1 UNIT – I ELECTROSTATICS (8) CHAPTER – 1
COULOB’S LAW
Frictional electricity: The electricity developed on objects, when they are rubbed with each other is called frictional electricity. This happens due to actual transfer of electrons from one objects to the other. The object which gains electrons becomes negatively charged while which loses electrons becomes positively charged. Electric charge: The additional property of electrons and protons which gives rise to electric forces between them is called electric charge. The electric charge is a scalar quantity. Its S.I. unit is coulomb(C). A proton posseses a positive charge +e and an electron a negative charge –e, where e = 1.6 x 10-19Coulomb. Properties of charge: 1) Like charges repel each other and unlike charge attract each other. 2) The total charge on an object is the algebraic sum of all the charges located at different points on the object. 3) Quantization of charge :- All charges are always an integral multiple of the charge on electron (e = 1.6 x 10 – 19 C) If q is the total charge on the object q =
+ −
ne where n = 0, 1, 2, 3, ……
Coulomb’s law: The first precise measurement of the force between two electric charges was performed by the French scientist Charles-Augustin de Coulomb in 1788. Coulomb concluded that: It states that the electrostatic force between any two charges is directly proportional to the product of the two charges and inversely proportional to the square of the distance between the charges. 1 q1 q2 F= i.e. 4π ∈0 r2 where F is the magnitude of the force, q 1 and q2 are the magnitudes of the two charges (with the appropriate signs), and r is the distance between the two charges. Where ∈0 is permittivity of free space. The universal constant ∈0 = 8.854x10-12 N-1m-2 C2 is called the permittivity of free space or the permittivity of the vacuum. We can also write Coulomb's law in the form q q F = k 122 r 1 where the constant of proportionality k = takes the value k = 8.988x 109 Nm2 C-2 4π ∈0 Relative permittivity (Dielectric constant): It is the ratio of the force between two charges placed at a certain distance in air to the force between the same charges placed at the same distance apart in that medium. ∈ Fvac ∈r or K = = ∈0 Fmed Coulomb’s law in vector form
r F12 =
1 q1 q2 rˆ 21 4π ∈0 r 2
r F21 =
2
1 q1 q2 rˆ 12 4π ∈0 r 2 Two charges q1 and q2 are placed at a distance r apart. The force on q1 due to q2 is given by, r 1 q1 q2 F= rˆ21 4π ∈0 r2 The force on q2 due to q1 is given by, r 1 q1 q2 F21 = rˆ21 4π ∈0 r2 Equation (i) and (ii) are called coulomb’s law in vector form.
--------- (i)
--------- (ii)
Principle of Superposition: It states that when a number of charges are interacting the total force on a given charge is the vector sum of all the forces exerted on the given charge by all other charges.
r r r If F12 , F13 ,....+, F1x are the forces on q1 due to q2, q3 ….., qx then the total force on q1 is given by r r r r F1 = F12 + F13 + L L + F1x r 1 q1 q2 q q q q F1 = rˆ21 + 1 2 2 rˆ31 + ... + 1 2 2 rˆx1 2 4π ∈0 r r r Linear charge density: It is defined as q λ= L Where q is the total charge over the length L. Its S.I. unit is cm – 1 or c/m. Surface charge density: It is defined as σ =
q where q is the total charge over the surface area A. Its S.I. A
unit is cm – 2 or c/m 2. Volume charge density: It is defined as δ = cm – 3
q where q is the total charge over the volume. Its S.I. unit is V
3
CHAPTER – 2 ELECTRIC FIELD Electric field Intensity: Electric field intensity at a point ur is defined as the force experience by a unit positive charge without disturbing the source. It is denoted by E r ur F E = lim qo → 0 q o Its S.I. unit is Newton per coulomb / Ne −1 Physical significance of Electric field: 1) It helps to give the methods for finding force on a charge due to another charge in terms of electric field. 2) It helps to determine the nature of interaction between charges kept at a distance apart. Electric field due to point charge: E=
F qo
q qo 1 4π ∈0 r 2 1 q qo 4π ∈0 r 2 q E= = qo 4π ∈0 r 2 A point charge + q is kept at p at a distance r from point p where the electric field is to be measured. According to coulomb’s law the force between the two charges is given by q qo 1 F= --------- (i) 4π ∈0 r 2 The electric field at point P is defined as F E= qo F=
4 1 q qo 4π ∈0 r 2 q q F= = = 2 qo 4π ∈0 r 4π ∈0 r 2 In vector rotation, where ur q E= rˆ 4π ∈0 r 2 Where rˆ is unit vector along AP direction. Electric lines of force: It is the path along which unit positive charge would move it is free to do so. Properties of electric lines of force: 1) Electric lines of force starts from positive and ends with negative. 2) Two lines of force never intersect each other. Q. Ans.
Explain why two electric lines of force never intersect each other? If two lines of force intersect each other, it would mean that the electric field has two directions at that point. Due to this, electric lines of force never intersect each other.
Electric Dipole: Two equal and opposite charges placed at a small distance between them is called an electric dipole.(or) A system of two equal and opposite charges separate by a certain distance is called an electric dipole.
Electric dipole moment: u r It is defined as the product of either charge or a distance between them. It is denoted by p, i.e. P = 2a × q Its S.I. unit is Coulomb meter (Cm). Electric field on axial line of electric dipole:
Consider an electric dipole consisting of charge + q and – q, separated by a distance 2a. Let O be the centre of the dipole and 2a be the length of the dipole. The electric field due to – q at point + p is given by, q EA = 4π ∈0 AP 2 q EA = OR 2 --------- (i) 4π ∈0 ( r + a ) The electric field due to +q at point P is given by q q EB = = 2 2 4π ∈0 BP 4π ∈0 ( r − a ) The resultant electric field can be written as q q E = EB − EA = − 2 2 4π ∈0 ( r − a ) 4π ∈0 ( r + a ) =
q 4π ∈0
1 1 q 4q − = . 2 2 2 ( r + a ) 4π ∈0 ( r 2 − a 2 ) ( r − a )
E=
5
( 2aq ) 2r
4π ∈0 ( r 2 − a 2 )
2
But 2aq = p, electric dipole moment 2 pa E= 2 4π ∈0 ( r 2 − a 2 ) If dipole is of short length, ‘a’ can be neglected 2p E= 4π ∈0 r 3
Electric field on equatorial line of electric dipole:
Consider an electric dipole consisting of charges – q and + q whose length is 2a. Let P be the point on equatorial line of the dipole at a distance r from O. The electric field at point p due to charge – q is given by
EA =
q 4π ∈0 AP
2
( or )
EA =
q 4π ∈0 ( r 2 + a 2 )
The electric field at point p due to charge + q is given by q q EB = (or) E B = 2 4π ∈0 BP 4π ∈0 ( r 2 + a 2 ) i.e. EA = EB By triangle law of vector addition we have, E EA EB = = AB AP BP AB E = EA× Or AP
6 = =
q
4π ∈0 ( r + a 2
)
2 2
×
2a 1/ c ( r + a2 ) 2
2aq
3 2 But 2aq = p, electric dipole moment P E= 4π ∈0 ( r 2 + a 2 ) 3 2 4π ∈0 ( r 2 + a 2 )
If dipole is of short length, a can be neglected. E=
P 4π ∈0 r 3
Torque on electric dipole placed in electric field:
Consider an electric dipole consisting of charge – q and + q whose length is 2a. Let O be the angle between the dipole and direction of electric field. Force on charge – q at point A = - qE Force of charge + q at point B=+qE These two force are equal and different line of action. So, they, form a couple giving rise to a torque. Torque, τ = force x ⊥ distance. = qE = BN = qE × 2 a sin θ = ( 2aq ) E sin θ τ = pE sin θ Where r u r 2aq ur = P is electric dipole moment. τ= P×E Electric field lines due to an isolated charge: Electric field lines due to a system of two charge: 1) for electric dipole (q1 q2 < O) 2) same charges (q1 q2 > O)
7
Q Ans:
Draw u the r lines of force to represent the uniform electric field. Here E represents the electric field.
Q Ans: Q Ans:
What is the net force on electric dipole placed in uniform electric field? It does not experience any net force. When is the torque acting on a dipole maximum? The torque is maximum when the electric dipole is placed perpendicular to the direction of electric field. When is the torque acting on a dipole minimum? The torque is minimum when the electric dipole is placed parallel to the direction of electric field.
Q Ans:
CHAPTER – 3 GAUSS THEOREM ELECTRIC FLUX: The electric flux through a small surface is define as the electric lines of force passing through that area, when held perpendicular to the lines of force. It is denoted by φ . ur uu r φ = E.ds uu r ur Where E the electric is field and ds is the area vector. Its S.I. unit is Nm2 / C. GAUSS THEOREM: Any close surface around a charge distribution so that the gauss theorem can be conveniently applied to find electric field due to it is called Gaussian surface. Q Ans:
Write the importance of Gaussian surface. To find electric field due to a charge distribution using gauss theorem, we have to evaluate the surface integral. This surface integral can be evaluated easily by choosing a suitable surface.
Electric field due to linear charge : Consider an infinitely long linear charge having linear charge density. To find electric field at point p at a distance r from O, we draw a Gaussian surface of radius r. According to gauss’ Theorem. ur uu r q E. ds = Ñ ∫ ∈0 q Ñ ∫ ∈ . ds cos O = ∈0 q q = λL We have λ = , L And Ñ ∫ ds = 2π rL
8 E × 2πrL = E=
λL ∈0 λ 2π ∈0 r
Electric field due to an infinite plane sheet of charge : Consider an infinite in plane sheet of charges having surface charge density on both sides. To find electric field, we draw Gaussian surface (cylinder) having cross sectional area A. According to gauss’ theorem r r q E. Ñ ∫ d S = ∈0 q
Ñ ∫ E dS cos θ = ∈
0
q EÑ ∫ ds = ∈0 But
Ñ ∫ ds = 2 A
And
σ= q= σA
q A
∴ E×2A =
σA ∈0
E=
σA 2 ∈0
If the plane sheet has finite thickness q σ= 2A 2σA ∴ E×2A = Then, ∈0 E=
σ ∈0
Electric field due to uniformly charge spherical shell: Consider a thin spherical shell of radius R centre O. Let +q be the charge of the spherical shell (1) When point P lies outside the spherical shell Consider small area ds at point p at a distance r from centre of the spherical shell. According to Gauss’s Theorem ur r q E Ñ ∫ . d S = ∈0
Ñ ∫ E ds cos 0
=
EÑ ∫ ds = But,
q ∈0
q ∈0
Ñ ∫ ds = 4πr
2
9 And, surface charge density, q 4πR 2 q = 4πσR 2
σ=
4πσR 2 ∈o σ E= R2 ∈o × 2 r
E × 4πr 2 =
(2) When P lies inside the spherical shell Let P be point inside the spherical shell at a distance r from O. According to gauss’ theorem ur uu r q E. ds = Ñ ∫ ∈0 But, the Gaussian surface does not enclose any charge. i.e. ur uqu r=0 Ñ ∫ E. ds = 0 E=O i.e. electric field inside a spherical shell is 0. CHAPTER – 4 ELECTRIC POTENTIAL
Q
Show that the work done in moving a unit charge along a close path is). (3m) OR Show that the integral of electric field.
