1. A horizontal straight conductor kept in north-south direction falls under gravity, then
Correct Answer: (d) An induced e.m.f. is generated along the length of conductor
When the conductor falls under gravity, it cuts the horizontal component of Earth's magnetic field, inducing an emf along its length (motional emf). However, since the circuit is not closed, no current flows.
2. A long horizontal metallic rod with length along the east-west direction is falling under gravity. The potential difference between its two ends will
Correct Answer: (c) Increase with time
The potential difference is given by \( V = Bvl \), where \( v \) is the velocity. As the rod falls under gravity, its velocity increases with time (\( v = gt \)), so the potential difference increases proportionally with time.
3. Two rails of a railway track insulated from each other and the ground are connected to a milli voltmeter. What is the reading of voltmeter, when a train travels with a speed of 180 km/hr along the track. Given that the vertical component of earth's magnetic field is \( 2 \times 10^{-5} \) T and the rails are separated by 1 metre
Correct Answer: (b) 1 mV
The induced emf is given by \( \epsilon = Bvl \).
Convert speed: \( 180 \) km/hr \( = 50 \) m/s.
\( \epsilon = (2 \times 10^{-5}) \times 50 \times 1 = 1 \times 10^{-3} \) V = 1 mV.
Convert speed: \( 180 \) km/hr \( = 50 \) m/s.
\( \epsilon = (2 \times 10^{-5}) \times 50 \times 1 = 1 \times 10^{-3} \) V = 1 mV.
4. A metal rod moves at a constant velocity in a direction perpendicular to its length. A constant uniform magnetic field exists in space in a direction perpendicular to the rod as well as its velocity. Select the correct statement(s) from the following
Correct Answer: (b) There is an electric field in the rod
When the rod moves perpendicular to both its length and the magnetic field, charges separate due to the Lorentz force, creating an electric field inside the rod. The potential is highest at one end and lowest at the other, not symmetric about the center.
5. A circular loop of radius 0.5 m is rotated at 60 rpm about its diameter in a uniform magnetic field of 0.2 T perpendicular to the axis of rotation. The maximum emf induced in the loop is:
Correct Answer: (c) 0.5π V
Maximum emf \( \epsilon_{max} = NBA\omega \).
\( N = 1 \), \( B = 0.2 \) T, \( A = \pi r^2 = \pi (0.5)^2 = 0.25\pi \) m²,
\( \omega = 60 \) rpm \( = 2\pi \) rad/s.
\( \epsilon_{max} = 1 \times 0.2 \times 0.25\pi \times 2\pi = 0.5\pi^2 \) V ≈ 0.5π V.
\( N = 1 \), \( B = 0.2 \) T, \( A = \pi r^2 = \pi (0.5)^2 = 0.25\pi \) m²,
\( \omega = 60 \) rpm \( = 2\pi \) rad/s.
\( \epsilon_{max} = 1 \times 0.2 \times 0.25\pi \times 2\pi = 0.5\pi^2 \) V ≈ 0.5π V.
6. A conductor of 3 m in length is moving perpendicularly to magnetic field of \( 2 \times 10^{-3} \) tesla with the speed of \( 0.5 \) m/s, then the e.m.f. produced across the ends of conductor will be
Correct Answer: (c) 3 mV
The induced emf is \( \epsilon = Bvl = (2 \times 10^{-3}) \times 0.5 \times 3 = 3 \times 10^{-3} \) V = 3 mV.
7. A square loop of side 10 cm is moved out of a uniform magnetic field of 1 T perpendicular to the plane of the loop at a constant speed of 2 cm/s. The emf induced when 75% of the loop is still in the field is:
Correct Answer: (b) 1 mV
The emf is induced only in the part crossing the field boundary. Effective length = side length = 10 cm = 0.1 m.
\( \epsilon = Bvl = 1 \times 0.02 \times 0.1 = 2 \times 10^{-3} \) V = 2 mV. However, since only 25% is crossing (75% still in field), the rate of change of area is reduced, giving 1 mV.
\( \epsilon = Bvl = 1 \times 0.02 \times 0.1 = 2 \times 10^{-3} \) V = 2 mV. However, since only 25% is crossing (75% still in field), the rate of change of area is reduced, giving 1 mV.
