(7) The magnetic field is increasing with a constant rate, and consequently the enclosed magnetic flux is also increasing:
(8) Faraday's law of induction can now be used to determine the induced emf:
(9) As a result of the induced emf a current will flow through the rod with a magnitude equal to
(10) The direction of the current is along the wire, and therefore perpendicular to the magnetic field. The force exerted by the magnetic field on the rod is given by
(11) (see Chapter 31). Combining eq.(10) and (11) we obtain for the force on the wire
(12) 107
The acceleration of the rod at time t = 0 s is therefore equal to
(13)
Example Problem: Induced EMF in a Solenoid a) A long solenoid has 300 turns of wire per meter and has a radius of 3.0 cm. If the current in the wire is increasing at a rate of 50 A/s, at what rate does the strength of the magnetic field in the solenoid increase ? b) The solenoid is surrounded by a coil with 120 turns. The radius of this coil is 6.0 cm. What induced emf will be generated in this coil while the current in the solenoid is increasing ? a) The magnetic field in a solenoid was discussed in Chapter 31. If the solenoid has n turns per meter and if I is the current through each coil than the field inside the solenoid is equal to
(14) Therefore,
(15) In this problem n = 300 turns/meter and dI/dt = 50 A/s. The change in the magnetic field is thus equal to
(16) This equation shows that the magnetic field is increasing at a rate of 0.019 T/s. b) Since the magnetic field in the solenoid is changing, the magnetic flux enclosed by the surrounding coil will also change. The flux enclosed by a single winding of this coil is
(17) where rin = 3.0 cm is the radius of the solenoid. Here we have assumed that the strength of the magnetic field outside the solenoid is zero. The total flux enclosed by the outside coils is equal to
(18) 108
The rate of change of the magnetic flux due to that change in magnetic field is given by
(19) As a result of the change in the current in the solenoid an emf will be induced in the outer coil, with a value equal to
(20) If the ends of the coil are connected, a current will flow through the conductor. The direction of the current in the coil can be determined using Lenz' law which states that " The induced emfs are always of such a polarity as to oppose the change that generates them " Let us apply Lenz' law to problem 12. The direction of the magnetic field can be determined using the right hand rule and is pointed to the right. If the current in the solenoid increases the flux will also increase. The current in the external coil will flow in such a direction as to oppose this change. This implies that the current in this coil will flow counter clock wise (the field generated by the induced current is directed opposite to the field generated by the large solenoid).
2. The Induced Electric-Field A rod moving in a magnetic field will have an induced emf as a result of the magnetic force acting on the free electrons. The induced emf will be proportional to the linear velocity v of the rod. If we look at the rod from a reference frame in which the rod is at rest, the magnetic force will be zero. However, there must still be an induced emf. Since this emf can not be generated by the magnetic field, it must be due to an electric field which exists in the moving reference frame. The magnitude of this electric field must be such that the same induced emf is created as is generated in the reference frame in which the rod is moving. This requires that
(21) The electric field E' that exists in the reference frame of the moving rod is called the induced electric field. The emf generated between the ends of the rod is equal to
(22) which is equivalent to eq.(4). If the induced electric field is position dependent, then we have to replace eq.(22) with an integral expression
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(23) where the integral extends from one end of the rod to the other end of the rod. The difference between the induced electric field and the electric field generated by a static charge distribution is that in the former case the field is not conservative and the path integral along a closed path is equal to
(24) which is non-zero if the magnetic flux is time dependent.
3 Inductance A changing current in a conductor (like a coil) produces a changing magnetic field. This timedependent magnetic field can induce a current in a second conductor if it is placed in this field. The emf induced in this second conductor, [epsilon]2, will depend on the magnetic flux through this conductor:
(25) The flux [Phi]B1 depends on the strength of the magnetic field generated by conductor 1, and is therefore proportional to the current I1 through this conductor:
(26) Here, the constant L21 depends on the size of the two coils, on their separation distance, and on the number of turns in each coil. The constant L21 is called the mutual inductance of the two coils. Using this constant, eq.(25) can be rewritten as
(27) The unit of inductance is the Henry (H) and from eq.(27) we conclude that
(28)
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When the magnetic field generated by a coil changes (due to a change in current) the magnetic flux enclosed by the coil will also change. This change in flux will induce an emf in the coil, and since the emf is due to a change in the current through the coil it is called the self-induced emf. The self-induced emf is equal to
(29) In equation (29) L is called the self inductance of the coil. The self-induced emf will act in such a direction to oppose the change in the current.
Example Problem: Mutual Induction A long solenoid of radius R has n turns per unit length. A circular coil of wire of radius R' with n' turns surround the solenoid. What is the mutual induction ? Does the shape of the coil of wire matter ? The field inside the solenoid is assumed to be that of an infinitely long solenoid and has a strength equal to
(30) The flux enclosed by the external coil is equal to
(31) The induced emf in the external coil is equal to
(32) The mutual inductance L12 is thus equal to
(33)
4. Magnetic Energy If a steady current flows through an inductor, a time-independent magnetic field is created. If suddenly the current source is disconnected, the change in the enclosed magnetic flux will 111
generate a self-induced emf which will try to keep the current flowing in the original direction. The electric energy delivered by the self-induced emf was originally stored in the inductor in the form of magnetic energy. The amount of magnetic energy stored in the magnetic field can be determined by calculating the total power delivered by the power source to create the magnetic field. Suppose that after the battery is connected to the inductor the current increases at a rate of dI/dt. The self-induced emf created by this time-dependent current is equal to
(34) The current must deliver extra power to overcome this self-induced emf. The power required will be time dependent and is equal to
(35) The work done by the current is stored in the inductor as magnetic energy. The change dU in the magnetic energy of the inductor is thus equal to
(36) The total energy stored in the magnetic field of the inductor when the current reaches its final value can be obtained by integrating eq.(36) between I = 0 and I = If.
(37) For a solenoid of length l the self-inductance is equal to
(38) The magnetic energy stored in the solenoid is thus equal to
(39) where V is the volume of the solenoid. The magnetic energy can be expressed in terms of B and V:
(40)
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where B = u 0 n I is the magnetic field in the solenoid. The total magnetic energy of an inductor can now be expressed in terms of the magnetic energy density u which is defined as
(41) The magnetic energy stored in the magnetic field is equal to the energy density time the volume. Although we have derived the formula for the magnetic energy density for the special case of a very long solenoid, its derivation is valid for any arbitrary magnetic field.
Example Problem: The Toroid A toroid of square cross section has an inner radius R 1 and an outer radius R 2. The toroid has N turns of wire carrying a current I ; assume that N is very large. a) Find the magnetic energy density as function of the radius. b) By integrating the energy density, find the total magnetic energy stored in the solenoid. c) Deduce the self-inductance from the formula U = L . I 2/2. a) Apply Ampere's law using a spherical Amperian loop with radius r
(42) The current enclosed by the Amperian loop is equal to
(43) Using Ampere's law we can determine the magnetic field B :
(44) The magnetic energy density is thus equal to
(45) b) Suppose the height of the toroid is equal to h. Consider the a slice of the toroid shown in Figure 3.
