Solutions and Explanations for HW Chapter 27 BB Prepared by Prof Omar Chmaissem and TA Abhilash Sajja Question 1 A 1 A circular loop of wire of radius radius 0.50 m is in a uniform magnetic field of 0.30 T. The current in the loop is 2.0 A. What is the magnetic torque when when the plane of the loop loop is perpendicular to the magnetic field?
0.47 m∙N 0.52 m∙N 0.59 m∙N zero
0.41 m∙N Explanation: Magnetic Torque is given by
= B sin where is the angle between the magnetic moment and the Magnetic field.
Since the direction of is perpendicular to plane of the loop and the direction of B is perpendicular to the plane,
should be parallel to B which results in to be 0°.
= Bsin0°
= 0 m.N
Question 2 A 2 A stationary proton proton is in a uniform uniform magnetic field of 0.20 T. What is the magnitude of the magnetic force on the proton? zero 3.2 × 10- N 1.6 × 10-20 N 1.6 × 10- N 3.2 × 10- N
Explanation:
Force on a particle moving in a magnetic field is given by F = q( v x B ) where q is the charge of the particle, v is the velocity of the particle entering magnetic field, and B is the magnetic field strength. The magnitude of this magnetic force can be written as F = qvBsin where is the angle between the velocity vector and the magnetic field. In this case, the proton velocity is 0m/sec so the magnetic force will also be 0N.
Question 3 A 3 A proton travels travels through a potential of of 1.0 kV and then moves into a magnetic field of 0.040 T. What is the radius of the proton's resulting orbit? 0.11 m 0.19 m 0.17 m 0.14 m 0.080 m
Explanation: Remember the equation we derived in class for the cyclotron radius, r = mv/Bq [eq(25.1)]. This was obtained by setting the net force which is centripetal (mv2 /r) to equal the magnetic force qvb. To solve this equation we need to figure out the proton’s velocity. Initially the proton is accelerated by a potential V = 1000V. Hence, the Electric potential energy gained by the proton is qV. This energy is converted into kinetic energy resulting in the proton moving with some velocity v. Hence, KE = PE (½)mv2 = qV eq(25.2) Substituting the values of mproton = 1.67 x 10 -27Kg, V = 1000V , q = 1.6 x 10 -19C in eq(25.2), we get v = 4.38 x 10 5 m/s. Hence, the proton enters the magnetic field with this velocity v. Substituting m= 1.6 x 10-27kg, v = 4.38 x 10 5 m/s, B = 0.04T, q = 1.6 x 10 -19 C in eq(25.1), we get r = 0.11m 5
Question 4 An electron traveling due north with speed 4.0 × 10 m/s enters a region where the Earth's magnetic field has -5 the magnitude 5.0 × 10 T and is directed downward at 45° below the horizontal. What force acts on the electron? 3.2 × 10 N 2.3 × 10-20 N 3.2 × 10- N 2.3 × 10- N 2.3 × 10-18 N
Explanation: Straightforward. The magnitude of the magnetic force on a particle moving in a magnetic field is given by F = qvB sin . We just need to clearly identify the angle . Draw a sketch for the direction of the electron’s velocity due north and the earth’s magne tic field downward and tilted, the angle between the two vectors is indeed 45 degrees. Other given parameters are q = 1.6 x 10-19C, v =4.0 105 ×
m/s, B = 5.0 10 5 T ×
-
Substituting these parameters in the equation, we get F = 2.3 x 10-18N. -6
-6
6
Question 5 A charge q = 3 × 10 C of mass m = 2 × 10 kg, and speed v = 5 × 10 m/s enters a uniform magnetic field. 4 2 The mass experiences an acceleration a = 3 × 10 m/s . What is the minimum magnetic field that would produce such an acceleration? 0.001 T 0.04 T 0.004 T 0.1 T 0.01 T
Explanation: The mass m experiences an acceleration a due to a net force F given by F = ma = 0.06 N
The only force acting significantly on this charge is the magnetic force. Therefore the net force found above is the same as F = qvBsin . Hence, to produce this force, the magnetic field B should be B = F/qvsin . Minimum magnetic field is obtained when sin is maximum because it’s in the denominator. Therefore, sin = 1 and Bmin= F/qv Plug F= 0.06N, q = 3 x 10 -6C, v = 5x10 6 m/s into the above equation, we get B = 0.004T Question 6 A beam of electrons is accelerated through a potential difference of 10 kV before entering a velocity selector. If the B-field of the velocity selector has a value of 0.010 T, what value of the E-field is required if the particles are to be undeflected? 7.2 × 10 V/m 6.0 × 105 V/m 2.3 × 10 V/m 5.9 × 10 V/m 7.9 × 103 V/m