A point charge +q is located at origin and let AP1, BP2, A be a closed path. Then represents the work done in moving a unit positive charge from point A to B. Bu u r r 1 1 1 E. Then, Ñ ∫ A dl = 4π ∈ o q rA − rB ÷ (along P1) Bu u r r 1 1 1 E. dl = q and − ÷ (along P2 Ñ ∫A 4π ∈ o rB rA Bu r r A ur ur 1 1 1 1 1 1 E. ∫A dl + ∫B E. d s = 4π ∈ o q rA − rB ÷ + 4π ∈ o q rB − rA ÷
B
ur r
A
10
ur r
∫ E.dl + ∫ E. dl = 0 A
B
The LHS of the above equation represents the line integral of electric field over the closed path. Therefore, ur r E ∫ . dl = O i.e. the line integral of electric field is O. ELECTROSTATIC POTENTIAL DIFFERENCE The electrostatic potential difference between two points B and A in electric field is defined as the amount of work done per unit positive test charge in moving it from point A to B against the electrostatic force due to the electric field. If WAB is work done in moving test charge from point A to B. Then the electrostatic potential difference is given by 1 W 1 1 VB - VA = AB = q − ÷ qo 4π ∈o rB rA It is a scalar quantity. Its S.I. unit is volt (V)
ELECTRIC POTENTIAL: The electric potential at a point in an electric field is define as the amount of work done per unit positive test charge from infinity to that point against the electrostatic force due to the field. If W ∞A is work done in moving test charge qo from infinity to point A, W ∞A 1 q V= = − ∈o 4π ∈o r Its S.I. unit is volt (V). Q Ans:
Define one volt. OR Define the S.I. unit of electrostatic potential. The electrostatic potential at a point is said to be one volt. If one Joule of work is done in moving a charge of 1 from infinity to that point against the electrostatic force.
Electric potential due to a point charge:
Point charge +q is placed at point O. Let V A is electric potential at point A, at a distance r A from O. If W ∞A is work done in moving a test charge from infinity to point A. W ∞A Then, qo ur For a test charge q0 we apply an external force −qo E . If q0 move through a small displacement dl r r r dW = F. dl = −qo E ×dl = − qo Edl cos180° then,
(
)
11 r r r dW = F. dl = −qo E ×dl = − qo Edl cos180°
(
)
= − qo Edl (−1) = qo Edl ∴ dW = − qo E dr We have,
∴ dl = − dr q E = 4π ∈o r 2
− q q0 dr 4 π ∈0 r 2 Hence, work done in moving test charge from infinity to point A is given by rA r A −q q0 − q q0 A −2 W∞A = ∫ dW = ∫ dr = r dr 4π ∈0 r 2 4π ∈0 ∞∫ ∞ ∞ dW = − q0 E dr =
rA
W∞A
or VA =
W∞A q0
− q q0 1 q q0 1 1 = × − = − ÷ 4π ∈0 rA ∞ 4π ∈0 rA ∞ q q0 = 4π ∈0 rA q q0 q = 4π ∈0 rA = VA = 4π ∈0 rA q0
ELECTRIC POTENTIAL AT ANY POINT DUE TO AN ELECTRIC DIPOLE:
Consider an electric dipole consisting of charge – q and + q whose length is 2a. Let P be the point at a distance R from O such that ∠ POB = θ The potential at point P due to – q is given by, −q VA = --------- (i) 4π ∈0 AP The potential at point P due to + q is given by, q VB = --------- (ii) 4π ∈0 BP To find AP & BP draw BN || OB and AN || OP such that AP = OP + ON = r + a COS θ And BP = OP – ON = r – a COS θ Now, the net potential at point P is given by
12 V = VA + VB q 1 1 = − 4π ∈0 BP AP 1 1 r − a cos θ − r + a cos θ r + a cos θ − r + a cos θ = r 2 − a 2 cos 2 θ =
q 4π ∈0
q 2 a cos θ 4π ∈0 r 2 − a 2 cos 2 θ But, 2 aq = P, electric dipole moment. P cos θ V= 4π ∈0 ( r 2 − a 2 cos 2 θ ) =
Cases 1. When p lies on axial line of dipole ( θ = O) P V= 4π ∈o ( r 2 − a 2 ) ° 2. When P lies on equitorial line of electric dipole ( θ = 90 )
V=O Q
Ans:
Show that electric field at a point is equal to negative gradient of electrostatic potential at that point. dV C E = − dr Consider a point charge +q at O. Let V and V + dV be the potential at point P and Q at a distance r and r – dr. If dW is small work done in moving test charge qo through distance dl dW ( V + dV ) − V = qo dV = Now,
dW qo
− − − − − − − − ( i)
r r dW = F . dl = − qo E dl cos 180° dW = qo E dl
But,
dl = − rd dW = − qo E dr Putting this value in equation (i) q E dr dV = − o qo
Q Ans:
dV dr Define equipotential surface. Any surface which has the same electrostatic potential at every point is called equipotential surface.
Q
Show that no work is done in moving a test charge over equipotentia surface.
E=
13
Ans:
Let point A and B be two points lying on equipotential surface. The potential difference is given by W AB VB − VA = qo But VA = VB (In equipotential surface) W AB O= qo = WAB = O i.e. no work is done in moving a test charge over equipotential surface.
EQUIPOTENTIAL SURFACE FOR UNIFORM ELECTRIC FIELD
EQUIPOTENTIAL SURFACE FOR ISOLATED POINT CHARGE
14 EQUIPOTENTIAL SURFACE FOR SYSTEM OF TWO CHARGES
POTENTIAL ENERGY OF AN ELECTRIC DIPOLOE PLACED IN UNIFORM ELECTRIC FIELD
The torque experienced by an electric dipole (having dipole moment p) placed inside uniform electric field E is given by τ = pE sin θ If the dipole is rotated through small angle dθ against the torque. Then, small work done is, dWτ= × dθ dW = pE sin θ dθ Thus, work done in rotating the dipole from its orientations making an angle θ1 to θ 2 is given by θ2
θ2
θ1
θ1
W = ∫ dW =
∫
pE sin θ dθ = pE
θ2
∫ sin θ dθ
θ1
= pE [ − cos θ ] θ2
W = − pE [ −cosθ 2 . cos θ1 ] This work done is stored as potential energy (U) of the dipole.
θ
1
μ = − PE ( cos θ 2 − cos θ1 )
15
if θ1 = 90oand θ 2 = θ
μ = − PE ( cos θ − cos 90o )
μ = − PE cos θ In vector notation u r ur μ = −P.E
CHAPTER – 5 POLAR AND NON-POLAR DIELECTRICS Dielectric in the atoms or molecules of which the centres of gravity of positive and negative charges coincide is called non-polar dielectric. A dielectric in the atoms or molecules of which the centres of gravity of +ve and – ve charges do not coincide is called polar dielectrics. Q. Ans:
Define polarization. The stretching of dielectric atoms due to displacement of charges in the atoms under the action of applied electric field is called polarization.
Polarization density: They induce dipole moment develop per unit volume in a dielectric slab on placing it inside the electric field is called polarization density. It is denoted by P.If P is induced dipole moment acquired by atoms of dielectric and N is the number of atoms per unit volume. Then, P – NP Let A = Area of capacitor, D = distance between plates of capacitor. Volume of dielectric slab = Ad ∴ -q1 and +q2 are developed on two face, the total dipole moment is
16 P=
qi d qi = Ad A
qi = σ P , polarization surface charge density. A ∴ P σ= P The reduce value of electric field is given by σ − σP σ σ P σ P E= = − = E0 − P = E0 − ∈0 ∈0 ∈0 ∈0 ∈0 Pχ= ∈E0 We know that When χ called electric susceptibility of the dielectric χ ∈0 E E = E0 − = Eχ0 −E ∈0 ∴
χ E = E0 − E χ E + E = E0 E E ( 1+ χ ) = 0 E E0 But = k, dielectric constant E ∴ k = 1χ+ i.e the relation between dielectric constant and electric susceptibility of the dielectric. Dielectric strand: The dielectric strand of a dielectric is defined as the maximum value of electric field that can be applied to the dielectric without its electric breakdown. Its SI unit is volt per metre. CAPACITANCE: Capacitance of a conductor may be defined as the ratio of the electric charge on it to its electric potential due to that charge. It is denoted by C. q C= V Where V is the electric potential due to the charge. Its SI unit is Farad (F) PRINCIPLE OF CAPACTOR A capacitor is a device for storing a large quantity of charge.
An insulated metallic plate A is given +ve charge till its potential becomes maximum. If another metallic plate B is placed near plate A 1 – ve charge will be induced on nearer face of plate B and +ve charge on farther face as shown in Fig (ii). The potential of plate A gets lower due to induced charge and to make the potential of plate A the same, additional charge has to be given, i.e. its capacitance increases.
17 If A plate B is connected to earth as shown in Fig (iii), +ve charge on plate B flows to the earth and -ve charge remains on the plate due to attraction from plate A. Thus, its potential is lower and a large amount of charge has to be given to make the potential the same, i.e. its capacitance is greatly increased. This is the principle of capacitor. CAPACITANCE OF PARALLEL PLATE CAPACITOR
If +ve charge (+q) is given on plate A, - ve charge (– q) is induced on plate B. The electric field is set up between the two plates which is given by dV E= (in magnitude) dr Since the electric field is uniform, we can write V E= d Where V is the potential between the two plates. ⇒ V = Ed --------- ( i ) If σ is surface charge density, then, q σ = A σ q also, E = = ∈0 ∈0 A Putting this value in equation (i) qd V= ∈0 A q We know that capacitance, C = V q C= qd ∈0 A C=
∈0 A d
ENERGY STORED IN A CAPACITOR Consider a capacitor of capacitance C, which does not contain any charge initially. If it is connected to a battery its potential rises to V and if q is the amount of charge on the capacitor then, q C= V ⇒ q = CV ---------- ( i )
18 If the battery supplies infinite simal charge dq to constant potential V. Then through the small q amount of work done is given by C = V q dW = Vdq = dq C Then, the total work done in delivering charge q is given by q
q
q
q
q 1 1 q2 1 1 W = ∫ dW = ∫ dq = ∫ qdq = = × q 2 − 0 C C0 C 2 0 2 C 0 0 1 q2 ∴W = × 2 C This work done is stored as electric potential energy (u) 1 q2 U= × − − − − − − −(ii) 2 C 1 q = qV QC = 2 V or, 1 u = CV 2 [Q q = CV ] 2 We know that, ∈ A C= 0 d Where A is area of the plate and d is distance between the two plates. σ E= An electric field ∈0 Where σ is surface charge density q σ= A Now, σA ⇒q= =∈0 EA Putting this value in eqn. (ii) 1 ∈02 E 2 A 2 m = × A 2 ∈0 d 1 U = ∈0 E 2 Ad 2 Where Ad = Volume of capacitor 1 Energy store per unit volume = ∈0 E 2 2 CAPACITANCE OF PARALLEL PLATE CAPACITOR WHEN DIELECTRIC IS INSERTED BETWEEN THE TWO PLATES: A parallel plate capacitor having plates of area A separated by a distance d is placed in vacuum and its capacitance is given by,
19 ∈0 A d When a dielectrics of dielectric constant k (kappa) is insulted between the two plates, the electric field reduces due to polarization of dielectrics. If V is the potential difference between the two plates then, C0 =
V = Et + E 0 (d − t ) E0 =k Since, E E ⇒E= 0 k E ∴ V = 0 t + E 0 (d − t ) k t V = E 0 (d − t + ) k σ q E0 = = ∈0 ∈0 A q t V= (d − t + ) But, ∈0 A k ∈0 A q q C= = = q t t V Capacitance, (d − t + ) d − t (1 − ) ∈0 A k k When dielectrics field the whole space between the two plates i.e t=d ∈0 A ∈0 A k ∈0 A ∴C = = = t d d d − d(1 − ) d − d − k k ∴ C = k C0 CAPACITANCE OF PARALLEL PLATE CAPACITOR WHEN CONDUCTING SLAB IS INSERTED BETWEEN TWO PLATES: A parallel plate capacitor having plates of area A and separated by a distance d is placed in vacuum and its capacitance is given by
20 ∈0 A d When conducting slab is insulted between the two plates, the potential difference between the two plates reduces and since the electric field inside the conducting slab is 0. V = E 0 (d − t ) C0 =
But, E 0 =
σ ∈0
q A q V= (d − t ) ∈0 A q C = V q = ∈ A q = 0 (d − t ) (d − t ) ∈0 A σ=
∴ Capacitance
When conducting slab fills the whole space between the two plates, t=d ∈ A ∈0 A C = 0 = (d − d) 0 C=∞ VAN-DE GRAFF GENERATOR PRINCIPLE: 1) Electric discharge takes place in air or gases readily at pointed conductor. 2) A hollow conductor continues excepting charge through its inner surface irrespective of how large its potential may grow. CONSTRUCTION AND WORKING: It consists of a large hollow metallic sphere placed on two insulating column. The metal comb (C 1) is kept at high potential and the collecting comb (C 2) is kept in contact with the inner sphere which is connected by two pulleys. Positive ions are sprayed on the belt due to repulsive action of C1, the comb C2 collects these ions and transfer them to the metallic sphere. As the belt goes moving the positive ions gets accumulated.