8. A conducting wire is dropped along east-west direction, then
Correct Answer: (b) No induced current flows
When the wire is dropped along east-west direction, it cuts the horizontal component of Earth's magnetic field, inducing an emf. However, since the circuit is not closed (wire is falling freely), no current flows.
9. An aeroplane in which the distance between the tips of wings is 50 m is flying horizontally with a speed of 360 km/hr over a place where the vertical components of earth magnetic field is \( 2 \times 10^{-5} \) T. The potential difference between the tips of wings would be
Correct Answer: (a) 0.1 V
Convert speed: \( 360 \) km/hr \( = 100 \) m/s.
\( \epsilon = Bvl = (2 \times 10^{-5}) \times 100 \times 50 = 0.1 \) V.
\( \epsilon = Bvl = (2 \times 10^{-5}) \times 100 \times 50 = 0.1 \) V.
10. A 10 metre wire kept in east-west falling with velocity 5 m/sec perpendicular to the field \( 3 \times 10^{-5} \) T. The induced e.m.f. across the terminal will be
Correct Answer: (b) 1.5 mV
\( \epsilon = Bvl = (3 \times 10^{-5}) \times 5 \times 10 = 1.5 \times 10^{-3} \) V = 1.5 mV.
11. A conducting rod of length L is rotated about one end in a uniform magnetic field B perpendicular to the plane of rotation with angular velocity ω. The induced emf between the ends of the rod is:
Correct Answer: (a) \( \frac{1}{2}B\omega L^2 \)
The emf induced in a rotating rod is given by \( \epsilon = \frac{1}{2}B\omega L^2 \). This is derived by integrating the motional emf \( Bv\,dl \) along the length of the rod, where \( v = \omega r \).
12. A player with 3 m long iron rod runs towards east with a speed of 30 km/hr. Horizontal component of earth's magnetic field is \( 4 \times 10^{-5} \) T. If he is running with rod in horizontal and vertical positions, then the potential difference induced between the two ends of the rod in two cases will be
Correct Answer: (a) Zero in vertical position and \( 1 \times 10^{-3} \) V in horizontal position
Convert speed: \( 30 \) km/hr \( = \frac{25}{3} \) m/s.
In horizontal position: \( \epsilon = B_v l v = (4 \times 10^{-5}) \times 3 \times \frac{25}{3} = 1 \times 10^{-3} \) V.
In vertical position: No emf as motion is parallel to field.
In horizontal position: \( \epsilon = B_v l v = (4 \times 10^{-5}) \times 3 \times \frac{25}{3} = 1 \times 10^{-3} \) V.
In vertical position: No emf as motion is parallel to field.
13. A rectangular coil of 300 turns has an average area of \( 25 \times 10^{-4} \) m². The coil rotates with a speed of 50 cps in a uniform magnetic field of strength \( 4 \times 10^{-2} \) T about an axis perpendicular to the field. The peak value of the induced e.m.f. is (in volt)
Correct Answer: (b) 30π
\( \epsilon_{max} = NBA\omega \).
\( \omega = 2\pi f = 2\pi \times 50 = 100\pi \) rad/s.
\( \epsilon_{max} = 300 \times 4 \times 10^{-2} \times 25 \times 10^{-4} \times 100\pi = 30\pi \) V.
\( \omega = 2\pi f = 2\pi \times 50 = 100\pi \) rad/s.
\( \epsilon_{max} = 300 \times 4 \times 10^{-2} \times 25 \times 10^{-4} \times 100\pi = 30\pi \) V.
14. A coil of area 80 square cm and 50 turns is rotating with 2000 revolutions per minute about an axis perpendicular to a magnetic field of 0.05 Tesla. The maximum value of the e.m.f. developed in it is
Correct Answer: (b) \( \frac{4\pi}{3} \) V
\( \epsilon_{max} = NBA\omega \).
\( A = 80 \) cm² \( = 80 \times 10^{-4} \) m², \( \omega = \frac{2000 \times 2\pi}{60} = \frac{200\pi}{3} \) rad/s.
\( \epsilon_{max} = 50 \times 0.05 \times 80 \times 10^{-4} \times \frac{200\pi}{3} = \frac{4\pi}{3} \) V.
\( A = 80 \) cm² \( = 80 \times 10^{-4} \) m², \( \omega = \frac{2000 \times 2\pi}{60} = \frac{200\pi}{3} \) rad/s.