Explanation: We already know that in a velocity selector we obtain v = E/B. I know the magnetic field but v and E are unknown. It’s obvious that I need to figure out the velocity in order to solve for the electric field magnitude. Initially proton is accelerated by a potential difference V = 10,000V Hence, KE = qV (½)mv2 = Vq eq(27.1) -31 Substituting the values of m = 9.1 x 10 Kg, V = 10000 V, q = 1.6 x 10 -19C in eq(27.1), we get v = 5.93 x 10 7 m/s Hence, the proton enters and exits undeflected the velocity selector with this velocity v. v = E/B E = vB E = 5.93 x 107 x 0.01 E = 5.93 x 105 V/m
Question 7 A loop of diameter d =12 cm, carrying a current I
T) j . The normal to the loop is parallel to the unit vector ˆ
n ˆ
=
0.4 A is placed inside a magnetic field
B
=
(0.2 T)
i + (0.4 ˆ
= 0.6 i – 0.8 j . What is the potential energy of the loop? ˆ
ˆ
+4.5 × 10- J +9.0 × 10- J -9.0 × 10- J -4.5 × 10-4 J -2.3 × 10- J
Explanation:
Magnetic loop potential energy is given by U = – μ . B (dot product of magnetic moment vector and magnetic field
vector B) A vector is defined as having two quantities: magnitude and direction. Applied to the magnetic moment vector μ , we have the magnitude (IA) and the direction which is the unit vector
n ˆ
μ
μ
= 0.4( * (0.06)2) * (0.6 i – 0.8 j ) ˆ
= (2.7 x 10-3
i
ˆ
ˆ
– 3.6 x 10-3 j ) A.m2 ˆ
B
= (0.2
i
ˆ
+ 0.4 j ) T ˆ
U = – μ . B – 3.6 x 10 -3 j ) . (0.2 i + 0.4 j ) = (–2.7 x 10-3 U = (2.7 x 10-3)(0.2)+( 3.6 x 10 -3)(0.4) = 0.9 x 10 -3 J = 9 x 10 -4 J U = –(2.7 x 10-3
i
ˆ
ˆ
ˆ
ˆ
i
ˆ
+ 3.6 x 10-3 j ) . (0.2 ˆ
i
ˆ
+ 0.4 j ) ˆ
Question 8 A wire, 0.60 m in length, is carrying a current of 2.0 A and is placed at a certain angle with respect to the magnetic field of strength 0.30 T. If the wire experiences a force of 0.18 N, what angle does the wire make with respect to the magnetic field? 35° 20° 25° 60° 30°
Explanation: Magnetic force on a current carrying wire is given by F = ILB sin( ) where is the angle between the wire length vector taken in the direction of the current (I) and the magnetic field (B). Substituting the parameters I = 2A, B = 0.3T, L = 0.6m, F = 0.18N in the above equation, we get 0.18 = 0.3 x 2 x 0.6 sin( ) sin( ) = 0.5 = 30° Note that if the angle were to be zero (meaning the wire is parallel to the field) then there would be no magnetic force present. 3
Question 9 A proton is projected with a velocity of 7.0 × 10 m/s into a magnetic field of 0.60 T perpendicular to the motion of the proton. What is the force that acts on the proton? 0N 6.7 × 10- N 13 × 10-16 N 3.4 × 10- N 4.2 × 10-16 N
Explanation: Magnetic force on a particle moving in a magnetic field is given by F = qvB sin where is the angle between the direction of the particle motion and the magnetic field. Substituting the parameters as v = 7000m/s, B = 0.6T q = 1.6 * 10 -19 (charge of proton), =90°, we get F = 6.7 x 10 -16 N
Question 10 A thin copper rod 1.0 m long has a mass of 0.050 kg and is in a magnetic field of 0.10 T. What minimum current in the rod is needed in order for the magnetic force to cancel the weight of the rod? 7.6 A 1.2 A 4.9 A 9.8 A 2.5 A
Explanation: Magnetic force on a current carrying wire is given by F = ILBsin where is the angle between the current direction (I) and the magnetic field (B). This force has to be canceled by gravity. Or, mg = ILBsin . Solving for I, we get I = mg/ LBsin . To get the minimum value of the current flowing in the wire we have to set ( maximizing the denominator – there are no other variables to change).
to be 90o
Substituting the parameters , B = 0.1T, L = 1.0m, =90° in the above equation, we get I = 4.9A 4.9 A is the minimum current required to generate a magnetic force which could cancel the force du e to gravity.