21
Electric discharge also takes place from the sphere. When the maximum potential is reach the charge leaks to the surrounding air inside the conductor and rate of laws of charge is equal to the rate of transfer of charge to the sphere to prevent this leakage, the generator is completely enclosed in each connected steal tank if projectile such as proton, deuteron etc. are generated in the discharge tube D. With lower and earth upper and inside the sphere. They accelerated in downward direction to hit a target with large kinetic energy to bring about nuclear disintegration. CAPACITOR IN SERIES:
1 1 1 1 = + + C C1 C2 C3 CAPACITOR IN SERIES:
C = C1 + C2 + C3 UNIT – II
CHAPTER – 1
ELECTRIC CURRENT RESISTANCE AND E.M.F. ELECTRIC CURRENT: It is defined as the rate of flow of charge through a conductor. It is denoted by q I= I. t Where q is the amount of charge flowing through a conductor in time t. It is a scalar quantity. It SI unit is ampere. Q Define one ampere. OR Define the SI unit of electric current. Ans: The current through a conductor is called one ampere. If 1 coulomb of charge flows through the conductor in 1 sec.
22 DRIFT VELOCITY When a potential difference V is applied to a conductor of length l. The electric field set up is given by
dV dr V E= l E=−
The force experienced by the electrons r ur is given by F = −e E The acceleration experienced by the electrons due to this force can be written as, ur r −e E a= m This acceleration of electron takes placed only for a short time due to deflection in their collision with +ve ions in the conductor and the time for which they accelerate is called relaxation time. It an electron having thermal u, accelerates for time t, and it will attain velocity. r ur velocity r v1 = u1 + aτ1 Velocity acquired by other electrons r uu ris r v 2 = u2 + aτ 2 r uu r r v 3 = u3 + aτ3 And The drift velocity can be written as ur ur uu r r v1 + v1 + ........ + vn vd = n ur r uu r r uu r r (u1 + aτ 1 ) + (u2 + aτ 2 ) + ........ + (un + aτ n ) = n ur uu r uu r uu r r u1 + u2 + ........ + un r τ 1 + τ 2 + ........ + τ n vd = +a n n uu r ur uu r uu r τ 1 + τ 2 + ........ + τ n Where u1 + u2 + ........ + un is called average thermal velocity which is equal to zero. And n is called average relaxation time uu r (τ). vd = 0 + aτ Therefore, ur uu r −e E ⇒ vd = τ m Q Ans:
Define drift velocity. Drift velocity is defined as the average velocity with which the free electrons in the conductor gets drifted under the influence of the applied electric field.
RELATION BETWEEN DRIFT VELOCITY AND ELECTRIC CURRENT: Consider a conductor length l and cross sectional area a. If n is the number of electrons per unit volume, then the total charge on the conductor is, q = nAle If vd is the drift velocity of the free electrons. Then,
23
vd =
l t
⇒t =
l vd
The current flowing through the conductor can be written as, q nAle I= = t l vd I = nAevd We know that, eE τ m nAe 2τ ∴I = E m vd =
ELECTROMOBILITY: The mobility of electrons in a conductor is defined as the drift velocity acquired per unit applied electric field. It is denoted by μ v μ= d E Its SI unit is m2 V – 1 S – 1 . OHM’S LAW: It states that physical conditions remaining unchanged, the current flowing through a conductor is directly proportional to the potential difference across its two ends. Mathematically V µ I ⇒ V = RI , Where, R is the proportionality constant known as resistance of the conductor. RESISTANCE: Resistance of a conductor is defining as the ratio of potential difference across the conductor to the current flowing through it. It is the opposition offered by the conductor to the flow of current. V R= I It SI unit is ohm (Ω) RESISTIVITY (SPECIFIC RESISTANCE): The resistively of the material of a conductor is define as the resistance offered by this material of unit length and unit area. It is denoted by ρ. A ρ=R l It SI unit is ohm metre.(Ω m) FACTORS AFFECTING ELECTRICLA RESISTIVITY: Consider a conductor having length (l) and cross sectional area A. If electric field E is applied across the conductor. Then the drift velocity is
24 eE τ m And the current flowing through the conductor is I = nAvd e vd =
− − − − − − − − (i)
nAe 2 Eτ m
∴I = We know that,
V l nAe 2 Vτ ∴ I= ml V ml ⇒ = 2 I ne τ A E=
According to ohm’s law,
V =R I m l ∴R = 2 . ne τ A If ρ is the resistively of the conductor, l R=ρ A Comparing equation (iii) and (iv)
ρ=
− − − − − − − − (ii)
− − − − − − − −(iii) − − − − − − − −(iv)
m ne 2τ
CONDUCTANCE: The reciprocal of the resistance of a conductor is called each conductance. It is denoted by G. 1 G= R – 1 (Ω −1 ) Its SI unit is ohm CONDUCTIVITY: The reciprocal of the resistively of a conductor is called each conductivity. It is denoted by ρ Its SI unit is ohm – 1 m – 1 (Ω −1m −1 ) 1 σ= ρ ELECTRICAL ENERGY: The total work done by the source of e.m.f. in maintaining the electric current in the circuit for a given time is called electrical energy consume in the circuit. W=VIt Its SI unit is Joule (J). Electric Power: The rate at which work is done by the source of e.m.f. in maintaining the electric current in the circuit is called electric power. V2 P = VI = I 2 R = R Its SI unit is Watt (W).
25 Q Define 1 kilowatt hour on board of trade unit ( B ∈0 ) Ans: The energy consume is called 1 kilowatt hour. It a device of electric power of 1 kilowatt is used for 1 hour. (1 kwh = 3.6 x 106J) COLOOUR CODE FOR CARBON RESISTOR : B B R O Y Great Britain Very Good Wife Latter as an aid to memory B B R O Y G B V G W Q Ans: Q Ans: Q
Colour Black Brown Red Orange Yellow Green Blue Violet Grey White
Figure
Multiplier
Colour
0 1 2 3 4 5 6 7 8 9
100 101 102 103 104 105 106 107 108 109
Gold Silver No colour
Tolerance 5% 10% 20%
A carbon resistor has three colors Blue, Yellow and Red respectively. What will be the resistance? 64 × 102 Ω ± 20% A carbon resistor is marked in colour bands of red, black, orange and silver. What is the reistance and tolerance value of the resistor ? 20 × 103 Ω ± 10% Draw the colour code scheme of 42 × 103 Ω + 10% carbon resistance.
4.5 × 103 Ω + 10% Ans: Q Ans:
Draw 4.5 × 103 kΩ ± 5% = 4.5 × 103 kΩ ± 5% = 45 × 102 kΩ ± 5%
RESISTORS IN SERIES: Three resistor R1, R2 and R3 are connected in series to a battery of e.m.f. E. Let V 1, V2 and V3 be the potential drop across the three resistors. Such that E = V1 + V2 + V3 ---------- (i) According to ohm’s law V1 = IR1, V2 = IR2, V3 = IR3 Therefore, E = IR1 + IR2 + IR3 E = I (R1 + R2 + R3) E = R1 + R 2 + R 3 ---------- (ii) I If R is the equivalent resistance of the combination. Then,
26 E = IR E ⇒ =R I Putting this value in equation (ii) R = R1 + R2 + R3. RESISTORS IN PARALLEL: Three resistors R1, R2 and R3 are connected in parallel to a battery of e.m.f.. Let I 1, I2 and I3 be the current flowing through its resistors. Such that I = I1 + I2 + I3 ----------- (i) According to ohm’s law, E E E I1 = , I2 = , I3 = R1 R2 R3 Putting this value in equation (i) E E E I = + + R1 R 2 R 3 I 1 1 1 = + + ---------- (ii) E R1 R 2 R 3 If R is the equivalent resistance of the combination, E=IR I 1 ⇒ = E R Putting this value in equation (ii) I 1 1 1 = + + R R1 R 2 R 3 Q Define internal resistance of a cell. Ans: The resistance offered by the electrolytes of the cell when electric current flows through it is known as its internal resistance. Its SI unit is ohm. It is denoted by r. Q Ans:
Define electromotive force (e.m.f.). The potential difference between the two poles of the cell when no current is drawn from the cell (open circuit) is called electromotive force of the cell.
Q Ans:
Define terminal potential difference. The potential difference between the two poles of the cell when current is drawn from the cell (closed circuit) is called terminal potential difference.