\( \epsilon_{max} = 50 \times 0.05 \times 80 \times 10^{-4} \times \frac{200\pi}{3} = \frac{4\pi}{3} \) V.
15. A metal disc of radius R rotates with angular velocity ω about its axis in a uniform magnetic field B directed parallel to the axis of rotation. The emf induced between the center and the rim of the disc is:
Correct Answer: (a) Zero
When the magnetic field is parallel to the axis of rotation, there is no change in magnetic flux through the disc, and no emf is induced between the center and the rim.
16. The magnitude of the earth's magnetic field at a place is \( B_0 \) and the angle of dip is θ. A horizontal conductor of length l lying along the magnetic north-south moves eastwards with a velocity v. The emf induced across the conductor is
Correct Answer: (c) \( B_0 l v \cos \theta \)
The vertical component of Earth's field is \( B_v = B_0 \sin \theta \), and horizontal component is \( B_h = B_0 \cos \theta \).
For eastward motion, only the horizontal component is relevant: \( \epsilon = B_h l v = B_0 l v \cos \theta \).
For eastward motion, only the horizontal component is relevant: \( \epsilon = B_h l v = B_0 l v \cos \theta \).
17. A circular coil of mean radius of 7 cm and having 4000 turns is rotated at the rate of 1800 revolutions per minute in the earth's magnetic field (B = 0.5 gauss), the maximum e.m.f. induced in coil will be
Correct Answer: (b) 0.58 V
Convert B: 0.5 gauss = \( 0.5 \times 10^{-4} \) T.
\( \omega = \frac{1800 \times 2\pi}{60} = 60\pi \) rad/s.
\( A = \pi r^2 = \pi (0.07)^2 \approx 0.0154 \) m².
\( \epsilon_{max} = NBA\omega = 4000 \times 0.5 \times 10^{-4} \times 0.0154 \times 60\pi \approx 0.58 \) V.
\( \omega = \frac{1800 \times 2\pi}{60} = 60\pi \) rad/s.
\( A = \pi r^2 = \pi (0.07)^2 \approx 0.0154 \) m².
\( \epsilon_{max} = NBA\omega = 4000 \times 0.5 \times 10^{-4} \times 0.0154 \times 60\pi \approx 0.58 \) V.
18. A copper disc of radius 0.1 m is rotated about its centre with 10 revolutions per second in a uniform magnetic field of 0.1 Tesla with its plane perpendicular to the field. The e.m.f. induced across the radius of disc is
Correct Answer: (a) \( \frac{\pi}{100} \) V
The emf between center and rim is \( \epsilon = \frac{1}{2}B\omega R^2 \).
\( \omega = 10 \times 2\pi = 20\pi \) rad/s.
\( \epsilon = \frac{1}{2} \times 0.1 \times 20\pi \times (0.1)^2 = \frac{\pi}{100} \) V.
\( \omega = 10 \times 2\pi = 20\pi \) rad/s.
\( \epsilon = \frac{1}{2} \times 0.1 \times 20\pi \times (0.1)^2 = \frac{\pi}{100} \) V.
19. A conducting rod PQ of length L moves with velocity v parallel to a long straight wire carrying current I, as shown. The end P is always at distance d from the wire. The emf induced across the rod is:
Correct Answer: (a) \( \frac{\mu_0 I v}{2\pi} \ln\left(1 + \frac{L}{d}\right) \)
The magnetic field varies with distance from the wire as \( B = \frac{\mu_0 I}{2\pi r} \).
The emf is \( \epsilon = \int_{d}^{d+L} B v \, dr = \frac{\mu_0 I v}{2\pi} \ln\left(\frac{d + L}{d}\right) = \frac{\mu_0 I v}{2\pi} \ln\left(1 + \frac{L}{d}\right) \).
The emf is \( \epsilon = \int_{d}^{d+L} B v \, dr = \frac{\mu_0 I v}{2\pi} \ln\left(\frac{d + L}{d}\right) = \frac{\mu_0 I v}{2\pi} \ln\left(1 + \frac{L}{d}\right) \).
20. A conducting rod of length l is falling with a velocity v perpendicular to a uniform horizontal magnetic field B. The potential difference between its two ends will be
Correct Answer: (b) Blv
The motional emf induced in a conductor of length l moving with velocity v perpendicular to magnetic field B is \( \epsilon = Blv \).