Question 11 ) A loop of diameter d = 10 cm, carrying a current I = 0.2 A is placed inside a magnetic field B
0.3 T k .
=
ˆ
= - 0.6 i – 0.8 j . Calculate the magnitude of the torque on the loop. (This is a nice but tedious problem to solve. You will need to determine vector A . This can be done by multiplying the magnitude of A by the unit vector direction because it is in the same direction. Once you figure it out, you do cross product according to the torque equation. The answer is in a vector notation form. You will need then to calculate the magnitude of the torque vector.) The normal to the loop is parallel to a unit vector
n
ˆ
ˆ
ˆ
1.2 × 10- Nm 0 0.6 × 10- Nm 2.8 × 10- Nm 4.7 × 10-4 Nm
Explanation: As described in the hint, magnetic moment vector can be obtained by multiplying its magnitude (IA) with the unit vector representing its direction Hence vector = 0.2* =
n ˆ
= – 0.6 i – 0.8 j ˆ
ˆ
= (I*A) *unit vector n
*(0.05)2 * (– 0.6 i – 0.8 j ) ˆ
ˆ
(–9.42 x 10-4 i –12.56 x 10-4 j ) A.m2 ˆ
ˆ
= x B
(cross product)
= = (–9.42 x 10-4 i –12.56 x 10-4 j ) x (0.3 k ) Please review the cross product rules
= (+2.83 x 10-4 j – 3.77 x 10-4 i ) Nm
ˆ
ˆ
Magnitude of =10-4
ˆ
ˆ
ˆ
(2.83) 2
(3.77) 2 = 4.7 x 10 -4 N.m
Question 12 What is the magnetic moment of a rectangular loop of 120 turns that carries 6.0 A if its dimensions are 4.0 cm × 8.0 cm?
2.3 A∙m2 230 A∙m 23 A∙m 0.023 A∙m2 0.23 A∙m2 Explanation: Magnetic moment of a rectangular loop is given by = NIA where N is the number of turns, I is the current in the loop, and A is the area of the loop. = 120*6*32x10-4 = 2.3 A.m2
The magnetic moment is a vector quantity with a direction perpendicular to the current loop according to the righthand rule. It is the same as the direction of the vector A representing the loop area A. -2
Question 13 An electron moving perpendicular to a magnetic field of 3.2 × 10 T moves in a circle of radius 0.40 cm. How fast is this electron moving? 3.0 × 109 m/s 1.9 × 10- m/s 0.80 × 10 m/s 2.3 × 107 m/s 1.9 × 10- m/s
Explanation: The centripetal force is balanced by the magnetic force resulting in a circular motion. Hence, mv2 /r = qvBsin( ) (with =90°) v = qBr/m v = (3.2 10 2 x 1.6 x 10 -19 x 0.004) / (9.1 x 10 -31) v = 2.3 x 107 m/s. ×
-
Question 14 A proton is accelerated from rest through 500 V. It enters a magnetic field of 0.30 T oriented perpendicular to its direction of motion. Determine the radius of the path it follows. 1.1 m 11 m 11 cm 1.1 mm 1.1 cm
Explanation:
We know this is going to be a circle. We need the speed. We can calculate the incoming kinetic energy since we know that the proton is accelerated in a potential difference V = 500V Hence, KE = qV (½)mv2 = qV eq(25.1) Substituting the values of m = 1.67 x 10 -27Kg, V = 500V , q = 1.6 x 10 -19C in eq(25.1), we get v = 3.1 x 105 m/s Hence, the proton enters the magnetic field with this velocity v. Remember the equation, the centripetal force balancing the magnetic force From that we derive, r = mv/qB eq(25.2) -27 5 Substituting m= 1.67 x 10 kg, v = 3.1 x 10 m/s, B = 0.3T, q = 1.6 x 10 -19C in eq(25.2), we get r = 1.1 cm Question 15 In a Hall experiment a conducting strip of thickness d = 100 μm is placed in a magnetic field B = 0.05 T. The magnetic field is perpendicular to the direction of the strip along which there is a 10-A current. What is the Hall voltage 28 3 measured across the strip if the charge carrier density is 2.5 × 10 /m in this material?