EXPRESSION FOR INTERNAL RESISTANCE OF A CELL: A source of E having internal resistance r is connected to an external resistance R. If I is the current flowing through the circuit, then E I= ---------- (i) R+r The terminal potential difference is equal to the potential drop across resistor R. V = IR ---------- (ii) From equation (i) and (ii)
27
V=
E R R+r
E R V E r = − 1÷ R V
R+r =
CELLS IN SERIES: 1) When the cells are of the same e.m.f. and internal resistance. A combination of n cells having e.m.f. E and internal resistance r. Each is connected in series to an external resistance R. Total e.m.f. of the cell = n E Total internal resistance of the cell = nr If I is the current flowing through the circuit, then
I=
total e.m.f total resistance
eN R + nr a) If R << nr, i.e R + nr ≈ nr E ∴I = = currenr due to a single cell. r b) If R >> nr, i.e R + nr ≈ R nE ∴I = = n times the currenr due to a single cell. R 2) When the cells are of different e.m.f. and external resistance. I=
Two cells having internal resistance r1 and r2 and e.m.f. E1 and E2 are connected in series to an external resistance R. The terminal potential difference due to the first cell is, V1 = E1, - I r1
28 For the second cell, V 2 = E 2 – I r2 The potential difference between two points A and B is given by V = V1 + V2 V 1 = E 1 – I r1 + E 2 – I r2 V = (E1 + E2) – I (r1 + r2) ---------- (i) If E and r are the equivalent e.m.f and internal resistance. Then, V=E–Ir ---------- (ii) Comparing equation (i) and (ii) E = E1 + E2 and R = r 1 + r2 total e.m.f ∴I = total resistance E = R+r E1 + E 2 = R + r1 + r2 CELLS IN PARALLEL: 1) When the cells are of same e.m.f and total internal resistance. A combination of n cells having e.m.f E and internal resistance r, each, is connected to an external resistance R. If r1 is the combined internal resistance of the cell then, 1 1 1 1 = + + + ......... n times r′ r r r n = r r r′ = n Total e.m.f of the cell = E Total resistance = R + r′ If I is the current flowing through the circuit. total e.m.f I= total resistance E = R + r′ E = R+r n a) If R ? r n , i.e R + r n ≈ R E ∴I = = currenr due to a single cell. R b) If R = r n , i.e R + r n ≈ r n E En ∴I = = n times the currenr due to a single cell. r n r 2) When the cell are of different e.m.f and total external resistance.
29
Two cells having e.m.f E1 and E2 and internal resistance R1 and R2 are connected in parallel to an external resistance R. Let I1 and I2 be the current from the two cells such that, I = I 1 + I2 ----------- (i) The potential difference between the two points A and B is given by V = E1 – I 1 r 1 I1 r1 = E1 – V E −V I1 = 1 r1 Similarly, E −V I2 = 2 r2 Putting this value in equation (i) E − V E2 − V I= 1 + r1 r2 E1 − V r1 E r − Vr2 + E 2 r1 − Vr1 Ir1r2 = 1 2 r1r2 ( E r + E 2 r1 ) − I r r V ( r1 + r2 ) = 1 2 ( 1 2) r1 + r2 ( E r + E 2 r1 ) − I ( r1r2 ) V= 12 ----------(ii) r1 + r2 r1 + r2 If E is the total e.m.f and r is the total internal resistance. Then V = E – Ir ---------- (iii) From equation (ii) and (iii) ( E r + E 2r1 ) E= 12 r1 + r2 ---------- (iv) r2 r1 r= r1 + r2 total e.m.f ∴I = total resistance E = R + r′ Where the values of E and r are given by equation (iv) I=
CHAPTER – 2
30 ELECTRICAL MEASUREMENT KIRCHOFF’S LAW: First Law: It states that the algebraic sum of a current meeting at a point in an electrical circuit is always zero. It is also known as function rule. Second Law: It states that in any close part of an electrical circuit, the algebraic sum of e.m.f is equal to the algebraic sum of the product of resistance and current flowing through them. PRINCIPLE OF WHEATSTONE BRIDGE: Four résistance PQR and X are connected across the arms of a Wheatstone Bridge. Let I be the current flowing through the circuit.
Applying Kirchhoff’s second law on the close loop ABDA, - I1 P + I 2 R – I 3 G = 0 ---------- (i) In the close loop BCDB, - (I1 – I3) Q + (I2 + I3) X + I3 G = 0 ---------- (ii) If points B and D are at the same potential. I3 = 0 And the Wheatstone bridge is said to be balanced. -I1 P + I2 R = 0 I1 P = I 2 R ---------- (iii) and – I1 Q + I2 X = 0 I1 Q = I 2 X ---------- (iv) Dividing equation (iii) and (iv) I1P I 2 R = I1Q I 2 X P R = Q X This is the condition for balanced Wheatstone bridge.
31 SLIDE WIRE BRIDGE (METRE BRIDGE) Metre Bridge is a device for measuring unknown resistance. Principle: It is based on the principle of Wheatstone bridge i.e. when the bridge is balanced. P R = Q X Where P and Q are called ratio arms. R is variable resistance and X is unknown resistance. Construction: It consists of a wooden board over which constantan wire AC (100cm length) and a metre scale are fitted. A copper strip is fitted on the board to provide two gaps for connecting resistance box R and unknown, resistance X. A galvanometer is connected to terminal D and a jockey. Applications: 1) To measure an unknown resistance: When the bridge is balanced . Let AB = l and BC = 100 – l. The resistance of the wire between AB is taken as P and between B and C is taken as Q. Then, Pµ l Q µ 100 − l P l ∴ = − − − − − − − − (i) Q 100 − l From the principle of Wheatstone bridge. P R = Q X From equation (i) and (ii) R l = X 100 − l Knowing the values of R and l, the unknown resistance X can be determine. 2) To compare two unknown resistance: First the resistance R1 is connected in place of X and the balance point is obtained by moving the jockey over the wire. Let AB = l1 and BC = 100 – l 1. The resistance of the wire between AB is taken as P and between BC is taken as Q. Such that P µ l1 Q µ 100 − l1
And
P l1 = Q 100 − l1 From the principle of Wheatstone bridge P R = Q R1 ∴
∴
R l1 = R1 100 − l1
l1 ⇒ R1 = ÷R 100 − l1
− − − − − − − − (i)
32 The experiment is repeated with R2 replacing R1 and let AB = l2 and BC = 100 – l2 l1 ∴R2 = − − − − − − − − (ii) ÷R 100 − l2 Dividing equation (i) by equation (ii) we get R1 l1 100 − l2 = × R 2 l2 100 − l1 Knowing the values of l1 and l2 the ratio R 1 R 2 can be determine.
POTENTIOMETRE: Potentiometer is a device for measuring internal resistance of the cell or comparing emfs of two cells. Principle: When a constant current is passed through a wire of uniform area of cross section, the potential drop across any portion of the wire is directly proportional to the length of that portion. Construction: It consists of a number of segments of wire (manganin or constantan) of uniform cross section (each 100cm long) stretched on a wooden board. A metre rod is fixed parallel to its length as shown in figure..
A constant current is maintained through the potentiometer wire with the help of the rheostat Theory: Let V be the potential difference across certain portion of wire of resistance R Then, V = IR l R=ρ Since A l ∴ V = Iρ A ρ Where is the resistivity of the wire. If a constant current is passed through the wire. Then, V = (constant × l ) i.e., V µ l i.e. potential drop along the wire is directly proportional to the length of that portion. Applications: 1) To measure e.m.f of the cell: The circuit connection is shown in the The potentiometer wire is first calibrated a standard cadmium cell of e.m.f 1.018 V. The
figure. by using jockey
33 is put at 203.6cm mark and current is adjusted by using rheostat so that galvanometer shows no deflection. The potential gradient along the potentiometer wire is given by dv 1.018V = = 5 × 10−3 Vcm −1 dl 203.6cm The cadmium cell is now replaced by the unknown cell of e.m.f ,say E. Keeping the current through the potentiometer wire constant, the balancing length l is found. Then the e.m.f of the unknown cell is then given by dV E =l× dl = l × 5 × 10−3 V 2) To measure internal resistance of a cell: The circuit connection is shown in the figure. A constant current I is maintain through the potentiometer wire by using Rheostat. The key K2 is kept out and the balancing length l1 is obtained. Then, Let x be the resistance per unit length of the wire. E = ( xl1 )I
---------- (i)
After introducing the resistance , say S, the key K2 inserted and the balancing length l2 is obtained to find terminal potential difference, V = ( xl2 )I ---------- (ii) Dividing equation (i) by (ii) E l1 = V l2 The internal resistance of a cell is given by E r = − 1÷S V l or , r = 1 − 1÷S l2
is
l −l r = 1 2 ÷S l2 Knowing the values of l1 and l2, the internal resistance of a cell can be measure. 3) To compare e.m.f of two cells: ∴
The circuit connection is shown in the figure. A constant current I is maintained through the potentiometer wire between points A and B by using Rheostat. When terminals 1 and 2 of two way key are connected, let l be the balancing length between point A and jockey J. If x is the resistance per unit length of the wire, then, E1 = ( xl1 )I ---------- (i) If terminals 2 and 3 are connected (after removing it from 1 and 3), let the balancing length be l2. Then E 2 = ( xl2 )I ---------- (ii) Dividing equation (i) by (ii), we get E1 l1 = E 2 l2
34 Knowing the values of l1 and l2, the ratio of the emfs of two cells can be found.
UNIT – V CHAPTER – I ELECTROMAGNETIC WAVES ELECTROMAGNETIC WAVE: It consists of sinusoidally time varying electric and magnetic field acting at right angle to each other and at right angle to the direction of the propagation of the waves. ELECTROMAGNETIC SPECTRUM: The orderly distribution of electromagnetic waves (according to wavelength or frequency) in the form of distinct groups having widely differing properties is known as electromagnetic spectrum. It is divided into the following main parts: 1) Radio waves 2) Microwaves 3) Infra-red 4) Visible light 5) Ultra-violet rays 6) X-rays
7) γ- rays
The speed of electromagnetic waves in free space is given by 1 c= µ 0 ∈0 6 -1 −12 2 −1 −2 Where μ 0 (= 1.257 ×10 TmA ) and ∈0 (= 8.854 ×10 C N m ) permeability and absolute permittivity of the free space. The speed of electromagnetic waves in a medium is given by
are respectively the absolute
35 1 μ∈ Where μ is the absolute permeability and is the absolute permittivity of that medium. v=
RADIO WAVES (3 × 103 - 3 × 1011 Hz) Properties: 1) Radio waves obey laws of reflection and refraction. 2) Radio waves are electromagnetic waves and travels with a speed of 3 ×108 ms-1 . Uses: 1) Radio waves are used in radio astronomy. 2) They are used to transmit radio and T.V. signals. Microwaves: (3 × 108 - 3 × 1011 Hz) Properties: 1) Microwaves are electromagnetic waves and travels with a speed of 3 ×108 ms-1 2) Microwaves produced heat, when absorbed by matter. Uses: 1) Microwaves are used in radar system. 2) Microwaves are used in long distance telephone communication systems. Infra-red ray: (3 × 1011 - 4 × 1014 Hz) Properties: 1) Infra-red rays obey laws of reflection and refraction. 2) Infra-red rays effect a photographic plate. Uses: 1) Infra-red rays are used in solar water heaters and cookers. 2) Infra-red rays are used for producing dehydrated fruits. Visible Light: The frequency of visible light ranges from 4 × 1014 - 8 ×1014 Hz Ultra-violet rays: (8 × 1014 - 8 × 1016 Hz) Properties: 1) They obey the laws of reflection and refraction. 2) They can effect a photographic plate. Uses: 1) Ultra violet rays are used for checking the mineral samples by making use of its property of causing fluorescence. 2) As ultra violet rays can cause photoelectric effect, they are used in burglar’s alarm. X-rays: (1016 - 3 × 1019 Hz) Properties: 1) X-rays are electromagnetic waves of very short wavelength ranging from 0.01Ao − 10A o 2) They affect the photographic plate very intensely. Uses: 1) X-rays are used in detecting charge in old oil paintings. 2) X-rays are used in surgery for the detection of fractures, diseased organs and foreign bullets. Gamma rays: (3 × 1019 - 5 × 10 20 Hz) Properties: 1) γ-rays are electromagnetic waves and have velocity equal to that of light. 2) γ-rays are not deflected by electric and magnetic fields. Uses: 1) γ-rays are used in radiotherapy. 2) γ-rays are used to produce nuclear reactions.