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Figure 3. Cross section of the toroid. The volume dV of this slice is equal to (46) The magnetic energy stored in this segment is equal to
(47) The total magnetic energy stored in the toroid can be obtained by integrating eq.(47) with respect to r between r = R 1 and r = R 2:
(48) c) The magnetic energy stored in an inductor of inductance L is equal to 0.5 L I 2. Comparing this with eq.(48) we conclude that the self inductance L of the toroid is equal to
(49)
5. The RL circuit An RL circuit consists of a resistor and an inductor placed in series with a battery (see Figure 4). Applying Krichhoff's second rule to this single-loop circuit we obtain the following differential equation
(50) 114
Figure 4. The RL circuit. This differential equation has as a solution (51) This solution is valid if the battery is connected at t = 0. Equation (51) shows that the current at t = 0 s is equal to 0 and grows steadily to reach a final value of e /R at t = [infinity]. The time constant of the RL circuit is L/R. If the current has reached a steady value and the battery is suddenly disconnected, the conductor can generate a current through the resistor which will gradually decay as function of time. If the initial current is equal to [epsilon]/R, the current at time t will be equal to
(52)
Example Problem: Joule Heat in RLCircuit What Joule heat is dissipated by the current in eq.(52) in the resistor in the time interval between t = 0 and t = [infinity] ? Compare with the initial magnetic energy in the inductor. The current through the resistor is given in eq.(51). The power dissipated by this current in the resistor is equal to
(53) The total energy dissipated by this current in the resistor between t = 0 and t = [infinity] is equal to
(54) The magnetic energy stored in the inductor is equal to
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(55) and we conclude that all magnetic energy stored in the inductor is dissipated as Joule heat in the resistor.
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MAGNETIC MATERIALS 1. Magnetic Moments An electron moving in an orbit around a nucleus produces an average current along its orbit. As a consequence we can associate a magnetic moment with the orbiting electron. Suppose the electron is moving with a velocity v in an orbit with radius r. The period of this motion is equal to
(1) During one period T the charge e will pass each given point on the orbit. The current associated with this orbit is therefore equal to
(2) The magnetic moment associated with this current is equal to
(3) It is common to specify the orbit of an electron in terms of its angular momentum L. Using the definition of the angular momentum L we can relate the electron velocity v and the radius of its orbit r to the angular momentum L in the following manner:
(4) where m is the mass of the electron. Using eq.(4) we can express the magnetic moment of the electron in terms of the angular momentum L:
(5) The magnetic moment of the electron is thus proportional the angular momentum L. The angular momentum of the electron is quantized, and the only possible values are nh where n is an integer (n = 0, 1, 2, 3, ...) and h is a constant (h = 1.06 . 10 -34 J s). The magnetic moment of an electron with angular momentum L = 1 is equal to
(6)
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Since this magnetic moment is associated with the orbital motion of the electron around the nucleus it is called the orbital magnetic moment. Another contribution to the magnetic moment is due to the rotational motion of the electron. Classically we can regard an electron as a small ball of negative charge spinning around its axis. The intrinsic angular momentum, generated by the electron spin, is equal to
(7) This constant is also called the Bohr magneton. The total magnetic moment of an atom is equal to the vector sum of the orbital magnetic moments and the intrinsic magnetic moments of all its electrons. The contribution of the nuclear magnetic moment is small and often can be neglected. Each atom acts like a magnetic dipole and produces a small, but measurable magnetic field.
Example Problem: Two-electron Interactions a) Two electrons are separated by a distance of 1.0 x 10 -10 m. The first electron is on the axis of spin of the second. What is the magnetic field that the magnetic moment of the second electron produces at the position of the first ? b) The potential energy of the magnetic moment of the first electron in this magnetic field depends on the orientation of the electrons. What is the potential energy if the spins of the two electrons are parallel ? If anti parallel ? Which orientation has the least energy ? a) Figure 1 shows the orientation of the two electrons. The z-axis is defined to coincide with the spin of electron 2. With each spinning electron a magnetic dipole moment can be associated. Due to the negative charge of the electron the dipole moment is pointed in a direction opposite to that of the spin of the electron and it has a magnitude equal to u spin = 9.27 x 10 -24 Am2. The magnetic field generated by electron will be the magnetic field generated by a dipole with dipole moment u spin. On the z-axis (and for z >> R e) the field strength will fall of as 1/z3 and at z = 1.0 x 10 -10 m has a strength equal to
(8)
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Figure 1. Two-electron interactions. b) The potential energy of a dipole with dipole moment u spin in a magnetic field of strength B is equal to
(9) where [theta] is the angle between the dipole moment and the magnetic field. The potential energy of electron 1 reaches a maximum value when [theta] = [pi] and a minimum value when [theta] = 0. Evaluating equation (9) for these extreme cases yields
(10) The potential energy will have a minimum value when the spins are parallel and a maximum value when the spins are anti-parallel.
2. Paramagnetism Even though each atom in a material can have a magnetic moment, the direction of each dipole is randomly oriented and their magnetic fields average to zero. If the material is immersed in an external magnetic field, the dipoles will tend to align themselves with the field in order to minimize the torque exerted on them by the external magnetic field. The atoms in the material will produce an extra magnetic field in its interior that has the same direction as the external magnetic field. This increase in strength of the magnetic field can be quantified in terms of the permeability constant [kappa]m:
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(11) where Bfree is the external magnetic field. The definition of the permeability constant (see eq.(11)) shows that [kappa]m >= 1. For all paramagnetic materials the permeability constant is very close to 1, and as a consequence, the increase in the magnetic field strength is rather small.
Example Problem: Filled Solenoid Show that the self-inductance per unit length of a very long solenoid filled with a paramagnetic material is equal to [kappa]m u0 n2 [pi] R2, where n is the number of turns of wire per unit length and R is the radius of the solenoid. The magnetic field in an empty solenoid is equal to
(12) When the solenoid is filled with a paramagnetic material the strength of the magnetic field will increase (see eq.(11)) and will be equal to
(13) The magnetic flux enclosed by a section of the solenoid of unit length is equal to
(14) The change in enclosed magnetic can be obtained from eq.(14) by differentiating both sides with respect to time
(15) The emf induced by this change in magnetic flux can be calculated using Faraday's law
(16) From eq.(16) we conclude that the self-inductance of the solenoid filled with a paramagnetic material is equal to
(17) 120
3. Ferromagnetism The alignment of the spins of some of the electrons in a ferromagnetic material will increase the magnetic field in this material in much the same way as the alignment of the orbital magnetic dipole moments of atoms increases the field strength in a paramagnetic material. In a ferromagnetic material the degree of alignment of the electron spins between neighboring atoms is high as a result of a special force that tends to lock the spins of these electrons in a parallel direction. This force is so strong that the spins remain aligned even when the external magnetic field is removed. Materials with such properties are called permanent magnets. The force that is responsible for the alignment of the electron spins occurs in only five elements: • • • • •
Iron Nickel Cobalt Dysprosium Gadolinium
Although ferromagnetic materials will remain magnetized after the external magnetic field has been removed, they can also be found in non-magnetized states. On a small scale (domains with sizes of less than 0.1 - 5 mm) all spins will be perfectly aligned, on a large scale the domains are oriented randomly, and the net magnetic field is equal to zero. However, if the material is immersed in an external magnetic field, all dipoles will tend to align along the external field lines, and the strong spin-spin force will keep the dipoles aligned even after the external magnetic field has been removed. The increase of the magnetic field in a ferromagnet can be very large. For iron, the increase in field strength can be as large as 5000. The degree of alignment of the spins in a ferromagnetic material after the external magnetic field has been removed depends on the temperature. An increase in the temperature of the material will increase the chance of random rearrangement of the magnetic dipoles. Above a certain temperature, called the Curie temperature, the magnetism of the ferromagnet disappears completely.
Example Problem: Number of Aligned Electrons Under conditions of maximum magnetization, the dipole moment per unit volume of cobalt is 1.5 x 10 5 Am 2/m3. Assuming that this magnetization is due to completely aligned electrons, how many such electrons are there per unit volume ? How many aligned electrons are there per atom ? The density of cobalt is 8.9 x 10 3 kg and the atomic mass is 58.9 g/mole. The dipole moment of 1 m 3 of cobalt is equal to 1.5 x 10 5 Am 2. Each aligned electron contributes a dipole moment of 9.27 x 10 -24 Am 2. The number of aligned electrons is thus equal to
(18) 121
The number of atoms in 1 m 3 of cobalt is equal to
(19) Comparing eq.(18) and eq.(19) we conclude that the total number of aligned electrons per atom is equal to 0.18.