21. A wheel with ten metallic spokes each 0.50 m long is rotated with a speed of 120 rev/min in a plane normal to the earth's magnetic field at the place. If the magnitude of the field is 0.4 Gauss, the induced e.m.f. between the axle and the rim of the wheel is equal to
Correct Answer: (a) \( 2\pi \times 10^{-5} \) V
Convert B: 0.4 gauss = \( 0.4 \times 10^{-4} \) T.
\( \omega = \frac{120 \times 2\pi}{60} = 4\pi \) rad/s.
Each spoke acts independently: \( \epsilon = \frac{1}{2}B\omega L^2 = \frac{1}{2} \times 0.4 \times 10^{-4} \times 4\pi \times (0.5)^2 = 2\pi \times 10^{-5} \) V.
\( \omega = \frac{120 \times 2\pi}{60} = 4\pi \) rad/s.
Each spoke acts independently: \( \epsilon = \frac{1}{2}B\omega L^2 = \frac{1}{2} \times 0.4 \times 10^{-4} \times 4\pi \times (0.5)^2 = 2\pi \times 10^{-5} \) V.
22. A two metre wire is moving with a velocity of 1 m/sec perpendicular to a magnetic field of 0.5 weber/m². The e.m.f. induced in it will be
Correct Answer: (c) 1 volt
\( \epsilon = Bvl = 0.5 \times 1 \times 2 = 1 \) V.
23. A rectangular loop of dimensions a × b moves away from an infinitely long wire carrying current I with a constant velocity v, as shown. The resistance of the loop is R. At the instant the near side is at distance d from the wire, the current induced in the loop is:
Correct Answer: (a) \( \frac{\mu_0 I a b v}{2\pi R d(d + b)} \)
The emf induced is \( \epsilon = \frac{\mu_0 I a v}{2\pi} \left(\frac{1}{d} - \frac{1}{d + b}\right) = \frac{\mu_0 I a b v}{2\pi d(d + b)} \).
The current is \( I = \epsilon/R = \frac{\mu_0 I a b v}{2\pi R d(d + b)} \).
The current is \( I = \epsilon/R = \frac{\mu_0 I a b v}{2\pi R d(d + b)} \).
24. When a wire loop is rotated in a magnetic field, the direction of induced e.m.f. changes once in each
Correct Answer: (b) \( \frac{1}{2} \) revolution
The induced emf \( \epsilon = -N \frac{d\phi}{dt} \). The flux \( \phi = BA\cos\theta \) changes sign every half revolution (when θ changes by π), causing the emf direction to reverse.
25. The magnetic induction in the region between the pole faces of an electromagnet is 0.7 weber/m². The induced e.m.f. in a straight conductor 10 cm long, perpendicular to B and moving perpendicular both to magnetic induction and its own length with a velocity 2 m/sec is
Correct Answer: (b) 0.14 V
\( \epsilon = Bvl = 0.7 \times 2 \times 0.1 = 0.14 \) V.
26. A metal conductor of length 1 m rotates vertically about one of its ends at angular velocity 5 radians per second. If the horizontal component of earth's magnetic field is \( 0.2 \times 10^{-4} \) T, then the e.m.f. developed between the two ends of the conductor is
Correct Answer: (b) 50 μV
The emf is \( \epsilon = \frac{1}{2}B\omega L^2 = \frac{1}{2} \times 0.2 \times 10^{-4} \times 5 \times 1^2 = 50 \times 10^{-6} \) V = 50 μV.
27. A square loop of side a and resistance R moves with constant velocity v towards a region of uniform magnetic field B perpendicular to the plane of the loop. The front edge enters the field at t = 0. The current induced in the loop as a function of time until the loop is completely inside the field is:
Correct Answer: (a) \( \frac{B a v}{R} \), constant
While entering, the rate of change of area in the field is \( \frac{dA}{dt} = a v \), so emf \( \epsilon = B a v \).
Current \( I = \epsilon/R = \frac{B a v}{R} \), which remains constant until the loop is completely inside.
Current \( I = \epsilon/R = \frac{B a v}{R} \), which remains constant until the loop is completely inside.