36
UNIT – III CHAPTER – 1 MAGNETIC EFFECTS OF CURRENT BIOT-SAVART’S LAW: According to Biot Savart’s law the magnetic field (in magnitude) at a point at a distance r from the current element of length el carrying a current I is given by μ Idl sin θ dB = × 4π r2 Where θ is the angle between direction of the current and the line joining the current element to the observation points and is the absolute permeability of free space. RIGHT HAND THUMB RULE: If we grasp the conductor in the palm of right hand so that the thumb points in the direction of flow of current, then the direction in which the fingers and represents the direction of magnetic field lines. MAGNETIC FIELD AT THE CENTRE OF THE CIRCULAR LOOP: Consider a circular loop of radius r carrying a current I whose centre is O. To find magnetic field at the centre of the loop, consider a small element AB = dl. According to Biot-Savart’s Law, the magnetic field due to the whole circular loop is given by
dB =
μ 0 Idl sin θ × 4π r2
Since θ = 90·, sin 90· = 1 μ Idl sin θ ∴ dB = 0 × 4π a2 The magnetic field due to the whole circular loop is given by μ Idl μ I B=Ñ ∫ dB = Ñ ∫ 4π0 ×a 2 = 4π0 ×a 2 Ñ ∫ dl But Ñ ∫ dl = 2π a, the circumference of the loop μ 0 2πI × 4π a If the loop has n number of turns, then μ 2πnI B= 0 × 4π a ∴B =
37 MAGNETIC FIELD AT A POINT ON THE AXIS OF CIRCULAR LOOP: A circular loop of radius a, centre o carries a current I. Let P be the point on the axis of the loop at a distance X from O. To find magnetic field at point P, consider a small element AB = dl and CP = r.
According to Biot Savart’s Law. μ Idl sin θ dB = 0 × 4π r2 μ Idl ⇒ dB = 0 × 2 (θ Q 90 = )o 4π r Consider a small element A′ B′ = dl ′ opposite to AB and dB = dB′. Both dB and dB1 can be resolved into two components: dB cosθ and dB′ cosθ cancel each other and the effective component is only dB sin θ. Thus the magnetic field at point P due to the whole circular loop is B=Ñ ∫ dB sin θ
But
μ Idl =Ñ ∫ 4π0 ×r 2 sinθ μ I = 0 × 2 sinθ Ñ ∫ dl 4π r Ñ ∫ dl = 2π a, the circumference of the loop of radius a B=
From VOCP,
μ 0 Isin θ × 2πa 4π r 2
− − − − − − − − − − (i)
r =a +x ⇒r =(a +x 2
2
sin θ = And
=
2
2
)
a r a
(a
2
+x
1 2 2
)
Putting this value in eqn. (i) μ I B= 0 × 2 × 4π ( a + x 2 )
(a
=
1 2 2
μ0 2πIa 2 × 3 4π 2 2 2 (a +x )
a 2
+x
1 2 2
)
×2πa
38 AMPERE’S CIRCUITAL LAW: Ampere’s circuital law states that the line integral of magnetic field around any close path or circuit is equal to m0 times the total current (I) threading the close circuit. r r Bμ Ñ ∫ ×dlI = 0 Where μ 0 is the absolute permeability of free space.
Application: Magnetic field due to infinitely long straight wire: Consider an infinitely long straight wire carrying current I. Consider a circular path of radius r, such that the wireuupass u r urthrough its centre. Let PQ=dl be a small element of circular path and B be msgnetic field at point P. According to ampere’s circuital law,
r r Bμ Ñ ∫ ×dlI = or or or But
0
Ñ ∫ B ×dl cos 0μ=I ( Ñ ∫ Bμ×dlI = Bμ Ñ ∫ dlI = o
0
cosQ 0
0
1) o =
0
Ñ ∫ dl = 2π a, circumference of circular path. ∴ B × 2πr = μ 0I
μ 0 2I × 4π r This equation gives strength of magnetic field due to a straight conductor. B=
Magnetic field due to a Toroidal Solenoid: Consider a toroidal solenoid consisting of an anchor of radius r. Let n be the number of turns per unit length of the solenoid. According to ampere’sr circuital law, r Bμ Ñ ∫ ×dlr =total0 ×current threading the ring r or Bμ Ñ ∫ ×dl =( 2π0 ×I)n r or
Ñ ∫ Bμ×dl =(
or
B 2μ π r(=2π0 I)n
or
Bμ= I0 n
2π 0 × I)n r
r (
cosQ0
(since
1) o = = )r Ñ ∫ dl 2π
39
Magnetic field due to a long straight solenoid:
Consider a long solenoid having n turns per unit length. Let I be the current flowing in the solenoid. Consider a close path ABCD and let AB = r L. The line integral of B along path ABCD is given by r r Br r Cr r Dr r Ar r B Ñ ∫ ×dl = ∫ B ×dl + ∫ B ×dl + ∫ B ×dl + ∫ B ×dl ABCD
A
B
C
D
……………. i ur r Let AB=L. For path AB, B and dl are along the same direction. B r r B × dl = B ∫ ∫ dl = BL B
A
……………. ii
A
ur r For path BC and DA, B and dl are perpendicular to each other. r r Ar r ∫ B ×dl =∫ B ×dl = 0 C
B
…………….iii
D
r For path CD, B = 0 since field outside the solenoid is zero. D
r
r
∫ B ×dl = 0
……………. iv
C
Combining i, ii, iii, and iv, we get r r Br r B Ñ ∫ ×dl = ∫ B ×dl = BL ABCD
A
According to ampere’s circuital law, r r Bμ Ñ 0 × current enclosed by the path ABC ∫ ×dl =Total ABCD
or
r
r
Ñ ∫ Bμ×dl =(
L I) n 0×
ABCD
or
BLμ= 0(× Ln I)
or
Bμ= I0 n
This equation gives the magnetic field due to a solenoid.
D
40
CHAPTER – 2 FORCE ON A CHARGE FORCE ON A CHARGE IN ELECTRIC FIELD: r r A charge q inside an electric field E experienced a force F given by r r F = qE FORCE ON A CHARGE MOVING INSIDE AN ELECTRIN FIELD: r r r A charge q moving with velocity v inside a magnetic field B experience a force F given by r r r F = q (v + B) F = Bqv sin θ or, LORENTZ FORCE: The total force experience by a charge moving inside the electric and magnetic field is called Lorentz force. r r r r F = qE + q (v + B) Q Define the SI unit of magnetic field. (Tesla) Ans: The strength of magnetic field is called Tasla if a charge of 1 C moving with a velocity of 1 ms – 1 (along the direction perpendicular to the magnetic field), experience a force of 1 Newton.
MOTION OF A CHARGE PARTICLE INSIDE A UNIFORM MAGNETIC FIELD: r r When a charge particle having charge q moves inside a magnetic field B with velocity v is experience a force given by r r r F = q (v × B) 1) When velocity of the charge particle is perpendicular to the magnetic field: When a charge particle moves perpendicular to the magnetic field, it moves in a circular path as it experience towards its centre. The necessary centripetal force is provided by the lorentz magnetic force. mv 2 = Bqv r mv ⇒r= Bq The period of the circular motion is given by
41 2πr v 2πr mv = × v Bq 2πm = Bq
T=
The angular frequency, ω =
2π T
2πr 2mv Bq Bq ω= m =
(Cyclotron frequency)
r r 2) When v and B are inclined to each other: A charge particle moves inside a magnetic field making an angle θ with the r direction of the magnetic field. The velocity v charge particle can be resolved into two components: V cos θ and V sin θ. As the particle experienced an inward force. The period of the circular path is mv 2 = Bqv r mv sin θ r= Bq
The period of the circular path is 2πr T= vn 2π mv sin θ × v sin θ Bq 2πm T= Bq v cos θ × 2πn Pith of helican path = Bq 2πnv cos θ = Bq =
of the charge
42 CYCLOTRON: Principle: A positive ions can acquire sufficiently large energy with cooperatively smaller alternating potential difference by making it to cross the same electric field time and again by making used of a strong magnetic field. Construction: It consists of two D-shaped hollow semi circular metal chambers D1 and D2 (called dees) separated by a small gap and connected to high frequency electric field. The whole arrangement in enclosed in a metallic box at low pressure between two poles of strong electromagnet. Positive ions are produced in the gap of ionization of gas. Theory: Positive ion is produced at the centre gap when D 1 is at positive potential and D2 at negative potential. Hence the ion will move from D1 to D2. The force on the positive ion provides the necessary centripetal force and it is deflected along a circular path . 2 mv = Bqv r mv ⇒r= − − − − − − − − (i) Bq When the ions reach the gap after passing through D 2, the polarity of the dees changes and it is attracted by D1 and this process repeats itself time and again. The time spent inside the dee, πr t= v πm = − − − − − − − − (ii) Bq T This time t is equal to half of the time period of the electric field ÷ 2 Tπ m ∴ = 2 Bq 2πm ⇒T= Bq 2π Angular frequency, ω = T 2π ω= 2πm Bq Bq ω= m 1 Cyclotron frequency, v = T 1 v= 2πm Bq Bq v= 2πm
43 If vmax is the maximum velocity acquired by the ions corresponding to radius R mv R = max Bq BqR ∴ vmax = 2πm Hence maximum kinetic energy 1 E max = mv 2 2 2
1 1 BqR = mvmax 2 = m ÷ 2 2 m 1 B2 q 2 R 2 = 2 m
44 CHAPTER – 3 FORCE ON A CURRENT FORCE ON A CONDUCTOR PLACED IN A MAGNETIC FIELD : A conductor of length l carries a current I and is placed inside a magnetic field. If the conductor has cross sectional area A, number of electrons per unit volume n and drift velocity vd then,
I = nAevd Multiplying both sides by l, we get Il = nAlevd r r ∴ Il and vd are opposite in direcrion r r Il = −nAlevd − − − − − − − − (i) The magnetic lorentz force can be written as r r r f = −e(vd + B) − − − − − − − − (ii) Each three electrons experience a force and let N be the total number of electrons then, r r F = nf − − − − − − − − (iii) ∴ N = nAl we have r r r F = nAl × −e(vd + B) r r = −nAle(vd + B) r r r ∴ F = Il × B F = BIl sinθ or, o a) If θ = 0 or 180o then, sinθ = 0 ∴F = 0 (minimum) o o b) If θ = 90 or 180 then, sinθ = 1 ∴ F = BI l (maximum) FLEMING’S LEFT HAND RULE: It states that if the four finger, central finger and thumb of left hand are stre mutually perpendicular to each other, such that the four fingers point in the direction of magnetic field and the central finger in the direction of current and the thumb points in the direction of the force experienced by the conductor.