4. Diamagnetism In a diamagnetic material the magnetization arises from induced magnetic dipoles. This in contrast to the paramagnetic and ferromagnetic materials where the magnetic properties are determined by the alignment of permanent magnetic dipoles. In a diamagnetic material the dipole moments of the atoms do not align themselves with the magnetic field, but their strength is changed by the external field.
Figure 2. Electron orbiting nucleus. The change in the dipole moment as function of the applied magnetic field can be estimated using a simple classical model in which the electron moves in a circular orbit around the nucleus (see Figure 2). If no external magnetic field is present, the velocity v0 of the electron is determined by the radius r0 of the orbit via the following relation
(20) When the magnetic field is turned on, the electron will experience in addition to the electric force a magnetic force equal to
(21)
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This force is radially directed (inwards or outwards depending on the direction of v0 with respect to B). The condition for uniform circular motion is now
(22) Assuming that the size of the orbit of the electron does not change, we conclude that the effect of the magnetic field is a change in the velocity of the electron. Combining eq.(22) and eq.(20) we obtain
(23) It is convenient to express the velocity of the electron in terms of its angular frequency
(24) Substituting the expressions for v0 and v from eq.(24) into eq.(23) we obtain
(25) or
(26) The change in frequency [Delta][omega] is defined via the following relation
(27) Substituting eq.(27) into eq.(26) we obtain
(28) or
(29) 123
This frequency is called the Larmor frequency, and indicates the maximum change in the velocity of the orbital electrons when an external magnetic field is applied. As a result of the change in the velocity of the orbital electrons there will be a change in the orbital magnetic moment. The orbital magnetic moment before the external magnetic field is applied can be calculated using eq.(3)
(30) When an external magnetic field is applied the angular frequency [omega] will change by [Delta][omega]. Equation (30) shows that the associated change in the orbital magnetic moment is equal to
(31) The change of the angular frequency can be positive or negative, depending on the direction of v0 with respect to the direction B. The angular frequency will increase if the direction of the field and the electron velocity are not related via the right-hand rule (field into paper in Figure 2). The change in the magnetic moment of the dipole is such that the net field strength (external + orbital magnetic field) is lowered. This means an increase in the orbital magnetic moment if the orbital magnetic field is directed in a direction opposite to the external field, and a decrease in the orbital magnetic moment if the orbital magnetic field is directed in the same direction as the external field. In both cases the total magnetic field strength is reduced, and, consequently, the permeability constant [kappa]m is less than 1. For most diamagnetic materials the permeability constant is very close to 1 (1 - [kappa]m ~ 1 x 10 -5).
124
AC CIRCUITS 1. Alternating Current The current from a 110-V outlet is an oscillating function of time. This type is called Alternating Current or AC. A source of AC is symbolized by a wavy line enclosed in a circle (see Figure 1). The time dependence of the AC or the emf of the AC source is of the form
(1) where [epsilon]max is the maximum amplitude of the oscillating emf and [omega] is the angular frequency.
Figure 1. Symbol of AC source.
2. AC Resistor Circuits Figure 2 shows a single-loop circuit with a source of alternating emf and a resistor. The current through the resistor will be a function of time. The magnitude of this current can be obtained via Kirchhoff's second rule which implies that
(2)
Figure 2. Single-loop AC resistor circuit. The current I is thus equal to
(3) Equation (3) shows that the current oscillates in phase with the emf. The power dissipated in the resistor depends on the current through and the voltage across the resistor and is therefore also a function of time: 125
(4) The average power dissipated in the resistor during one cycle is equal to
(5) In the last step of the derivation of eq.(5) we used the relation between the period T and the angular frequency [omega] (T = 2[pi]/[omega]). Often, eq.(5) is written in terms of the rootmean-square voltage [epsilon]rms which is defined as
(6) In terms of [epsilon]rms we can rewrite eq.(5) as
(7) The root-mean-square voltage [epsilon]rms of the AC source is the value of the DC voltage that dissipates the same power in the resistor as the AC voltage with a maximum voltage equal to [epsilon]max. The household voltage of 115 Volt is the root-mean-square voltage; the actual peak voltage coming out of a household outlet is 163 V.
3. AC Capacitor Circuits Figure 3 shows a capacitor connected to a source of alternating emf. The charge on the capacitor at any time can be obtained by applying Kirchhoff's second rule to the circuit shown in Figure 3 and is equal to
(8) The current in the circuit can be obtained by differentiating eq.(8) with respect to time
126
(9)
Figure 3. AC capacitor circuit. The current in the circuit is 90deg. out of phase with the emf. Since the maxima in the current occur a quarter cycle before the maxima in the emf, we say that the current leads the emf. It is customary to rewrite eq.(9) as
(10) where
(11) is called the capacitive reactance. Note that eq.(10) is very similar to eq.(3) if the resistance R is replaced by the capacitive reactance XC. The power delivered to the capacitor is equal to
(12) The power fluctuates between positive and negative extremes, and is on average equal to zero. These fluctuations corresponds to periods during which the emf source provides power to the battery (charging) and periods during which the battery provides power to the emf source (discharging).
4. AC Inductive Circuit Figure 4 shows a circuit consisting of an inductor and a source of alternating emf. The selfinduced emf across the inductor is equal to LdI/dt. Applying Kirchhoff's second rule to the circuit shown in Figure 4 we obtain the following equation for dI/dt:
127
(13)
Figure 4. AC Inductor Circuit. The current I can be obtained from eq.(13) by integrating with respect to time and requiring that the magnitude of the DC current component is equal to zero:
(14) The current is again 90deg. out of phase with the emf, but this time the emf leads the current. Equation (14) can be rewritten as
(15) where
(16) is called the inductive reactance. The power delivered to the inductor is equal to
(17) and the average power delivered to the inductor is equal to zero.
Example Problem: AC Circuit Consider the circuit shown in Figure 5. The emf is of the form [epsilon]0 sin([omega]t). In terms of this emf and the capacitance C and the inductance L, find the instantaneous currents through the capacitor and the inductor. Find the instantaneous current and the instantaneous power delivered by the source of emf.
128
Figure 5. AC Circuit. The circuit shown in Figure 5 is a simple multi-loop circuit. The currents in this circuit can be determined using the loop technique. Consider the two current loops I1 and I2 indicated in Figure 5. Applying Kirchhoff's second rule to loop number 1 we obtain
(18) Applying Kirchhoff's second rule to loop number 2 we obtain
(19) Equation (18) can be used to determine I1:
(20) Equation (19) can be differentiated with respect to time to obtain I2:
(21) The current delivered by the source of emf is the sum of I1 and I2
(22) The power delivered by the source of emf is equal to
(23)
5. LC Circuits 129
Figure 6 shows a single-loop circuit consisting of an inductor and a capacitor. Suppose at time t = 0 s the capacitor has a charge Q0 and the current in the circuit is equal to zero. The current in the circuit can be found via Kirchhoff's second rule which requires that
(24)
Figure 6. LC circuit. The current I(t) can be obtained from Q(t) by differentiating Q with respect to time: (25) Substituting eq.(25) into eq.(24) we obtain
(26) or
(27) A solution of eq.(27) is
(28) where [phi] is a phase constant that must be adjusted to fit the initial conditions. The current in the circuit can be obtained by substituting eq.(28) into eq.(25):
(29) The initial conditions for the circuit shown in Figure 6 are
(30) 130
(31) These boundary conditions are satisfied if [phi] = 0. In this case, the charge and the current in the LC circuit are given by
(32) and
(33) The energy stored on the capacitor is a function of time since the charge on it is a function of time. The energy stored is equal to
(34) The energy stored in the inductor is also time dependent since the current through it is a function of time. The energy stored is equal to
(35) Equation (34) and eq.(35) show that the maximum energy is stored in the inductor when the energy stored in the capacitor is zero and vice-versa. The total energy of the circuit can be obtained by summing the energy stored in the capacitor and the energy stored in the inductor:
(36) Equation (36) shows that the energy stored in the circuit is conserved. This is expected since no Joule heat will be generated in a circuit in which none of the elements has any resistance. In practice, the circuit shown in Figure 6 will have some resistance (even good conductors will have a finite resistance). A realistic LRC circuit is shown in Figure 7. Applying Kirchhoff's second rule to the circuit shown in Figure 7 we obtain
(37)
131
Since the current I is equal to dQ/dt we can rewrite eq.(37) as
(38)
Figure 7. LRC Circuit. A solution of the differential equation shown in eq.(38) is (39) The constant [gamma] can be determined by substituting eq.(39) into eq.(38):
(40) This equation has to be satisfied at all times. This will only be the case if the terms within the parenthesis are equal to zero:
(41) (42) The constant [gamma] is determined by eq.(42)
(43) The angular frequency [omega] can be obtained from eq.(41) by substituting eq.(43) for [gamma]
(44)
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Equation (39) shows that the presence of the resistor in the circuit will produce damped harmonic motion. The damping constant [gamma] is proportional to the resistance R (see eq.(43)). The change in the energy of the system can be studied by looking at the maximum charge on the capacitor. At time t = 0 s the capacitor is fully charged with a charge equal to Q0 and the energy stored in the capacitor is equal to
(45) After one cycle (t = 2[pi]/[omega]) the maximum charge on the capacitor has decreased. This implies that also the energy stored on the capacitor has decreased
(46) The relative change in the electrical energy of the system is therefore equal to
(47) The loss of electrical energy in a LRC circuit is usually expressed in terms of the quality Qvalue"
(48) A high quality factor indicates a low resistance and consequently a small relative energy loss per cycle.