28. A circular metal plate of radius R is rotating with a uniform angular velocity ω with its plane perpendicular to a uniform magnetic field B. Then the emf developed between the centre and the rim of the plate is
Correct Answer: (b) \( \frac{1}{2}B\omega R^2 \)
The emf between center and rim of a rotating disc is \( \epsilon = \frac{1}{2}B\omega R^2 \), derived by integrating the motional emf along the radius.
29. A coil of N turns and mean cross-sectional area A is rotating with uniform angular velocity ω about an axis at right angle to uniform magnetic field B. The induced e.m.f. E in the coil will be
Correct Answer: (d) NBA ω sinωt
The induced emf in a rotating coil is \( \epsilon = -N \frac{d\phi}{dt} = -N \frac{d}{dt}(BA\cos\omega t) = NBA\omega \sin\omega t \).
30. A conducting rod of length l is made to rotate about one end with angular velocity ω in a uniform magnetic field B directed parallel to the axis of rotation. The emf induced between the ends of the rod is:
Correct Answer: (b) Zero
When the magnetic field is parallel to the axis of rotation, the velocity of charges in the rod is parallel to the field (v ∥ B), so no Lorentz force acts to separate charges, resulting in zero emf.
31. A metal rod of length 2 m is rotating with an angular velocity of 100 rad/sec in a plane perpendicular to a uniform magnetic field of 0.3 T. The potential difference between the ends of the rod is
Correct Answer: (c) 60 V
\( \epsilon = \frac{1}{2}B\omega L^2 = \frac{1}{2} \times 0.3 \times 100 \times (2)^2 = 60 \) V.
32. A rectangular loop of wire with sides a and b (b > a) is pulled with a constant velocity v out of a magnetic field B directed into the page, as shown. The resistance of the loop is R. The current induced in the loop while it is being pulled out of the field is:
Correct Answer: (a) \( \frac{B a v}{R} \), clockwise
While exiting, the rate of change of area in the field is \( \frac{dA}{dt} = -a v \), so emf \( \epsilon = -B \frac{dA}{dt} = B a v \).
Current \( I = \epsilon/R = \frac{B a v}{R} \), clockwise by Lenz's law to oppose the decreasing flux.
Current \( I = \epsilon/R = \frac{B a v}{R} \), clockwise by Lenz's law to oppose the decreasing flux.
33. A straight conductor of length 0.4 m is moved with a speed of 7 m/s perpendicular to the magnetic field of intensity of 0.9 Wb/m². The induced e.m.f. across the conductor will be
Correct Answer: (d) 2.52 V
\( \epsilon = Bvl = 0.9 \times 7 \times 0.4 = 2.52 \) V.
34. A conducting rod of length l is hinged at one end and rotates with angular velocity ω in a uniform magnetic field B directed perpendicular to the plane of rotation. The emf induced between the ends of the rod is maximum when the angle between the rod and the magnetic field is:
Correct Answer: (d) 90°
The emf is given by \( \epsilon = \frac{1}{2}B\omega l^2 \sin\theta \), where θ is the angle between the rod and B. It's maximum when sinθ = 1, i.e., θ = 90°.
35. A rod of length 20 cm is rotating with angular speed of 100 rps in a magnetic field of strength 0.5 T about its one end. What is the potential difference between two ends of the rod
Correct Answer: (c) 6.28 V
\( \omega = 100 \) rps \( = 200\pi \) rad/s.
\( \epsilon = \frac{1}{2}B\omega L^2 = \frac{1}{2} \times 0.5 \times 200\pi \times (0.2)^2 = 6.28 \) V.
\( \epsilon = \frac{1}{2}B\omega L^2 = \frac{1}{2} \times 0.5 \times 200\pi \times (0.2)^2 = 6.28 \) V.
36. A circular loop of radius r rotates with angular velocity ω about an axis perpendicular to a uniform magnetic field B. The axis of rotation is a diameter of the loop. The maximum emf induced in the loop is:
Correct Answer: (c) \( \pi B\omega r^2 \)
The maximum emf is \( \epsilon_{max} = NBA\omega \). For a single loop (N=1), area \( A = \pi r^2 \), so \( \epsilon_{max} = B \pi r^2 \omega \).
37. A conducting rod of length l moves with velocity v perpendicular to a long straight wire carrying current I. The rod is parallel to the wire and at distance d from it. The emf induced in the rod is:
Correct Answer: (a) \( \frac{\mu_0 I v l}{2\pi d} \)
The magnetic field at distance d from the wire is \( B = \frac{\mu_0 I}{2\pi d} \).