45 FORCE BETWEEN INFINITELY LONG PARALLEL CONDUCTOR:
Consider two infinitely long parallel conductors X1Y1 and X2Y2 separated by a distance r and carrying currents I1 and I2 . The magnetic field at point P due to current I2 flowing through the conductor X2 Y2 is μ 2I B2 = 0 × 2 4π r r The direction of B2 at point P is perpendicular to the plane of paper and in inward direction. Hence the conductor X1 Y1 lies inside the magnetic field B2 and experienced a force given by F = B2 × I1 μ 0 2I1I 2 × (towards left) 4π r Similarly the conductor X2 Y2 experience an equal force but towards right. Torque on a current loop placed in a magnetic field. F=
TORQUE ON A CURRENT LOOP PLACED IN A MAGNETIC FIELD:
A rectangular coil ABCD carrying current I is suspended in a magnetic field B. Let l be the length r r r r and b be the breadth of the coil. Let F1 , F2 , F3 and F4 and be the force acting on the arm DA, BC, AB and CD of the coil. Now,
uuur r r F1 = I(DA) × B
46
= I(DA)Bsin ( 90o + θ )
and,
= IbBcos θ r ∴ F1 = BIb cos θ uuu r r r F2 = I(BC) × B
− − − − − − −(i)
= I(BC)Bsin ( 90o − θ )
= IbBcos θ r ∴ F2 = BIb cos θ − − − − − − −(ii) The forces and are equaluuin and opposite in direction. Hence they cancel each other. ur magnitude r r F3 = I(AB) × B = I(AB)Bsin 90o = IbB
Then,
r ∴ F3 = BIb uuur r r F4 = I(CD) × B
− − − − − − −(iii)
= I(CD)Bsin 90o = IbB
r ∴ F4 = BIb − − − − − − −(iv) r r The forces F3 and F4 are equal in magnitude, opposite in direction and have different line of action. Hence the form a torque given by τ = Force × ⊥ er distance = BIl × b sin θ = BIA sin θ Where A is area of the coil In vector notation r r r ∴τ = M×B r r Since IA = M , magnetic dipole moment of the loop 1) If the coil has n turns τ = nBIA sin θ 2) If is the angle between the coil and the magnetic field, we can write. τ = nBIA cos ∞
47 MOVING COIL GALVANOMETER: Principle: When a current carrying coil is placed in a magnetic field it experienced a torque. Construction: It consists of a coil wounded on a non-metallic frame and has a central soft iron core. The coil suspended between poles of permanent magnet (whose poles made concave so as to produce radial magnetic field)from a torsion head. The other end of the coil is attached to a hair spring and a small concave mirror is attached to a small suspension wire. The whole arrangement is enclosed in a non-magnetic case which can be leveled by leveling screws. Theory: The coil ABCD of length l and breath b having n turns is suspended inside a magnetic field. If current passed through the coil in the direction of ABCD. Force on arm AB, F = nBIl (normally outwards) Force on arm CD, F = nBIl (normally inwards) The two forces are equal, parallel and opposite and acting different points. So they given rise to a torque. In case of radial magnetic field, Deflecting torque = nBIl ´ b = nBIA If k is restoring torque per unit twist. Restoring torque = ka In equilibrium, Defecting torque = restoring torque.
is are
is
at
nBIA = k a k I= a nBIA I = Ga
or , or , or ,
k is called Galvanometer constant. nBIA i.e., I ¥ a Deflection produced is directly proportional to the current passed through the Galvanometer.
Where G =
a)
b)
Current sensitivity: It is defined as the deflection produced in the galvanometer on α nBA passing unit current through its coil current sensitivity, = I k Voltage sensitivity: It is defined as the deflection produced in the galvanometer when a unit voltage is applied across its coil. α α nBA = = V IR kR
48 TO CONVERT A GALVANOMETRE INTO AMMETR:
To convert an galvanometer into an ammeter of range 0 to I, a small resistance S is connected in parallel to the coil of the galvanometer. Let G be the resistance of the galvanometer. Potential difference across shunt = potential difference across galvanometer. ⇒ ( I − I g ) S = Ig G S=
Ig
( I−I )
G
g
Knowing the value of Ig , I and G, the value of S can be find out. TO CONVERT GALVANOMETRE INTO VOLT METRE: To convert a galvanometer into volt meter of range 0 to V volt, a resistance R is connected in series with the galvanometer. When a current Ig flows through the coil, the galvanometer gives full scale deflection. If G is the resistance of the galvanometer, then,
V = Ig (G + R) ∴R =
V −G Ig
Knowing the value of Ig , I and G the value of R van be found out.
49 Magnetism is a force generated in matter by the motion of electrons within its atoms. Magnetism and electricity represent different aspects of the force of electromagnetism, which is one part of Nature's fundamental electroweak force. The region in space that is penetrated by the imaginary lines of magnetic force describes a magnetic field. The strength of the magnetic field is determined by the number of lines of force per unit area of space. Magnetic fields are created on a large scale either by the passage of an electric current through magnetic metals or by magnetized materials called magnets. The elemental metals—iron, cobalt, nickel, and their solid solutions or alloys with related metallic elements—are typical materials that respond strongly to magnetic fields. Unlike the all-pervasive fundamental force field of gravity, the magnetic force field within a magnetized body, such as a bar magnet, is polarized—that is the field is strongest and of opposite signs at the two extremities or poles of the magnet. Magnetism arises from two types of motions of electrons in atoms—one is the motion of the electrons in an orbit around the nucleus, similar to the motion of the planets in our solar system around the sun, and the other is the spin of the electrons around its axis, analogous to the rotation of Earth about its own axis. The orbital and the spin motion independently impart a magnetic moment on each electron causing each of them to behave as a tiny magnet. The magnetic moment of a magnet is defined by the rotational force experienced by it in a magnetic field of unit strength acting perpendicular to its magnetic axis. In a large fraction of the elements, the magnetic moment of the electrons cancel out because of the Pauli exclusion principle, which states that each electronic orbit can be occupied by only two electrons of opposite spin. However, a number of so-called transition metal atoms, such as iron, cobalt, and nickel, have magnetic moments that are not cancelled; these elements are, therefore, common examples of magnetic materials. In these transition metal elements the magnetic moment arises only from the spin of the electrons. In the rare earth elements (that begin with lanthanum in the sixth row of the periodic table of elements), however, the effect of the orbital motion of the electrons is not cancelled, and hence both spin and orbital motion contribute to the magnetic moment. Examples of some magnetic rare earth elements are: cerium, neodymium, samarium, and europium. In addition to metals and alloys of transition and rare earth elements, magnetic moments are also observed in a wide variety of chemical compounds involving these elements. Among the common magnetic compounds are the metal oxides, which are chemically bonded compositions of metals with oxygen. Earth's geomagnetic field is the result of electric currents produced by the slow convective motion of its liquid core in accordance with a basic law of electromagnetism which states that a magnetic field is generated by the passage of an electric current. According to this model, Earth's core should be electrically conductive enough to allow generation and transport of an electric current. The geomagnetic field generated will be dipolar in character, similar to the magnetic field in a conventional magnet, with lines of magnetic force lying in approximate planes passing through the geomagnetic axis. The principle of the compass needle used by the ancient mariners involves the alignment of a magnetized needle along Earth's magnetic axis with the imaginary south pole of the needle pointing towards the magnetic north pole of the earth. The magnetic north pole of Earth is inclined at an angle of 11° away from its geographical north pole.
You were introduced to the concept of an electric charge and studied some properties of charges at rest. You learnt that a static distribution of charge produces a static electric field. Similarly, steady flow of charge (i.e., a steady current) produces a static magnetic field, which is, infact, the topic of this unit. However, there are some major differences between the two fields which you will discover in this unit. In the science laboratory, during your school days, you must have been fascinated with magnets. Recall, when you tried to push two magnets together in a way they didn't want to go, you felt a mysterious force! In fact, magnetic fields or the effect of such fields have been known since ancient times when the effect of the naturally occurring permanent magnet (Fe304) was first observed. The north and south seeking properties of such materials played a large role in early navigation and exploration. Except for this application, magnetism was a little known phenomenon until the 19th century, when Oersted discovered that an electric current in a wire deflects a compass needle. This discovery showed that electric current has something to do with the magnetic field because a compass needle gets deflected and finally points inathe north-south
50 direction only when placed in a magnetic field. In this Unit, we shall consider in detail the production of the magnetic fields due to steady currents, and the forces they exert on circuits carrying steady currents and on isolated moving charge. A good way of gaining a better understanding of the nature of fields is to know how they affect the charged particles on which they act. Hence, in the next unit, you will study the behaviour of charged particles in both electric and magnetic fields CHAPTER – 4 MAGNETIC DIPOLE AND EARTH MAGNETISM MAGNETIC DIPOLE: An arrangement of two magnetic poles of equal and opposite strength separated by a finite distance is called magnetic dipole. MAGNETIC DIPOLE MOMENT: The product of the strength of either magnetic pole and the magnetic r length of the magnetic dipole. It is denoted by M r r M = m(2l ) Its SI unit is Ampere metre square. CURRENT LOOP AS A MAGNETIC DIPOLE: A circular coil carries current I as shown in fig. The magnetic field lines due to and portion will be the shape of circular loop. Using right hand thumb rule it follow that the lower face behaves as south pole while the upper face behaves as north pole. The magnetic dipole moment of the current loop of area A carrying current I is given by r r M = IA
If the coil has ‘n’ turns
r r M = nIA
ATOM AS MAGNETIC DIPOLE (Magnetic dipole moment of a revolving electron): In an atom electron revolves around the nucleus and may be considered as small current loop. Due to this an atom posses magnetic dipole moment and behaves as a magnetic dipole.
The angular momentum of electron on due to its orbital motion is given by L = M e vr − − − − − − −(i) Where M e is the mass of electron and r is the radius of the orbit. The orbital motion of electric is equivalent to current.
51 1 I = e ÷ T Where I is the time period of the electron 2πr T= v ev ∴I = 2πr 2 Area of the loop Aπ= r Magnetic dipole moment of the atom ev M = IAπ= × r2 2πr ev = ×r 2 e = × me vr 2me
− − − − − − −(ii) − − − − − −(iii) −
e M= ÷L 2me In vector rotation r e M = − 2me
r ÷L
− − − − − − −(iv)
According to Bohr’s theory, the angular momentum of an electron is an integral multiple of the nh 2π Where n =1, 2, 3, …. and so on…. From equation (iv) and (v), we have ev nh M= ÷× 2me 2π L=
n 2π
− − − − − − −(v)
eh ∴M = n ÷ 4πme Q. Define Bohr magnetron. Ans: It is define as the magnetic dipole moment associated with an atom due to orbital motion of an electron in the first orbit of hydrogen atom. It is denoted by μ B . ehv μB = = 9.27 ×10−24 Am 2 4πme MAGNETIC FIELD ON AXIAL LINE OF A BAR MAGNET: Consider a bar magnet NS having pole strength m 1 magnetic length 2l and centre O. Let P be the point on the axial line at a distance r from O.