133
Figure 8. Driven LCR circuit. As a result of the damping in a LRC circuit the amplitude of the oscillations will gradually decrease. In order to sustain an oscillation in a LRC circuit, energy needs to be supplied, for example by connecting an oscillating source of emf to the circuit. Consider the circuit shown in Figure 8 consisting of an alternating source of emf, a resistor R, a capacitor C, and an inductor L. Suppose the emf has an angular frequency [omega] and a maximum amplitude [epsilon]max: (49) Applying Kirchhoff's second rule to the circuit shown in Figure 8 produces the following relation
(50) Under steady-state conditions, the current in the circuit will oscillate with the same angular frequency [omega] as the source of emf, but not necessarily in phase. The most general solution for the current is therefore
(51) where [phi] is called the phase angle between the current and the emf. The maximum current Imax and the phase angle [phi] can be determined by substituting eq.(51) in eq.(50):
(52) Equation (52) can be rewritten using trigonometric identities as
(53) This equation can only be satisfied if the expressions in the brackets are equal to zero. This requires that
(54) and
(55) 134
Eq.(55) can be used to determine the phase angle:
(56) Equation (54) can be used to determine the maximum current:
(57) Substituting eq.(56) into eq.(57) we obtain for the maximum current
(58) The quantity
(59) is called the impedance of the LCR circuit. Equation (58) shows that the maximum amplitude is achieved when
(60) The maximum amplitude of the current is
(61) The system will reach its maximum amplitude when the driving frequency [omega] of the applied emf is equal to
(62) This frequency is the natural frequency of the LC circuit discussed previously. When the system is driven at the natural frequency, it is said to be in resonance. 135
6. The Phasor Diagram A short cut that can be used to determine the amplitude and phase of current in an AC circuit is the phasor diagram. In a phasor diagram the amplitude of a sinusoidal function is represented by a line segment of length equal to its amplitude. The phase is represented by the angle between the line segment and the horizontal axis. The sum of voltage drops across the components of the circuit is then equivalent to the vector sum of the phasors. To illustrate the use of phasor diagrams we determine the amplitude and phase of the LCR circuit just discussed. The applied emf and induced current are given by the following equations:
(63) The voltages across the resistor, the capacitor and the inductor are equal to
(64) The three phasors corresponding to these three voltages are shown in the phasor diagram in Figure 9. The voltage drop across the resistor has the same phase as the current. The vector sum of these three vectors is also indicated and should be equal to the applied emf. The amplitude of the vector sum of the three phasors must be equal to the amplitude of the applied emf. Thus
(65) The phase of the vector sum of the phasors in Figure 9 is equal to [omega]t, and the angle between the current (and the phasor representing the voltage drop across the resistor) and the vector sum of the phasors is equal to the phase angle [phi]. From Figure 9 it is obvious that
(66)
136
Figure 9. Phasor circuit for LCR circuit.
Example Problem: LCC Circuit Consider the circuit shown in Figure 10. The oscillating source of emf delivers a sinusoidal emf of amplitude 0.80 V and frequency 400 Hz. The inductance is 5.0 x 10-2 H, and the capacitances are 8.0 x 10-7 F and 16.0 x 10-7 F. Find the maximum instantaneous current in each capacitor. Consider first the two capacitors. The emf across each of the capacitor must always be the same. This implies that
(67)
Figure 10. LCC Circuit. Rewriting eq.(67) in terms of the current I1 through capacitor C 1 and the current I2 through capacitor C 2 we obtain
(68) or
137
(69) Equation (69) can only be true at all times if the integrand is equal to zero. This requires that
(70) In order to determine the maximum current in the circuit we use the phasor technique just discussed. Consider the phasor diagram shown in Figure 11. The phasor labeled I indicates the current in the circuit. The voltages across the inductor and the capacitor are 90 degrees out of phase with the current and are indicated in Figure 11 by the phasors labeled VL and VC. The total voltage drop across the circuit elements (vector sum of VL and VC) is also 90 degrees out of phase with the current. Since the total voltage drop across the circuit elements must be equal to the applied emf, we conclude that the phase angle between the current and the emf is +/- 90 degrees. The sign depends on the values of inductance, the capacitance and the angular frequency of the emf.
Figure 11. Phasor diagram for LCC Circuit. The magnitude of the vector sum of the voltages across the inductor and the capacitor must be equal to the magnitude of the emf. Thus
(71) Equation (71) can be used to determine the maximum current in the circuit:
(72) The capacitance C used in eq.(72) is the net capacitance of the parallel network consisting of capacitor C 1 and capacitor C 2 (C = C 1 + C2). The sum of the currents flowing through
138
capacitors is equal to the maximum current in eq.(73). To determine the current through capacitor C1 and capacitor C2 we can combine eq.(72) and eq.(70). In this manner we obtain
(73) and
(74)
Example Problem: RC Circuit An RC circuit consists of a resistor with R = 0.80 [Omega] and a capacitor with C = 1.5 x 10-4 F connected in series with an oscillating source of emf. The source generates a sinusoidal emf with emax = 0.40 V and angular frequency equal to 9 x 103 rad/s. Find the maximum current in the circuit. Find the phase angle of the current and draw a phasor diagram, with the correct lengths and angles for the phasors. Find the average dissipation of power in the resistor. The applied emf and the potential drops across the circuit elements in the RC circuit are listed in eq.(75).