0.72 μV 4.2 μV 2.5 μV 1.3 μV 6.2 μV Explanation: Hall voltage is given by VH = IB/ned where I is the current through the strip, B is the magnetic field, d is the strip thickness, n is the carrier density, and e is electron charge. Substituting I = 10A, B = 0.05T, n = 2.5 x 10 28 /m3, e = 1.6 x 10 -19 C, d=10-4 m in the above equation, we get VH = 1.3 x 10-6 V (approx).
Question 16 A circular loop of wire of radius 0.50 m is in a uniform magnetic field of 0.30 T. The current in the loop is 2.0 A. What is the magnetic torque when the plane of the loop is parallel to the magnetic field?
0.59 m∙N zero
0.47 m∙N 0.41 m∙N 0.52 m∙N Explanation: Magnetic Torque is given by
= Bsin where is the angle between the magnetic moment and the Magn etic field.
Since the direction of is perpendicular to the plane of the loop and the direction of B is parallel to the plane, should be perpendicular to B which results in to be 90°. = Bsin90° ( = IA)
= IABsin90° = (2) x (3.14 x0.52) x (0.3) x sin90° = 0.47 m.N
Question 17 A straight wire 20 cm long, carrying a current of 4 A, is in a uniform magnetic field of 0.6 T. What is the force on the wire when it is at an angle of 30° with respect to the field? 0.5 N 0.3 N 0.4 N 0.6 N 0.2 N
Explanation: Magnetic force on a current carrying wire is given by F = ILBsin where is the angle between the current direction (I) and the magnetic field (B). Substituting the parameters I = 4A, B = 0.6T, L = 0.2m, = 30° in the above equation, we get F = 0.6 x 4 x 0.2 sin 30° F = 0.2 N Question 18 A wire carries a current of 11.4 A in a direction that makes an angle of 11.4° with a magnetic field of -3 magnitude 11.4 × 10 T. The magnitude of the force on 11.4 cm of this wire is 1.48 × 10-2 N. 2.93 × 10- N. 0.130 N. 11.4 × 10-3 N. 0.013 N.
Explanation: Magnetic force on a current carrying wire is given by F = ILBsin where is the angle between the current direction (I) and the magnetic field (B). Substituting the parameters I = 11.4A, B = 11.4mT, L = 0.114m, = 11.4 in the above equation, we get F = 11.4x10-3 x 11.4 x 0.114x sin 11.4 F = 2.93 x 10-3 N. Question 19 An electron moving with a velocity
and a magnetic fields are present. The electric field is through, undeflected? = + (2.0 × 10-4 T) = + (2.0 × 10-4 T) =- (2.0 × 10-4 T) = - (2.0 × 10-4 T) = + (5.0 × 10-4 T)
7
= 5.0 × 10 m/s enters a region of space where perpendicular electric 4
= - 10
. What magnetic field will allow the electron to go
Explanation: This is the velocity selector device. v = E/B We did not use this name in class but the sum of the electric and magnetic forces is called the Lorentz force. Lorentz force = Electric force + Magnetic force When the charge goes undeflected it’s because the electric force is canceled out by the magnetic force and vice versa. This is essentially saying that the Lorentz force is zero. 0 = Electric force + Magnetic force Magnetic force = - Electric force Using vector notations because we also need to figure out the magnetic field direction:
q v B
vB
qE
E
(where the x sign represents the cross product)
7
v B = (5.0 × 10 m/s ) x (Bx +By +Bz k )
E 104 j so ˆ
ˆ
E
= (5.0 × 107) By k + (5.0 × 107)Bz (- ) ˆ
10 4 j ˆ
Equating the two equations v B E show that (By) must be zero. We get the following relationship: (-5.0 × 7 4 4 7 -4 10 )Bz = (10 ) or Bz = 10 /(-5.0 × 10 ) = - 2.0 x 10 T. Bz obviously represents a magnetic field that is fully pointing in the negative z direction with the magnitude we just calculated.
B
- (2.0 10-4 T) k
=
×
ˆ
2
Question 20 A circular loop of wire of cross-sectional area 0.12 m consists of 200 turns, each carrying 0.50 A. It is placed in a magnetic field of 0.050 T oriented at 30° to the plane of the loop. W hat torque acts on the loop?