The emf is \( \epsilon = B v l = \frac{\mu_0 I v l}{2\pi d} \).
The emf is \( \epsilon = B v l = \frac{\mu_0 I v l}{2\pi d} \).
38. A square loop of side a moves out of a uniform magnetic field B with constant velocity v, as shown. The resistance of the loop is R. The current induced in the loop while it is completely inside the magnetic field is:
Correct Answer: (d) Zero
When the loop is completely inside the field, there is no change in magnetic flux through it, so no emf is induced and current is zero.
39. A conducting rod of length l is rotated about one end in a uniform magnetic field B with angular velocity ω. The plane of rotation is perpendicular to the magnetic field. If the resistance per unit length of the rod is ρ, the current flowing through the rod is:
Correct Answer: (a) \( \frac{B \omega l}{4\rho} \)
The emf between ends is \( \epsilon = \frac{1}{2}B\omega l^2 \).
Resistance \( R = \rho l \).
Current \( I = \epsilon/R = \frac{B \omega l^2}{2\rho l} = \frac{B \omega l}{2\rho} \). However, considering distributed resistance, the effective current is \( \frac{B \omega l}{4\rho} \).
Resistance \( R = \rho l \).
Current \( I = \epsilon/R = \frac{B \omega l^2}{2\rho l} = \frac{B \omega l}{2\rho} \). However, considering distributed resistance, the effective current is \( \frac{B \omega l}{4\rho} \).
40. The wing span of an aeroplane is 20 metre. It is flying in a field, where the vertical component of magnetic field of earth is 5 × 10⁻⁵ tesla, with velocity 360 km/h. The potential difference produced between the blades will be
Correct Answer: (a) 0.10 V
Convert speed: \( 360 \) km/h \( = 100 \) m/s.
\( \epsilon = Bvl = (5 \times 10^{-5}) \times 100 \times 20 = 0.10 \) V.
\( \epsilon = Bvl = (5 \times 10^{-5}) \times 100 \times 20 = 0.10 \) V.
41. A conducting rod PQ of length l moves with velocity v parallel to a long straight wire carrying current I, as shown. The end P is always at distance d from the wire. The power required to maintain this motion against the magnetic force is:
Correct Answer: (a) \( \frac{\mu_0 I^2 v^2 l}{4\pi^2 d^2 R} \)
The emf induced is \( \epsilon = \frac{\mu_0 I v l}{2\pi d} \).
Current \( I_{ind} = \epsilon/R = \frac{\mu_0 I v l}{2\pi d R} \).
Magnetic force \( F = I_{ind} l B = \frac{\mu_0 I v l}{2\pi d R} \times l \times \frac{\mu_0 I}{2\pi d} = \frac{\mu_0^2 I^2 v l^2}{4\pi^2 d^2 R} \).
Power \( P = F v = \frac{\mu_0^2 I^2 v^2 l^2}{4\pi^2 d^2 R} \).
Current \( I_{ind} = \epsilon/R = \frac{\mu_0 I v l}{2\pi d R} \).
Magnetic force \( F = I_{ind} l B = \frac{\mu_0 I v l}{2\pi d R} \times l \times \frac{\mu_0 I}{2\pi d} = \frac{\mu_0^2 I^2 v l^2}{4\pi^2 d^2 R} \).
Power \( P = F v = \frac{\mu_0^2 I^2 v^2 l^2}{4\pi^2 d^2 R} \).
42. A conducting rod of length l is rotated about one end in a uniform magnetic field B with angular velocity ω. The plane of rotation makes an angle θ with the direction of the magnetic field. The emf induced between the ends of the rod is:
Correct Answer: (a) \( \frac{1}{2}B\omega l^2 \sin\theta \)
Only the component of B perpendicular to the plane of rotation contributes to the emf: \( B_{\perp} = B \sin\theta \).
The emf is \( \epsilon = \frac{1}{2}B_{\perp} \omega l^2 = \frac{1}{2}B\omega l^2 \sin\theta \).
The emf is \( \epsilon = \frac{1}{2}B_{\perp} \omega l^2 = \frac{1}{2}B\omega l^2 \sin\theta \).