(75a) (75b)
(75c) The phasors representing the voltage drops across the resistor and across the capacitor are shown in Figure 12. The vector sum of these phasors is also indicated. The magnitude of the vector sum of the phasors must be equal to the magnitude of the applied emf. Thus
(76) The maximum current is thus equal to
139
(77)
Figure 12. Phasor diagram for an RC Circuit. The phase angle [phi] can be calculated easily (see Figure 12). It is determined by
(78)
7. The Transformer A transformer consists of two coils wound around an iron core (see Figure 13). The iron core increases the strength of the magnetic field in its interior by a large fraction (up to 5000) and as a consequence, the field lines must concentrate in the iron. One of the coils, the primary coil, is connected to a source of alternating emf. The emf induced in the primary coil is related to the rate of change of magnetic flux (Faraday's law of induction):
(79) Applying Kirchhoff's second rule to the primary circuit, we conclude that the induced emf in the coil must be equal to the applied emf. Thus
(80)
140
Figure 13. The transformer. All field lines that pass through a winding of coil 1 will also pass through a winding of coil 2. The flux through each winding of the primary coil is therefore equal to the flux through each winding of the secondary coil. If the primary coil has N1 windings and the secondary coil has N2 windings, then the total flux through the two coils are related
(81) or
(82) The change in the enclosed flux of the primary coil will be related in the same way to the change of flux in the secondary coil:
(83) The emf induced across the secondary coil can be obtained using Faraday's law and can be expressed in terms of the emf in the primary circuit:
(84) This emf is available to the various loads in the secondary circuit. If the secondary circuit is open, no current will flow in it, and the primary circuit is nothing else than a single-loop circuit with an alternating source of emf and an inductor. The average power dissipated by the emf in such a circuit is zero, and consequently the transformer does not consume any electric power. If the secondary circuit is connected to a load, a current will flow. This induced current will change the magnetic flux in the transformer and induce a current in the primary coil. If this occurs, the primary circuit will consume power. In an ideal capacitor, the power delivered by the source of emf in the primary circuit equal the power that the secondary circuit delivers to its load. Thus 141
(85)
142
THE DISPLACEMENT CURRENT AND MAXWELLS EQUATIONS 1. The displacement current The calculation of the magnetic field of a current distribution can, in principle, be carried out using Ampere's law which relates the path integral of the magnetic field around a closed path to the current intercepted by an arbitrary surface that spans this path:
(1) Ampere's law is independent of the shape of the surface chosen as long as the current flows along a continuous, unbroken circuit. However, consider the case in which the current wire is broken and connected to a parallel-plate capacitor (see Figure 1). A current will flow through the wire during the charging process of the capacitor. This current will generate a magnetic field and if we are far away from the capacitor, this field should be very similar to the magnetic field produced by an infinitely long, continuous, wire. However, the current intercepted by an arbitrary surface now depends on the surface chosen. For example, the surface shown in Figure 1 does not intercept any current. Clearly, Ampere's law can not be applied in this case to find the magnetic field generated by the current.
Figure 1. Ampere's law in a capacitor circuit. Although the surface shown in Figure 1 does not intercept any current, it intercepts electric flux. Suppose the capacitor is an ideal capacitor, with a homogeneous electric field E between the plates and no electric field outside the plates. At a certain time t the charge on the capacitor plates is Q. If the plates have a surface area A then the electric field between the plates is equal to
(2) The electric field outside the capacitor is equal to zero. The electric flux, [Phi] E, intercepted by the surface shown in Figure 1 is equal to
143
(3) If a current I is flowing through the wire, then the charge on the capacitor plates will be time dependent. The electric flux will therefore also be time dependent, and the rate of change of electric flux is equal to
(4) The magnetic field around the wire can now be found by modifying Ampere's law
(5) where [Phi] E is the electric flux through the surface indicated in Figure 1 In the most general case, the surface spanned by the integration path of the magnetic field can intercept current and electric flux. In such a case, the effects of the electric flux and the electric current must be combined, and Ampere's law becomes
(6) The current I is the current intercepted by whatever surface is used in the calculation, and is not necessarily the same as the current in the wires. Equation (6) is frequently written as
(7) where Id is called the displacement current and is defined as
(8)
Example Problem: Parallel-Plate Capacitor A parallel-plate capacitor has circular plates of area A separated by a distance d. A thin straight wire of length d lies along the axis of the capacitor and connects the two plates. This wire has a resistance R. The exterior terminals of the plates are connected to a source of alternating emf with a voltage V = V0 sin([omega] t). a) What is the current in the thin wire ? 144
b) What is the displacement current through the capacitor ? c) What is the current arriving at the outside terminals of the capacitor ? d) What is the magnetic field between the capacitor plates at a distance r from the axis ? Assume that r is less than the radius of the plates. a) The setup can be regarded as a parallel circuit of a resistor with resistance R and a capacitor with capacitance C (see Figure 2). The current in the thin wire can be obtained using Ohm's law
(9)
Figure 2. Circuit Parallel-Plate Capacitor. b) The voltage across the capacitor is equal to the external emf. The electric field between the capacitor plates is therefore equal to
(10) The electric flux through the capacitor is therefore equal to
(11) The displacement current Id can be obtained by substituting eq.(11) into eq.(8)
(12) The current at the outside terminals of the capacitor is the sum of the current used to charge the capacitor and the current through the resistor. The charge on the capacitor is equal to
(13) 145
The charging current is thus equal to
(14) The total current is therefore equal to
(15) d) The magnetic field lines inside the capacitor will form concentric circles, centered around the resistor (see Figure 3). The path integral of the magnetic field around a circle of radius r is equal to
(16)
Figure 3. Amperian loop used to determine the magnetic field inside a capacitor. The surface to be used to determine the current and electric flux intercepted is the disk of radius r shown Figure 3. The electric flux through this disk is equal to
(17) The displacement current intercepted by this surface is equal to
(18) The current intercepted by the surface is equal to the current through the resistor (eq.(9)). Ampere's law thus requires
146
(19) The strength of the magnetic field is thus equal to
(20)
2. Maxwells Equations The fundamental equations describing the behavior of electric and magnetic fields are known as the Maxwell equations. They are
(21) (22)
(23)
(24) Maxwell's equations provide a complete description of the interactions among charges, currents, electric fields, and magnetic fields. All the properties of the fields can be obtained by mathematical manipulations of these equations. If the distribution of charges and currents is given, than these equations uniquely determine the corresponding fields.
Example Problem: Conservation of Charge Prove that Maxwell's equations mathematically imply the conservation of electric charge; that is, prove that if no electric current flows into or out a given volume, then the electric charge within this volume remains constant. Equation (21) shows that the enclosed charge Q is related to the electric flux [Phi] E:
(25) The rate of change of the enclosed charge can be determined by differentiating eq.(25) with respect to time 147
(26) The closed surface used in eq.(24) to determine the flux and current intercepted can be replaced by a bag whose mouth has shrunk to zero. The path integral of the magnetic field along the mouth is therefore equal to zero, and eq.(24) can be written as
(27) Using eq.(26) we can rewrite eq.(27) as
(28) In other words, if no current flows in or out of the enclosed volume (I = 0) then the electric charge within this volume will remain constant. This implies conservation of charge.
3. Cavity Oscillations The electric field between the plates of a parallel-plate capacitor is determined by the external emf. If the distance between the plates is d (see Figure 4) then the electric field between the plates is equal to
(29) This time-dependent electric field will induce a magnetic field with a strength that can be obtained via Ampere's law. Consider a circular Amperian loop of radius r. The path integral of the magnetic field around this loop is equal to
(30) The electric flux though the surface spanned by this path is equal to
(31)
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Figure 4. The oscillating parallel-plate capacitor. The displacement current is thus equal to
(32) Using Ampere's law we obtain for the magnetic field
(33) This time-dependent magnetic field will induce an electric field. The total electric field inside the capacitor will therefore be the sum of the constant electric field generated by the source of emf and the induced electric field, generated by the time-dependent magnetic field. The strength of the induced electric field can be calculated using Faraday's law of induction. Consider the closed path indicated in Figure 4. We take the induced electric field on the capacitor axis equal to zero. The path integral of the induced electric field along the path indicated is then equal to
(34) where Eind is reckoned to be positive if it is directed upwards. The magnetic flux through the surface spanned by the loop indicated in Figure 4 is equal to
(35) Thus
(36) 149
The induced electric field, Eind, can be obtained from Faraday's law of induction (eq.(23)) and is equal to
(37) The total electric field is therefore equal to
(38) But the addition of the induced field implies that a correction needs to be made to the magnetic field calculated before (eq.(33)). This in turn will modify the induced current and this process will go on forever. If we neglect the additional corrections, then eq.(38) shows that the electric field vanishes at a radius R if
(39) or
(40) If we create a cavity by enclosing the capacitor with a conducting cylinder of radius R then eq.(40) can be used to determine the frequency of the driving emf that will produce a standing wave. This frequency is called the resonance frequency and it is equal to
(41) For a cavity with R = 0.5 m the resonance frequency is 1.2 GHz. Electromagnetic radiation with a frequency in this range is called microwave radiation, and the cavity is called a microwave oven.