The magnetic field axial r at point P is given by
then and r r Since B1 > B2
r r r Baxail = B1 + B2 r μ μ m m B1 = 0 × 2 = 0 × 4π NP 4π ( r - l ) 2 r μ μ m m B2 = 0 × 2 = 0 × 4π SP 4π ( r + l ) 2
52 (outwards) (inwards)
r r r Baxail = B1 − B2 =
μ0 μ m m × − 0× 2 4π ( r - l ) 4π ( r + l ) 2
r +l 2 − r -l 2 1 μ0 ( ) ( ) 1 μ0 = ×m − = ×m 2 2 2 4π ( r - l ) ( r + l ) 4π ( r 2 - l 2 ) μ 4rl = 0 ×m 4π ( r 2 - l 2 ) 2 Since, m (2l) = M, magnetic dipole moment. μ 2Mr ∴B = 0 × 4π ( r 2 - l 2 ) 2 If the bar magnet is of very short length ‘l’ can be neglected. r μ 2M Baxail = 0 × 3 4π r
MAGNETIC FIELD ON EQUITORIAL LINE OF A BAR MAGNET: Consider a bar magnet NS having pole strength m magnetic length 2 l and centre O. Let P be the point on the equitorial line at a distance r from O.
88 r r Let B1 and B2 be magnetic field at point P due to N-pole and S-pole of the bar magnet.
53
and
B1 =
μ0 m μ m × 2 = 0× 2 2 4π NP 4π ( r + l )
B1 =
μ0 m μ m × 2 = 0× 2 2 4π SP 4π ( r + l )
Using triangle law of vector, we have, SN Bequi = ×B1 NP μ 2l m = × 0× 2 2 1 2 ( r 2 + l 2 ) 4π ( r + l ) =
μ0 M × 2 4π ( r + l 2 ) 3 2
(Q m (2l) = M, magnetic dipole moment)
If the bar magnet is of short length, l can be neglected. μ μ M M Bequi = 0 × = 0× 3 3 2 4π ( r 2 ) 4π r TORQUE ON A BAR MAGNET IN A MAGNETIC FIELD: r A bar magnet NS having pole strength m and of length 2l is placed in uniform magnetic field B making an angle θ.
Force on N-pole of magnet = mB (along) Force on S-pole of magnet = mB (opposite to) The two forces are equal, parallel and opposite and constitute a torque. (τ) τ = Force × ⊥ distance = mB × KN V NKS In KN KN sinθ = = NS 2l KN = 2l sinθ Since, m (2l) = M, magnetic dipole moment. ∴ τ = MBsin θ In vector rotation r r r τ = M×B
54
BAR MAGNET AS EQUIVALENT SOLENOID: For a solenoid having cross sectional area A and carrying current I. Its turns behaves as small magnetic dipole having magnetic dipole moment I A. Therefore, solenoid can be treated as an arrangement of small magnetic dipole placed in line which can be treated as a bar magnet having magnetic dipole moment n I A.
CHAPTER – 5 CLASSIFICATION OF MAGNETIC MATERIAL MAGNETIC INTENSITY (Also called H-field or Magnetic field strength)
55 Consider that inside vacuum, a magnetic field Bo exist. The magnetic intensity is given by the relation B H= 0 μ0 - 7 - 1 Where m0 = 4p´ 10 TmA Its SI unit is Ampere per metre (Am-1)
INTENSITY OF MANETISATION: It is defined as the magnetic moment developed per unit volume when a magnetic specimen is subjected to magnetizing field. It is denoted by I. Its SI unit is Am −1 . M I= V MAGNETIC INDUCTION: The magnetic induction is defined as the number of magnetic lines of induction (magnetic force lines inside the material) crossing per unit area normally through the magnetic substance. It is denoted by B. Its SI unit is Tesla (T) Magnetic induction B is the sum of B0 and the magnetic field µ0 I produced due to magnetization. B = B0 +µ0 I or , B = µ0 H+µ0 I or , B = µ0 (H+I) MAGNETIC SUSCEPTIBILITY: It is defined as the ratio of the intensity of magnetization to the magnetic intensity. I χm = (It has no unit) H MAGNETIC PERMEABILITY: It is defined as the ratio of the magnetic induction (B) to the magnetic intensity (H) B μ= H Its SI unit is Tesla metre per Ampere (TmA-1) Q. Ans:
Derive the relation mr = 1 + c m where symbols have their usual meaning. We have or , B = µ0 (H+I) Dividing both sides by H, we get B I = µ0 I+ ÷ …………………………1 H H I B and µ0 = Also, χ m = H H ∴ µ = µ0 ( 1 + χ m ) or , or ,
µ = ( 1 + χm ) µ0
µr = ( 1 + χ m ) Where µ r =
......................................2
µ is called relative permeability of the magnetic substance. µ0
CLASSIFICATION OF MAGNETIC SUBSTANCE
56 DIAMAGNETIC SUBSTANCE: A substance, which when placed in a magnetic field is feebly magnetized in a direction opposite to that of the magnetizing field is called diamagnetic substance. e.g.: Copper, Zinc, water, Bismuth, etc Properties: 1) It does not obey Curie’s law. 2) The magnetic susceptibility of a diamagnetic substance has a small negative value. 3) For a diamagnetic substance, the intensity of magnetization (I) has small negative value. PARAMAGNETIC SUBSTANCE: A substance, which when placed in a magnetic field is feebly magnetized in the direction of the magnetizing field is called paramagnetic substance. e.g.: Aluminum, Manganese, Sodium, antimony, etc. Properties: 1) A magnetic substance is feebly attracted by a magnet. 2) It obey Curie’s law i.e. the susceptibility of paramagnetic substance is inversely proportional to its absolute temperature. 3) The magnetic susceptibility of a paramagnetic substance has small positive value. FERROMAGNETIC SUBSTANCE: Those substances, which when placed in a magnetic field are strongly magnetized in the direction of magnetizing field are called ferromagnetic substances. E.g.: Iron, Nickel, Cobalt, etc. Properties: 1) A ferromagnetic substance is strongly attracted by a magnet. 2) The magnetic susceptibility ( c m ) of a ferromagnetic substances has a large positive. 3) The ferromagnetic substance does not obey Curie’s law. CURIE LAW IN MAGNETISM: For paramagnetic material, intensity of magnetization of the material is 1) directly proportional to magnetic intensity i.e., I a H 2) inversely proportional to the absolute temperature of the material, 1 i.e., I a T Combining, we get, H Ia T H or , I = C T Where C is called curie’s constant. I C or , c m = = H T This is called Curie’s law in magnetism. i.e., The magnetic susceptibility of paramagnetic material is inversely proportional to its absolute temperature. CURIE’S POINT (CURIE’S TEMPERATURE): The Curie point (or Curie temperature) of a ferromagnetic material is the temperature above which it loses its ability to posses magnetism in the absence of an enternal magnetic field. EFFICIENCY OF TRANSFORMER: It is the ratio of output power to input power. It is denoted by y,
57 y=
output power E s I s = input power E p I p
ENERGY LOSS IN A TRANSFORMER: 1) Copper losses: Due to resistance of the copper coil, some electrical energy is converted into heat energy. 2) Humming losses: When a.c. is supplied to the primary coil, the iron core starts vibrating and producing humming sound thereby losing some electrical energy in the form of sound. 3) Hysteresis losses: When the ion core is magnetized, the iron core gets heated due to hysteresis. The energy loses in the form of heat is equal to the area of hysteresis loop. USES OF TRANSFORMER: 1) They are used as voltage regulator and stabilized water supply. 2) Small transformer are used in radio sets, TV and telephones etc.
Q. Ans:
Which is the material used in making the core of transformer or a moving coil galvanometer? Soft iron is used in making the core of galvanometer. It makes the magnetic field strong as magnetic field lines tend to cross through the paramagnetic.
Q. Ans:
Why soft iron is used in making the core of a transformer? The area of the hysteretic loop for soft iron is very small. Since energy dissipated during a complete cycle of magnetization and demagnetization is proportional to the area of the hysteresis loop, a small amount of energy will be wasted, when the core of the transformer is made of soft iron.
Q. Ans:
State two methods to destroy the magnetism of a magnet? 1) By applying magnetizing field on the reverse direction. 2) By heating the magnet.
58
UNIT – IV CHAPTER – 1 ELECTROMAGNETIC INDUCTION MAGNETIC FLUX: The number of magnetic field lines crossing the surface normally is called magnetic flux. (φ ) linked with that surface. Mathematically, r r φ = B ×A = BA cosθ Where B is the magnetic field, A is area of the surfaces; Q is the link between the field and the normal to the surfaces. Its SI unit is Webes (Wb). ELECTROMAGNETIC INDUCTION: It is the phenomenon of production of e.m.f in a coil. When the magnetic flux linked with the coil is changed. FARADAY’S LAW OF ELECTROMAGNETIC INDUCTION: 1) Whenever magnetic flux linked with a circuit changes, induced e.m.f is produced. 2) The induced e.m.f lasts as long as the change in the magnetic flux continues. 3) The magnitude of the induced e.m.f is directly proportional to the rate of change of the magnetic flux. Mathematically, −dφ φ2 − φ1 = Induced e.m.f, e = dt t LENZ’S LAW:
59 It states that the induced current produced in a circuit always flows in such a direction that opposes the change or causes produced by it. LENZ’S LAW AND PRINCIPLE OF CONSERVATION OF ENERGY:
Consider a solenoid connected to electric lamp. When a magnet is moved towards the solenoid as shown in fig., induced current is produced and the lamp grows. All the energy conservation principle work in equivalent to electric energy which must be done by the agent in pushing the magnet. As one magnet experiences opposing sources from the solenoid, work has to be done. Hence Lenz’s law is in accordance with law of conservation of energy. EDDY CURRENT: The current induced in the body of a conductor when the magnetic flux link with the conductor changes are called Eddy current. Q How are Eddy’s current produced ? Give two application of Eddy’s current. Ans: Eddy currents are produced in a metallic conductor when the magnetic flux link with the conductor changes. Applications: 1) In energy matter. 2) Dead beat galvanometer. Q. Ans:
How can Eddy current be reduced? Eddy currents produced in the core of transformer is formed by joining identical iron strips together and coating than with a varnish.
Q. Ans:
Define self induction. The phenomenon according to which an opposing induced e.m.f is produced in a will as a result of change in current on magnetic flux linked with it is called self induction.
Q. Ans:
Define co-efficient of self induction. The co-efficient of self induction is equal to the magnetic flux linked with the coil when a unit current flows through it. It is denoted by L. φ = LI Its SI unit is Henry (H). The induced e.m.f can be written as − dφ − d e= = (LI) dt dt dI ∴ e = −L dt
SELF INDUCTION OF LONG SOLENOID: The magnetic field inside a solenoid carry current I is given by B = µ0 nI Where n is the no. of turns per unit length. In case of a long solenoid, magnetic flux passing through its turns
60 = B × area of each turnμ= 0 nIB× If l is the length of the solenoid, total magnetic flux, μ = IA0 n × nl φ ⇒ φ = μ 0 n 2lAI If L is self inductance of the solenoid φ = LI Comparing equation (i) and (ii) Lμ= 0A n2 l
− − − − − − −(i) − − − − − − −(ii)
Q Ans:
Define mutual induction. The phenomenon according to which an opposing emf is produced in a coil as a result of change in current or magnetic flux linked with the neighboring coil.