5.2 m∙N 0.52 m∙N 0.25 m∙N 2.5 m∙N 25 m∙N Explanation: Since the magnetic field is oriented 30° to the plane of the loop, B should be 60° (90° – 30°) to the magnetic moment . Draw a sketch and check the actual angle. Hence, = 60° Magnetic Torque is given by =NIABsin = 200x0.5x0.12x0.05xsin60° = 0.52 m.N 6
Question 21 A proton is moving at a speed of 3.2 × 10 m/s at an angle of 80° to a uniform magnetic field of strength 1.5 -4 × 10 T. What is the distance moved by the proton along the spiral path as it completes one rev olution? (This distance is called the "pitch" of the helix.)
240 m 1390 m 1370 m 220 m cannot be determined
Explanation: We already know that any travel angle different than 0 o, 90 o or 180o should produce a spiral path for a charge moving in a magnetic field. In this problem, we can calculate two components for the velocity: Parallel to the magnetic field v // = v cos = 0.555 × 106 m/s and Perpendicular to the field v ┴ = v sin = 0.985 × 106 m/s. The pitch distance is p = v // T. T is the time required to make a full revolution. When the proton completes one full revolution in one period T due to the perpendicular velocity component, it finds itself propelled forward by the pitch distance due to the parallel velocity component. To emphasize: the forward distance is solely due to the parallel velocity component as the distance equation shows. It’s obvious that we need to calculate the period T. We just mentioned that it is the time for the time to complete a full revolution. This means that the distance travelled around the circular path (due to the perpendicular component of the velocity) is 2 r = v ┴ T . r is not known so we have to use r = mv ┴ /qB. Plug and cancel the v ┴ terms to get: 2 m/qB = T T = 2 (1.672 x 10-27 kg)/[(1.602 x 10 -19 ) (1.5 x 10 -4)]= 4.37 x 10-4 seconds Notice that the period (or frequency) is actually independent of the velocity. Now that we know the time, let’s calculate the pitch distance: p = (0.555 × 106)( 4.37 x 10-4) = 242 m 3
Question 22 A proton moving eastward with a velocity of 5.0 × 10 m/s enters a magnetic field of 0.20 T pointing northward. What is the magnitude and direction of the force that acts on the proton? 1.1 × 10-16 N eastwards 0N 1.6 × 10- N downwards 1.6 × 10-16 N upwards 4.4 × 10- N westwards
Explanation: The proton is moving eastwards and can be considered to be moving in the positive x direction ( i ). ˆ
In vector notation v = 5000 i m/s. ˆ
The magnetic field is pointing in the north direction and can be considered to be in the positive y direction ( j ). ˆ
In vector notation B = 0.2 j T. ˆ
F = q(v x B) F = 1.6x10-19 (5000
i
ˆ
X 0.2 j ) ˆ
F = 1.6x10-19 x 5000 x 0.2( i X j ) (cross product of ˆ
ˆ
i
ˆ
x j = k ) ˆ
ˆ
F = 1.6x10-16 k ( k represents the positive z direction or upwards) ˆ
ˆ
Alternately, we can determine the magnitude F = qvBsin90o (field and velocity vectors are perpendicular to each other) and figure out the direction of the force using the right hand rule. Fingers towards east, curl them towards the north, this would give a thumb direction pointing up. We have a positive charge so this is the direction of the magnetic force.
Question 23 A force F
=
(6 × 10-3 N ) i
uniform magnetic field with a velocity
(8 × 10-3 N) k is acting on a charge q = 2x10-9 C moving inside a
(4 × 10-3 N) j
-
ˆ
v
-
ˆ
ˆ
= (2x106 m/s) i + (3x106 m/s) j . Calculate the components of the magnetic field ˆ
ˆ
knowing that the x -component of the field is equal to zero. . =- (4T) + (0.5 T) = - (2 T) + (1 T) = (4 T) - (0.5 T) =- (3T) + (2T) = (2 T) - (1 T)
Explanation: F = q*(v x B) (4 10-3) j (8 10-3) k . All we need is to calculate the right side vector and compare the components from which we can get the magnetic fields components. F is known in a vector form F = (6 10-3) ×
i
ˆ
-
×
ˆ
-
ˆ
×
Consider the general magnetic field B = Bx i +By j +Bz k ˆ
ˆ
ˆ
but the x- component is zero so: B = By j +Bz k ˆ
ˆ
i
j
k
3 10 6
0
B y
B z
ˆ
ˆ
ˆ
q( v B ) = 2 10 -9 2 10 6
0
i 6 10 3 B z
ˆ
j 4
ˆ
10 3 B z
k 4
ˆ
10 3 B y
Now equate the two sides:
6
10
3
i 4 ˆ
10
3
j 8 ˆ
10
3
k i 6 ˆ
ˆ
10 3 B z
j 4
ˆ
10 3 B z
k 4
ˆ
10 3 B y
This equality leads to:
6 10 6 10 B or B 1 4 10 4 10 B same solution 8 10 4 10 B or B 2 3
3
3
3
z
z
3
z
3
y
y
Hence, magnetic field vector is (-2 j + 1 k ) Tesla. ˆ
ˆ
Question 24 A current is flowing in a wire in direction 3 i
ˆ
on the wire is 0.78 i
ˆ
0.58 j
-
ˆ
0.24 k
+
ˆ
4 j where the direction of the magnetic field is 5
+
ˆ
j + 12 k . The force
ˆ
ˆ
j . ˆ
0.58 j
0.78 i
+
ˆ
0.71( i
+
0.78 i
+
ˆ
ˆ
ˆ
j ).