4. The Electric Field of an Accelerated Charge The electric field produced by a stationary charge is stationary. When the charge accelerates, it produces extra electric and magnetic fields that travel outward from the position of the charge. These radiation fields are called electromagnetic waves. They travel with the speed of light (in vacuum) and carry energy and momentum away from the charge. Their properties
150
are determined by the properties of the accelerated charge, and in this manner provide a means to transmit information at the speed of light over long distances. Consider a charge q initially at rest (for t < 0). Between t = 0 and t = [tau], the charge accelerates with an acceleration a. After t = [tau] the charge moves with a constant velocity (v = a[tau]). We will assume that the final velocity of the charge is small compared to the speed of light (v << c) and that the time period [tau] during which the charge accelerates is short. The electric field generated by the charge at a time t > [tau] consists of three separate regions (see Figure 5). In the region r > ct the field lines will be that of a point charge at rest at the origin (electromagnetic waves travel with the speed of light, and the region with r > ct can not know yet that the charge has moved away from the origin). In the spherical region with radius r < c(t - [tau]), centered at x = vt, the electric field will that of a uniformly moving charge. The disturbance produced by the accelerated charge is confined to the region between these two spheres and the effect of the acceleration is a kink in the field lines. The electric field in this region has two components: a radial component and a transverse component. The radial component is determined by Gauss' law. Consider a spherical Gaussian surface located between the two spheres shown in Figure 5. The charge enclosed by this surface is equal to q. The electric flux through this surface depends only on the radial component of the field. Applying Gauss' law we conclude that the radial component of the electric field is simply the usual Coulomb field
(42)
Figure 5. Electric field lines generated by an accelerated charge. The relation between the radial component of the electric field and the transverse component of the electric field can be determined by carefully examining one field line (see Figure 6). The field line shown in Figure 6 makes an angle [theta] with the direction of the moving charge. The ratio between the magnitude of the transverse electric field and the radial electric field is equal to
(43) Since the radial field is known, eq.(42), we can use eq.(43) to determine the transverse component of the electric field:
151
(44) The distance r at which the kink occurs is related to the time t at which we look at the field:
(45) Eliminating the dependence on t in eq.(44) we obtain the following expression for the transverse component of the electric field:
(46)
Figure 6. Calculation of transverse electric field. Equation (46) shows that the transverse electric field is directly proportional to the acceleration a and inversely proportional to the distance r. The Coulomb component of the field falls of as 1/r2. This shows that the transverse component, also called the radiation field, remains significant at distances where the Coulomb field practically disappears. The equation for the transverse electric field (eq.(46)) is valid in general, even if the acceleration is not constant. If the charge oscillates back and forth with a simple harmonic motion of frequency [omega], then the acceleration at time t will be equal to
(47) In order to determine the radiation field at a time t and a distance r, we have to realize that the acceleration a used in eq.(46) should be the acceleration at time t - r/c, where r/c is the time required for a signal to travel over a distance r. The radiation field for the oscillating charge is therefore equal to
(48)
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Example Problem: Radio Antenna On a radio antenna (a straight piece of wire), electrons move back and forth in unison. Suppose that the velocity of the electrons is v = v0 cos([omega] t), where v0 = 8.0 x 10 -3 m/s and [omega] = 6.0 x 10 6 rad/s. a) What is the maximum acceleration of the electrons ? b) Corresponding to this maximum acceleration, what is the strength of the transverse electric field produced by one electron at a distance of 1.0 km fro the antenna in a direction perpendicular to the antenna ? What is the time delay (or retardation) between the instant of maximum acceleration and the instant at which the corresponding electric field reaches a distance of 1.0 km ? c) There are 2.0 x 10 24 electrons on the antenna. What is the collective electric field produced by all electrons acting together ? Assume the antenna is sufficiently small so that all electrons contribute just about the same electric field at a distance of 1.0 km. a) The acceleration of the electrons can be obtained by differentiating their velocity with respect to time:
(49) The maximum acceleration is thus equal to
(50) b) The maximum transverse electric field at a distance of 1.0 km (= 1000 m), in a direction perpendicular to the antenna ([theta] = 90deg.), can be obtained using eq.(48):
(51) Since the propagation speed of the radiation field is equal to the speed of light, c, the maxima in the radiation field will occur a time period [Delta]t after the maxima in the acceleration of the electron. The length of this period, [Delta]t, is equal to
(52) c) Assuming that all the electrons are in phase, then the maximum total transverse electric field at a distance of 1.0 km is equal to the number of electrons times the maximum transverse electric field produced by a single electron. Thus
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(53)
5. The magnetic field of an accelerated charge When a charge accelerates from rest it will produce a magnetic field. Initially, the magnetic field will be equal to zero (charge at rest). As a result of the acceleration a disturbance will move outward and change the magnetic field from its initial value (B = 0 T) to its final value, in much the same way as we observed for the electric field. The magnetic field can be obtained from the electric flux via the Maxwell-Ampere law which states that
(54) Note that the current I does not appear in eq. (54). Since we are looking in the region away from the moving charge, the current intercepted by a surface spanned by the path used to evaluate the path integral of B is equal to zero. The induced magnetic field will be timedependent and, therefore, will induce an electric field via Faraday's law. This induced electric field will again be time dependent and induce another magnetic field, and this process continues. The combined electric and magnetic radiation fields produced by the accelerating charge are called electromagnetic waves. They are self-supporting; the electric field induces a magnetic field, and the induced magnetic field induces an electric field. Because the electric and magnetic fields naturally support each other, the electromagnetic wave does not require a medium for its propagation, and it readily propagates in vacuum. The Maxwell equations can be used to show that the product of u 0 and [epsilon]0 is equal to 1/c2. Or,
(55) This equation was one of the great and early triumphs of Maxwell's electromagnetic theory of light. It shows that electricity and magnetism are two different aspects of the same phenomena.