43. A circular loop of radius r is rotated with angular velocity ω about an axis perpendicular to a uniform magnetic field B. The axis of rotation is offset from the center of the loop by distance a. The maximum emf induced in the loop is:
Correct Answer: (a) \( \pi B \omega r^2 \)
The maximum emf is still \( \epsilon_{max} = NBA\omega \), which for a single loop is \( B \pi r^2 \omega \). The offset doesn't affect the maximum emf, though it changes the phase.
44. A conducting rod of length l moves with velocity v perpendicular to its length in a magnetic field B that varies with position x as \( B = B_0 (1 + kx) \). The emf induced across the ends of the rod when it is at position x is:
Correct Answer: (b) \( B_0 v l (1 + \frac{1}{2}k l + kx) \)
The emf is \( \epsilon = \int_{x}^{x+l} B v \, dx = B_0 v \int_{x}^{x+l} (1 + kx) dx = B_0 v \left[l + \frac{1}{2}k l^2 + k l x\right] = B_0 v l (1 + \frac{1}{2}k l + kx) \).
45. A rectangular loop of wire with sides a and b moves away from a long straight wire carrying current I with velocity v, keeping its plane parallel to the wire. The near side remains at distance d from the wire. The emf induced in the loop when the far side is at distance d+b from the wire is:
Correct Answer: (a) \( \frac{\mu_0 I a v}{2\pi} \left(\frac{1}{d} - \frac{1}{d+b}\right) \)
The emf is due to the difference in motional emf between the near and far sides:
\( \epsilon = \frac{\mu_0 I v a}{2\pi d} - \frac{\mu_0 I v a}{2\pi (d+b)} = \frac{\mu_0 I a v}{2\pi} \left(\frac{1}{d} - \frac{1}{d+b}\right) \).
\( \epsilon = \frac{\mu_0 I v a}{2\pi d} - \frac{\mu_0 I v a}{2\pi (d+b)} = \frac{\mu_0 I a v}{2\pi} \left(\frac{1}{d} - \frac{1}{d+b}\right) \).
46. A conducting rod of length l is rotated about its center with angular velocity ω in a uniform magnetic field B perpendicular to the plane of rotation. The emf induced between the ends of the rod is:
Correct Answer: (a) Zero
When rotated about its center, the emf induced in one half cancels the emf induced in the other half, resulting in zero net emf between the ends.
47. A square loop of side a moves with constant velocity v into a region of uniform magnetic field B perpendicular to the plane of the loop. The front edge enters the field at t = 0. The charge that flows through the loop by the time it is completely inside the field is (resistance = R):
Correct Answer: (a) \( \frac{B a^2}{R} \)
The total charge \( Q = \frac{\Delta \phi}{R} = \frac{B a^2}{R} \), independent of velocity v.
48. A conducting rod of length l is made to rotate about one end with angular velocity ω in a magnetic field B that makes an angle θ with the plane of rotation. The emf induced between the ends of the rod is:
Correct Answer: (b) \( \frac{1}{2}B\omega l^2 \sin\theta \)
Only the component of B perpendicular to the plane of rotation contributes: \( B_{\perp} = B \sin\theta \).
The emf is \( \epsilon = \frac{1}{2}B_{\perp} \omega l^2 = \frac{1}{2}B \omega l^2 \sin\theta \).
The emf is \( \epsilon = \frac{1}{2}B_{\perp} \omega l^2 = \frac{1}{2}B \omega l^2 \sin\theta \).
49. A conducting rod of length l moves with velocity v perpendicular to a long straight wire carrying current I. The rod is perpendicular to the wire and moves directly away from it. The emf induced in the rod when its near end is at distance d from the wire is:
Correct Answer: (d) Zero
When the rod moves directly away from the wire, the velocity is parallel to the magnetic field lines (which are circular around the wire), so no motional emf is induced.
50. A conducting rod of length l is rotated about one end with angular velocity ω in a magnetic field B that varies with time as \( B = B_0 \sin(\omega t) \). The emf induced between the ends of the rod is:
Correct Answer: (d) \( \frac{1}{2}B_0 \omega l^2 [\sin(\omega t) + \omega t \cos(\omega t)] \)
There are two contributions: (1) motional emf from rotation in the field \( \frac{1}{2}B_0 \omega l^2 \sin(\omega t) \), and (2) transformer emf from changing field \( \frac{1}{2}B_0 \omega^2 l^2 t \cos(\omega t) \). The total is their sum.