ˆ
0.58 j
ˆ
0.24 k
+
ˆ
0.24 k . ˆ
-
Explanation: The direction of force on the wire can be obtained using the unit vectors of the current and magnetic field. F = I(L x B) Since, L is a scalar quantity, the direction of the force is determined by the current and magnetic fields (cross product using the right hand rule). direction of L = (3 i + 4 j )/ (3) 2 (4) 2 ˆ
ˆ
;
L = (3/5) i + (4/5) j ; ˆ
ˆ
direction of B = (5 j + 12 k )/ (5)2 (12) 2 ; B = (5/13) j + (12/13) k ; ˆ
ˆ
ˆ
ˆ
i
ˆ
Direction of force is given by directions of L x B = 0.6 0
j
k
0.8
0
0.38
0.92
ˆ
ˆ
L x B = (0.8*0.92) i –(0.6*0.92) j +(0.6*0.38) k ˆ
ˆ
ˆ
L x B = 0.78 i – 0.58 j + 0.24 k (approx) ˆ
ˆ
ˆ
6
Question 25 In a mass spectrometer, a single-charged particle (charge e) has a speed of 1.0 × 10 m/s and enters a uniform magnetic field of 0.20 T. The radius of the circular orbit is 0.020 m. What is the m ass of the particle? 1.7 × 10-27 kg 3.2 × 10- kg 4.5 × 10-27 kg 3.1 × 10- kg 6.4 × 10- kg
Explanation: Mass spectrometer is an application which uses the magnetic force on a moving particle. Remember the equation, the centripetal force balancing the magnetic force From that we derive m = qBr/v Substituting B = 0.2T, q=1.6 x 10 -19C, r= 0.02m, v = 10 6 m/s in the above equation, we get m = 6.4 x 10 -28 kg.
Question 26 An electron enters a region of space where there is a magnetic field B = 7 mT k with a velocity ˆ
m/s)
i
ˆ
( 9 × 106
v =
( 7 × 106 m/s) j + ( 5 × 106 m/s) k . What is the spacing of the circles of the helix on which the electron moves?
+
ˆ
7.1 cm 5.7 cm 1.5 cm 2.6 cm
ˆ
9.5 cm
Explanation: Please check problem #21 for important conclusions we made there: 1Pitch depends on the velocity component that is parallel to the magneti c field direction. Here, the magnetic
field is in the positive z direction ( B = 7 mT k ) so the parallel velocity component is ( 5 × 106 m/s) k . 2The time registered during travel is the period. 3The period is independent of the perpendicular velocity component(s). T = 2 m/qB So, the pitch of the spiral is Pitch (p) = (v // )(2 m/qB) p = (5 x 10 6 * 2 *9.1 x 10-31) / (1.6 x 10 -19 * 7 x 10-3) x = 2.6 cm ˆ
ˆ
Question 27 A rectangular loop of wire of width 10 cm and length 20 cm has a current of 2.5 A flowing through it. Two sides of the loop are oriented parallel to a uniform magnetic field of strength 0.037 T, the other two sides being perpendicular to the magnetic field. The magnitude of the torque on the loop is
0.038 m∙N. 0.050 m∙N. 0.0019 m∙N. 0.025 m∙N. 0.0093 m∙N. Explanation: Draw a sketch and convince yourself that the angle between vectors
and B is 90o.
Magnetic Torque is given by = Bsin where is the angle between magnetic moment and Magnetic field =IABsin Substituting the values I = 2.5A, A = 0.02, B = 0.037, = 90° = 0.0019 m.N