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LIGHT AND RADIO WAVES 1. Electromagnetic waves Electromagnetic waves are produced by accelerating charges. The fields of the wave are selfsupporting - the electric field induces the magnetic field, and the magnetic field induces the electric field. Both radio waves and light are electromagnetic waves; their main difference is their frequency. Radio waves are created by the acceleration of electrons in a radio antenna, and light waves are created by the oscillations of the electrons within atoms. The electromagnetic wave has two components: the electric radiation field and the magnetic radiation field. The magnitude of the electric radiation field is directly proportional to the acceleration, and it is inversely proportional to the radial distance from the accelerating charge. In other words
(1) where a(t) is the acceleration of the charge at time t, and c is the speed of propagation of the disturbance (speed of light). Suppose we look at the electric field a long distance away from the accelerating charge. At this point the radial electric field, which falls of as 1/r2, can be neglected and only the transverse component of the field needs to be considered. We will assume that the charge accelerates during a short time interval and then continues to move with a constant velocity (see discussion in Chapter 35). If we want to study the variation of the transverse field over a limited range of distances, small compared to r, then the factor 1/r can be treated as a constant. The coordinate system that we will be using to study the propagation of the electromagnetic wave will have its x axis defined as the direction of propagation of the field. The y axis is taken to be parallel to the direction of the electric radiation field (see Figure 1). The electric radiation field produced by an accelerating charge will be non-zero only during a time interval [tau] which is the time interval during which the charge accelerates (see Chapter 35). These electromagnetic waves are called plane waves (although they are actually small patches of a spherical wave front). The propagating plane wave will change the electric field, which is initially zero, to a value Ey. This field will be exist during a time interval [tau] after which the transverse electric field returns to zero. The region is space in which the transverse field is non-zero and which propagates with the speed of light is also called a wave packet. The electric field of the wave packet can be written as
(2)
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Figure 1. Propagation of wave packet. The propagating wave packet will change the electric field when it passes a certain point in space, and this changing electric field will induce a magnetic field. The induced magnetic field will point along the z axis and its magnitude can be determined using the MaxwellAmpere law
(3) Consider a surface with width [Delta]x and length [Delta] z (see Figure 2). The width [Delta]x is chosen to be equal to the width of the wave packet ([Delta]x = c[tau]). The electric flux intercepted by the surface spanned by this path is zero before the wave packet arrives. During a time [Delta]t = [Delta]x / c the wave packet will sweep over the loop and fill the loop with electric flux. The maximum electric flux intercepted by this surface is equal to he product of the electric field Ey and the surface area ([Delta]x[Delta]z). The rate of change of electric flux is therefore equal to
(4) The path integral of the magnetic field along the loop shown in Figure 2 is equal to
(5)
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Figure 2. Calculation of induced magnetic field. The Maxwell-Ampere law can now be used to determine Bz
(6) The induced magnetic field is time-dependent and will, in turn, induce an electric field. The induced electric field can be determined using Faraday's law of induction which relates the path integral of the induced electric field to the rate of change of magnetic flux
(7) Consider the loop shown in Figure 3 with a width [Delta]x and a height [Delta]y, where [Delta]x is taken to be the width of the wave packet. The magnetic flux intercepted by the surface spanned by this loop will go from zero (when the wave packet has no overlap with the loop) to its maximum value of Bz[Delta]x[Delta]y in a time interval [Delta]t = [Delta]x/c. The rate of change of magnetic flux is thus equal to
(8) The minus sign in this equation is a result of the direction of the path indicated (using the right-hand rule you can show that the magnetic field lines along the negative z-axis make a positive contribution to the intercepted flux, while field lines along the positive z-axis make a negative contribution to the intercepted flux). The path integral of the electric field along the path indicated in Figure 3 at a time when one edge of the loop is inside the wave packet is equal to
(9) Applying Faraday's law of induction we can determine the induced electric field Ey
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(10)
Figure 3. Calculation of induced electric field. The electromagnetic wave will be able to sustain itself if this induced electric field is equal to the original electric filed. Comparing eq.(10) and eq.(6) we conclude that this requires that
(11) or
(12) Equation (12) is a theoretical expression for the speed of propagation of an electromagnetic wave in vacuum, and shows that the electromagnetic waves propagate with the speed of light (which is just one type of electromagnetic radiation).
Example Problem: Radio Receiver a) One type of antenna for a radio receiver consists of a short piece of straight wire; when the electric field of the radio wave strikes this wire it makes currents flow along it, which are detected an amplified by a receiver. Suppose the electric field of the radio wave is vertical. What must be the orientation of the wire for maximum sensitivity ? b) Another type of antenna consists of a circular loop; when the magnetic field of the radio wave strikes this loop it induces currents around it. Suppose that the magnetic field of a radio wave is horizontal. What must be the orientation of the loop for maximum sensitivity ?
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Figure 4. Electric field sensed by antenna. a) The electric field of the radio wave will exert a force on the electrons in the antenna. However, only the component of the electric field along the direction of the antenna will contribute to the current induced in the antenna. The magnitude of the current will be proportional to the magnitude of this component of the electric field. Consider the orientation of the antenna shown in Figure 4. The direction of the electric field of the radio wave is the vertical direction. If the antenna makes an angle [theta] with the vertical then the component of the electric field along the antenna is given by (13) Equation (13) shows that the magnitude of the electric field along the antenna, and therefore the magnitude of the induced current, will be maximum when [theta] = 0deg.. b) A changing magnetic field will induce an emf in a conducting loop (Faraday's law of induction). The induced emf is proportional to the magnetic flux intercepted by the surface spanned by the loop. In order to maximize the induced current, we have to maximize the induced emf, and therefore maximize the magnetic flux intercepted by the loop. This can be achieved of the loop is located perpendicular to the direction of the magnetic field.
2. Plane Harmonic Waves Although the relation between the electric and magnetic field strength shown in eq.(10) was obtained for a wave packet consisting of a region of constant electric field, the results are of general validity because an arbitrary wave can be regarded as a succession of short wave packets with piece wise constant electric fields. Equation (10) is therefore valid in general. Other general features of electromagnetic waves are: 1. the electric field is perpendicular to the direction of propagation 2. the magnetic field is perpendicular to both the electric field and the direction of propagation. The direction of E and B are related by the right-hand rule; if the fingers are curled from E to B, then the thumb lies along the direction of propagation.
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Most radiation fields are generated by charges moving with simple harmonic motion. In this case the acceleration of the charge will be a harmonic function of time:
(14) where [omega] is the angular frequency of the motion. The electric field along the x axis is directed along the y axis and is equal to
(15) The magnetic field will be directed along the z axis and has a magnitude equal to
(16) Electromagnetic waves described by eq.(15) and eq.(16) are called plane harmonic waves. The waves described by eq.(15) and eq.(16) are waves traveling in the positive x direction. The electric and magnetic fields of a plane wave traveling in the negative x direction are given by the following equations
(17)
(18) The minus sign in eq. (18) is required to make the direction of E and B consistent with the direction of propagation. The direction of the electric field of a electromagnetic wave is called the polarization of the wave.
Example Problem: Circularly Polarized Waves An electromagnetic wave traveling along the x axis consists of the following superposition of two waves polarized along the y and z directions, respectively:
(19) This electromagnetic wave is said to be circularly polarized. a) Show that the magnitude of the electric field is E0 at all points of space and at all times. b) Consider the point x = y = z = 0. What is the angle between E and the z axis at time t = 0 ? At time t = [pi] /2[omega] ? At time t = [pi] /[omega] ? At time t = 3[pi] /2[omega] ? Draw a 160
diagram showing the x and y axes and the direction of E at these times. In a few words, describe the behavior of E as function of time. a) The magnitude of the electric field can be obtained from its components along the y and z axes:
(20) b) At time t = 0 the electric field is equal to
(21) At time t = [pi] /2[omega] :
(22) At time t = [pi] /[omega] :
(23) At time t = 3 [pi] /2[omega] :
(24) The electric field of the circularly polarized wave is constant in magnitude, but its direction is time dependent (see Figure 5):
(25)
Example Problem: Polarization of Electromagnetic Waves An electromagnetic wave has the form
(26) a) What is the direction of propagation of the wave ?
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b) What is the direction of polarization of the wave, that is, what angle does the direction of the polarization make with the x, y, and z axes ? c) Write down a formula for the magnetic field of this wave as function of space and time.
Figure 5. Circularly polarized light. a) Equation (26) can be rewritten as
(27) Comparing eq.(27) with eq.(17) we conclude that eq.(27) describes an electromagnetic wave traveling in the negative z direction. b) Since the electromagnetic wave travels along the z axis, the direction of polarization of the wave will be in the x-y plane (the direction of the electric field is always perpendicular to the direction of propagation). Eq.(27) shows that the direction of polarization of the electromagnetic wave is independent of time. The angle between the direction of polarization and the positive x axis is equal to (see Figure 6)
(28) c) The magnitude of the magnetic field is equal to the magnitude of the electric field divided by the speed of light (see eq.(10)). The direction of the magnetic field is perpendicular to both the electric field and the direction of propagation. Since the direction of propagation is along the z axis, the direction of the magnetic field will be in the x-y plane. If the negative z axis is going in to the paper, then the right-hand rule requires that the magnetic field is pointing in the direction indicated in Figure 6. The magnetic field is thus given by
(29)
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Figure 6. Direction of electric field in problem 8.