Q. Ans:
Define co-efficient of mutual induction. The co-efficient of mutual induction of two coils is numerically equal magnetic flux linked with one coil, when a unit current flows through the neighboring coil. It is denoted by M. φ = MI Its SI unit is Henry. In terms of induced e.m.f. dI e = −M dt
Q Ans:
Define one Henry. One Henry is the rate of change of current of 1 ampere per second in one coil induces an e.m.f of 1 volt in the neighboring coil.
61
UNIT – IV CHAPTER – 2 ALTERNATING CURRENTS ALTERNATING CURRENT: The electric current, whose magnitude changes with time and direction reverse periodically, is known as alternating current. It can be represented by a sine curve or a cosine curve .i.e., I = I0 sin ωt I = I0 cos ωt or Where I0 is peak value of current (Amplitude) and I is called instantaneous value of AC. 2π ω= = 2πf T Where T is period of A.C. and f is frequency of A.C. Alternating emf can be represented by E = E 0 sin ωt E = E 0 cos ωt
62 MEAN OR AVERAGE VALUE OF AC: It is the value of steady current which would send the same amount of charge through the circuit in the time of half cycle ( i.e., T2 ) as is sent by A.C. through the same circuit in the same time. It is denoted by I m or I a The instantaneous value of a.c is given by, I = I0 sinω t − − − − − − −(i) If the current remains constant for small time, at then the amount of charge flowing is dq = Idt dq = Isin ωt dt − − − − − − −(ii) T The amount of charge flowing in time is given by, 2 T2
q=
∫
T2
dq =
0
∫
T 2
∫
I0 sin ωt dt = I 0
0
sin ωt dt
0
T 2
I T 2 cos ωt = I0 − = − 0 [ cos ωt ] 0 ω 0 ω I 2π T 2π 2π = − 0 cos × − cos ×0ω = Q ÷ 2π T T 2 T T IT IT = − 0 [ cosπ −cos 0 ] = − 0 [ −1 −1 ] 2π 2π T q = I0 − − − − − − −(iii) π If I m is mean value of current, then, I Tπ q Im = = 0 T2 T2 2I I m = 0 = 0.636 I0 π = 0.636 ×100% I0 = 63.6% I0 The mean value of a.c. during half cycle is 63.6 % of its peak value. During the next half cycle, the mean value of a.c. will be equal in magnitude but opposite in direction. Hence, the mean value of a.c. over a complete cycle is always zero. ROOT MEAN SQUARE (RMS) VALUE OF A.C.: It is a steady current which when passed through a resistance for a given time will produce the same amount of heat as the a.c. does the same through the same resistance and in the same time. It is denoted by I rms or Iv The instantaneous value of a.c. is given by, I = I0 sinω t − − − − − − −(i) If the current through the resistance remain constant for a small time dt small amount of heat produced is dH = I 2 Rdt − − − − − − −(ii) The amount of heat produced in time T 2 is given by T2
H=
∫
T 2
I0 2 R sin 2 ωt dt = I 0 2 R
0
sin 2 ωt dt
0
T 2
= I0 2 R
∫
∫ 0
T2 1T2 1 − cos 2ωt 1 dt = I 0 2 R ∫ dt − ∫ cos 2ωt dt ÷ 2 ÷ 2 2 0 0
63 Since
1 2
T2
∫
dt =
0
T 1 and 2 2
T 2
∫
cos 2ωt dt = 0
0
1 T ∴ H = I0 2 R × 2 2 R T H = I0 2 × 2 2 If I rms is the RMS value of current, then T H = I rms 2 R × 2 Comparing equation (iii) and (iv) I2 I rms 2 = 0 2 I ⇒ I rms = 0 2 I rms = 0.707 I0 The RMS value of a.c. is 70.7 % of its peak value.
− − − − − − −(iii)
− − − − − − −(iv)
A.C. THROUGH LCR: An alternating current emf is connected to a series combination of inductance L, capacitance C and resistance R. Let V L , V C and V R be the instantaneous across L, C and R. I L = I X L ; IC = I X C ; VR = I R 1C ωC = Where XωL L = and X Where XL = XC and X C = 1ωC are the resistance of inductor and capacitor The relation between E, V L , V C and VR are shown in the phases diagram. From phases diagram in VABD OB = OA 2 + AB2 E = OA 2 + OC2 = VR 2 + ( VL − VC ) = I 2 R 2 + ( I X L − IX C )
2
= I2 R 2 + I2 ( X L − X C ) E = I R 2 + ( XL − XC ) I= ⇒I=
2
2
E R 2 + ( XL − XC )
2
E Z
Where Z = R 2 + ( X L − X C ) = R 2 + ( ωL − 1 ω C ) is called impedance of LCR circuit. 2
2
64 AB OA OC = OA V − VC = L VR
tanφ =
From VOAB
e
=
I XL − I XC VR
=
I XL − I XC IR
X L − XC R ωL − 1 ω C φ = tan −1 ÷ R
tanφ =
RESONANCE CONDITION: If the inductive and capacitive reactants (XL and XC) are equal i.e. ωL and 1 ω C cancel out and the current in the circuit becomes very high. This condition is called resonance. If ω0 is the resonant frequency, then ω0 L = 1 ω C 1 LC 1 ⇒ ω0 = LC 1 2π f 0 = LC 1 ∴ f0 = 2π LC ⇒ ω0 2 =
POWER OF AN A.C. CIRCUIT: The instantaneous e.m.f and current in an a.c. circuit are given by E = E 0 sin ωt I = I0 ( sin ωt + φ )
Where φ is called the phase angle and, 2π ω= T If the e.m.f and current in the a.c. circuit remains constant for a small time dt than small amount of electrical energy is given by d ω = EIdlE 0 sin ωt + I 0 ( sin ωt + φ ) d ω = E 0 I0 sin ωt ( sin ωt cos φ + cos ωt sin φ ) dt
1 − cos ωt sin 2ωt and ∴ sin ωt cos ω t = 2 2 2 1 − cos ωt sin 2ω t ∴ d ω = E 0∂ I 0 cos φ + sin φ dt 2 2 The energy consumed in time T is ∴ sin 2 ωt =
2
− − − − − − −(i)
65 T
ω = ∫ dω = 0
T
∫ dt = [ t ]
And
T 0
E0I0 2
T T cos φ dt − cos φ cos ω t dt + sin φ sin 2 ω t dt ∫0 ∫0 ∫0 T
= ( T − 0) = T
0
T
T
0
0
∫ cos ωt dt =∫ sin ωt dt = 0 E 0 I0 [ cos φ (T) − cos φ (0) + sin φ (0)] 2 EI ω = 0 0 cos φ (T) 2 The average power of a.c. circuit is ω P= T E I 2 cos φ T E 0 I 0 P= 0 0 = × cos φ T 2 2 P = E rms I rms cos φ i) a.c having R only ∴ω =
φ =0 ∴ P = E rms I rms cos 0 P = E rms I rms ii) a.c having L only φ =π 2 ∴ P = E rms I rms cos π 2 iii) a.c having C only φ = −π 2 ∴ P = E rms I rms cos −π 2 P=0 A.C. CONTAINING LCR CIRCUIT: ωL − 1 ω C tan φ = ÷ R R ⇒ cos φ = 2 RωL + ( 1 ω−C ∴
P = E rms I rms
)
2
R 2 RωL + ( 1 ωC −
)
2
66
Fleming’s Right Hand Rule: It states that if the thumb, fore finger and the central finger of right hand are kept perpendicular to each other, fore finger points in the direction of the field and the thumb in the direction of motion of the conductor, then the induced current flows in the direction of central finger. TRANSFORMER: It is a device used to convert low alternating voltage at high current to high alternating voltage at low current or vice versa. Principle: It is based on the principle of mutual induction i.e. if two coils are inductively coupled with changing magnetic flux or current in one coil, induced e.m.f is produced in another coil. Construction: It consists of two coils, primary P and secondary coil S wounded on soft iron core. The a.c. input is applied across the primary coil P and the transformed output is obtained across the secondary coil S. The soft iron core is laminated to reduce eddy current. The transformer are of two types: 1) Step-up transformer: The number of turns in the coil S is greater than that in the coil P. The primary coil is made of thick insulated copper wire while the secondary coil is of thin wire. It converts a low voltage at high current to a high voltage at low current. 2) Step-down transformer: The number of turns in the coil S is less than that in the coil P. The primary coil is made of thin wire while the secondary coil is of thick wire. It converts a high voltage at low current into a low voltage at high current Working and Theory: When the primary coil is connected to a.c. source, induced e.m.f ‘ es ’is produced in secondary coil and also causes induced e.m.f ‘ e p ’ in itself. dφ p
d φs ………..1 dt dt Where φ p and φs are the fluxed linked with primary and secondary coil. ep = −
and es = −
self
67 Also, magnetic flux linked with the coil is simply proportional to their number of turns, then φs N s = φp N p or ,
d φs N s d φ p = dt N p dt
................2
From eqns 1 and 2, we get es = or , Where k =
Ns ep Np
es = k e p
Ns is called transformer ratio. Np
If k > I i.e. N s >N p , the transformer is called step up transformer. If k < I i.e. N s >N p , the transformer is called step down transformer.
EFFICIENCY OF TRANSFORMER: It is the ratio of output power to input power. It is denoted by y, output power E s I s y= = input power E p I p ENERGY LOSS IN A TRANSFORMER: 4) Copper losses: Due to resistance of the copper coil, some electrical energy is converted into heat energy. 5) Humming losses: When a.c. is supplied to the primary coil, the iron core starts vibrating and producing humming sound thereby losing some electrical energy in the form of sound. 6) Hysteresis losses: When the ion core is magnetized, the iron core gets heated due to hysteresis. The energy loses in the form of heat is equal to the area of hysteresis loop. USES OF TRANSFORMER: 3) They are used voltage regulator and stabilized water supply. 4) Small transformer are used in radio sets, TV and telephones etc. A.C. GENERATOR (AC Dynamo) It is a device used to convert mechanical energy into electrical energy by virtue of electro magnetic induction. Principle: It is based on the principle of magnetic induction. An induced emf is produced in a coil whenever there is a change in the magnetic flux linked with it. Construction; It consist of a rectangular coil of wire ABCD wound on a soft iron core. It is mounted on a shaft between the poles of a powerful magnet. It can be rotated about an axis perpendicular to the direction of magnetic field lines. The ends of the coil are connected to two metal rings R 1 and R2 which rotate with the coil. Contact is maintained with the external circuit by means of two carbon brushes B1 and B2 which touch the rings. working: During the motion from t = 0 and t = T/2, the arm AB move and CD moves up. Using Fleming’s Right Hand Rule, the direction of current will be DCBA.
down
68 During the motion from t = T/2 the arm AB move up and CD move up. Fleming Right Hand Rule, the direction will be ABCD. If A is the area of the n be number of turns in the coil, then magnetic flux = nBA cos ωt. Differentiating with respect to t, dφ = −nBA ω sin ωt dt If e is the induced emf, then
dφ dt = −nBA ω sin ωt e = e0 sin ωt e=−
e0 = nBA ω is caled the peak value of induced e.m.f .
and t = T, Using of current coil, and the we get,