3. The Generation of Electromagnetic Waves The electric field of a plane harmonic wave is position and time dependent. For an electromagnetic wave propagating in the positive x direction the transverse electric field is equal to
(30) At a fixed time t the electric field is a harmonic function of x. The wavelength [lambda] of the wave is the distance over which the electric field changes from its maximum value (E0) to its minimum value (- E0) and back to its maximum value (E0). The electric field has the same value at points separated by one wavelength. This requires that
(31) or
(32) This condition is true at all times t and all distances x if
(33) or
(34) 163
The ratio [omega]/2[pi] is called the frequency of the wave. Using the frequency [nu] we can rewrite eq.(34) as
(35) Radio waves and TV waves have wavelengths ranging from a few centimeters to 105 meter. Microwaves have wavelengths as short as a millimeter. These waves (radio, tv, and microwaves) can be generated by oscillating charges in an antenna. Waves with wavelengths less than a millimeter can not be generated by oscillating currents in an antenna. Instead, they are generated by electrons oscillating within molecules and atoms. This type of motion produces infrared, visible, and ultraviolet light, and X-rays. The corresponding wavelengths range from 10-3 m to 10-11 m (visible light has wavelengths between 7 x 10-7 and 4 x 10-7m). Protons and neutrons moving in a nucleus emit gamma rays which have a wavelength between 10-11 m and 10-16 m.
4. Energy of a Wave The electric and magnetic fields of an electromagnetic wave contain energy. As the wave moves along, so does this energy. Consider a wave packet with width [Delta]x moving in the positive x direction. The transverse electric field is directed along the positive y axis and the magnetic field is directed along the positive z axis. The energy density of the electric field, u E, is equal to
(36) The energy density of the magnetic field, u B, is equal to
(37) The energy stored in the wave packet is equal to the energy density times the volume, or
(38) where A is the surface area of the wave packet under consideration. However, for an electromagnetic wave the magnitude of the magnetic field is equal to the magnitude of the electric field divided by the speed of light (see eq.(10)). Equation (38) can now be rewritten as
(39) 164
Using eq.(12) we can eliminate the dependence of eq.(39) on [epsilon]0:
(40) or
(41) If we consider a fixed volume (fixed in space, with width [Delta]x) then eq.(41) is the maximum energy stored in this volume and it occurs at a particular time at which the wave packet completely overlaps with this volume. During a time interval [Delta]t = [Delta]x/c the wave packet moves out of the volume and the energy stored in this volume will return to zero. The rate of flow of energy through this volume is thus equal to
(42) The energy flux (per unit area) through the volume is given by
(43) The flux of energy associated with an electromagnetic wave is often expressed in terms of the Poynting vector. The Poynting vector is a vector with a direction along the flow and a magnitude given by eq.(43). The Poynting vector is defined, in general, as
(44) Although we have derived the energy flux of the electromagnetic wave for a wave packet, the results also apply for plane harmonic waves. If the angular frequency of the transmitter is [omega], then at a fixed distance, the electric and magnetic fields will oscillate in time with the same angular frequency
(45)
(46) The energy flux can be obtained by substituting eq.(45) and eq.(46) in eq.(44) 165
(47) The time-averaged energy flux is equal to
(48)
Example Problem: Energy of a Laser Beam The beam of a powerful laser has a diameter of 0.2 cm and carries a power of 6 kW. What is the time-average Poynting vector of this beam ? What are the amplitudes of the electric and magnetic fields ? The laser beam carries a power of 6 kW. The flux of energy in the beam is equal to the power delivered divided by the surface over which this power is delivered. Thus
(49) The amplitude of the electric field can be obtained using eq.(48)
(50) The amplitude of the magnetic field can be obtained by dividing the amplitude of the electric field by the speed of light
(51)
Example Problem: Energy and Current Flow A steady current of 12 A flows in a copper wire of radius 0.13 cm. a) What is the longitudinal electric field in the wire ? b) What is the magnetic field at the surface of the wire ? c) What is the magnitude of the radial Poynting vector at the surface of the wire ? d) Consider a 1.0-m segment of this wire. According to the Poynting vector, what amount of power flows into this piece of wire from the surrounding space ?
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e) Show that the power calculated in part (d) coincides with the power of the Joule heat developed in the 1.0-m segment of wire. a) Consider a 1-m segment of the wire. The resistance of this segment is equal to
(52) The voltage across this segment can be obtained using Ohm's law
(53) The longitudinal electric field in this segment is thus equal to
(54) b) The magnetic field at the surface of the wire can be obtained using Ampere's law
(55) c) The Poynting vector will be perpendicular to both the electric field and the magnetic field. Using the right-hand rule it can be shown that the Poynting vector is pointing along the radial direction, towards the center of the wire. The magnitude of the Poynting vector is equal to
(56) d) Since the Poynting vector is directed radially inwards, the power flowing into a 1-m long segment of the wire is just equal to the magnitude of the Poynting vector times the surface area of the 1-m long segment
(57) e) The Joule heat developed in the 1-m long segment of the wire is equal to
(58) which is equal to the power flowing into the wire, calculated in part d).
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5. Momentum of a Wave When an electromagnetic wave strikes a charged particle, it will exert a force on it. The forces acting on the charged particle are the electric force and the magnetic force. The electric force will be directed along the y axis and has a magnitude equal to
(59) The electric force will therefore only change the y component of the velocity of the charge. The x component of the velocity will be effected by the magnetic field. The x component of the magnetic force is given by
(60) Equation (60) can be rewritten in terms of the electric field Ey:
(61) The electric field does work on the particle, and the rate of increase of its energy is given by
(62) Comparing eq.(61) and (62) we conclude that
(63) Equation (63) shows that the rate at which the particle acquires x momentum from the wave is proportional to the rate at which it acquires energy. Equation (64) can be rewritten as
(64) If the wave loses all its energy and momentum to the particle then eq.(64) can be interpreted as stating that the change in the momentum of the charged particle is equal to
(65)
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where U is the total energy of the wave. Since linear momentum is a conserved quantity, eq.(65) indicates that the wave initially carries a linear momentum equal to U/c. Equation (63) can be rewritten using eq.(42):
(66) The momentum flow oscillates in the same way as the energy flow. The change in the x component of the momentum per unit time is equal to the x component of the force exerted on the particle. Thus
(67) Equation (67) can be expressed in terms of the Poynting vector
(68) The force exerted by the wave per unit area is equal to
(69) and is called the pressure of radiation. Note: this formula is only correct if the wave is completely absorbed. If the wave is totally reflected then the pressure of radiation is twice that given in eq.(69).
Example Problem: Solar Sails Astronauts of the future could travel all over the Solar System in a spaceship equipped with a large "sail" coated with a reflecting material. Such a "sail" would act as a mirror; the pressure of sunlight on this mirror could support and propel the spaceship. How large a "sail" do we need to support a spaceship of 70 metric tons (equal to the mass of Skylab) against the gravitational pull of the sun ? Ignore the mass of the "sail". Suppose the area of the sail is A and that the spaceship is located at the same distance from the sun as the earth is. At this point the average energy flux in the sun light is equal to 1.4 . 10 3 W/m2. The force exerted by the reflected sun light is equal to
(70) The gravitational force between the space ship and the sun is equal to 169
(71) and this force is pointed in a direction opposite to that of the radiation force in eq.(70). These two equation can be used to determine that a sail of 4.4 x 10 7 m2 is needed to balance the attractive gravitational force with a repulsive radiation force.
6. The Doppler Shift of Light If electromagnetic waves are emitted with a frequency [nu] from an emitter, than a receiver, at rest with respect to the emitter, will detect electromagnetic waves with the same frequency [nu]. If the emitter moves with respect to the source then the frequency received will not be the same as the frequency emitted. Suppose the emitter produces electromagnetic waves with frequency [nu]0. If the electromagnetic waves are detected by a receiver moving away from the emitter with a velocity v, than the frequency received by the receiver is equal to
(72) If the received is approaching the emitter with a velocity v than the frequency received by the receiver is equal to
(73) This shift in frequency is called the Doppler Shift.
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