physics : class12 Conducting Loop in Magnetic Field : Induced Current : Electromagnetic Induction and Alternating Currents MCQ Questions and Answers
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Conducting Loop in Magnetic Field
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MCQ Questions
1. A rectangular conducting loop of length 0.20 m moves with constant speed 3.0 m/s into a region of uniform magnetic field of magnitude 0.50 T. The magnetic field is perpendicular to the plane of the loop, and the total resistance of the loop is 0.40 ohm. While the loop is partly entering the field, what are the induced current and the magnetic force opposing its motion?
A.
0.75 A and 0.075 N
B.
0.75 A and 0.150 N
C.
1.50 A and 0.075 N
D.
1.50 A and 0.150 N
Correct Answer:
0.75 A and 0.075 N
Explanation:
The changing area inside the field gives an induced emf of Bℓv = 0.50 × 0.20 × 3.0 = 0.30 V. Thus I = emf/R = 0.30/0.40 = 0.75 A. The magnetic force on the side in the field is B I ℓ = 0.50 × 0.75 × 0.20 = 0.075 N, directed opposite to the motion by Lenz's law.
2. A conducting loop of area A and resistance R rotates with constant angular velocity omega in a uniform magnetic field B. At time t = 0, the plane of the loop is perpendicular to the field, and the loop rotates about an axis in its plane perpendicular to the field. Which expression correctly gives the induced emf for t greater than or equal to 0?
A.
BA omega cos(omega t)
B.
BA omega sin(omega t)
C.
-BA omega cos(omega t)
D.
-BA sin(omega t)
Correct Answer:
BA omega sin(omega t)
Explanation:
The magnetic flux is Phi = BA cos(omega t), since the loop's area normal initially aligns with the field. Faraday's law gives emf = -dPhi/dt = BA omega sin(omega t).
3. A square conducting loop is stationary in a magnetic field perpendicular to its plane. The field magnitude decreases uniformly from B0 to zero in time T. If the loop has side a and resistance R, what is the magnitude of the induced current?
A.
B0a/(RT)
B.
B0a^2/(RT)
C.
B0a^2T/R
D.
B0/(Ra^2T)
Correct Answer:
B0a^2/(RT)
Explanation:
The magnetic flux changes by B0a^2 in magnitude over time T, so the induced emf is B0a^2/T. Dividing by the loop resistance R gives the current B0a^2/(RT).
4. A conducting loop moves with constant velocity entirely within a sufficiently large region of uniform, time-independent magnetic field. The loop does not rotate, and its plane remains perpendicular to the field. What is the induced current in the loop?
A.
BA/R
B.
BAv/R
C.
Zero
D.
Bv/R
Correct Answer:
Zero
Explanation:
Although charges in different parts of the moving conductor experience magnetic forces, the total magnetic flux through the loop remains constant because its area and orientation in a uniform field do not change. Therefore the net induced emf and current are zero.
5. A square conducting loop of side 0.10 m and resistance 2.0 ohm enters a uniform magnetic field of 0.80 T with speed 5.0 m/s. The field is perpendicular to the loop and occupies a half-plane. During entry, what is the magnitude of the induced current?
A.
0.10 A
B.
0.20 A
C.
0.40 A
D.
2.00 A
Correct Answer:
0.20 A
Explanation:
During entry, the area inside the field changes at the rate av, where a is the side length. The emf is B av = 0.80 × 0.10 × 5.0 = 0.40 V. Hence the current is 0.40/2.0 = 0.20 A.
6. A conducting loop of resistance R and self-inductance L is placed in a region where a perpendicular magnetic field increases at a constant rate dB/dt. Neglecting mutual effects and assuming the loop area is A, which equation describes the induced current i(t)?
A.
L(di/dt) + Ri = A(dB/dt)
B.
L(di/dt) + Ri = -A(dB/dt)
C.
R(di/dt) + Li = -A(dB/dt)
D.
L(di/dt) - Ri = A(dB/dt)
Correct Answer:
L(di/dt) + Ri = -A(dB/dt)
Explanation:
An increasing external flux produces an emf of -A(dB/dt). Applying Kirchhoff's law to the loop gives the induced emf equal to L(di/dt) + Ri, yielding L(di/dt) + Ri = -A(dB/dt).
7. A circular conducting loop of radius r is placed in a nonuniform magnetic field perpendicular to its plane. The field varies radially as B(rho) = B0(1 - rho/R) for 0 less than or equal to rho less than or equal to R, where the loop radius is R. If B0 increases at a constant rate beta, what is the magnitude of the induced emf?
A.
pi R^2 beta
B.
pi R^2 beta/2
C.
pi R^2 beta/3
D.
2pi R^2 beta/3
Correct Answer:
pi R^2 beta/3
Explanation:
The flux is the integral of B(rho) over concentric area elements: Phi = integral from 0 to R of B0(1 - rho/R) 2pi rho d rho = pi B0 R^2/3. Since dB0/dt = beta, the emf magnitude is dPhi/dt = pi R^2 beta/3.
8. A conducting loop is pulled at constant speed through a localized region of uniform magnetic field. The loop has resistance R and is fully outside the field before entering and after leaving it. During which stage is the net work done by the external agent nonzero, and why?
A.
Only while the loop is fully inside, because its flux is maximum
B.
Only while entering and leaving, because the flux changes during those stages
C.
Throughout the motion, because the loop has a velocity
D.
Never, because magnetic forces cannot affect a conducting loop
Correct Answer:
Only while entering and leaving, because the flux changes during those stages
Explanation:
An induced current exists only when the magnetic flux through the loop changes. This occurs during entry and exit, producing magnetic forces that oppose the motion and requiring external work. When the loop is fully inside a uniform field, the flux is constant and no current or opposing magnetic force exists.
9. Two identical conducting loops are placed side by side with their planes parallel. A current in the first loop is increasing. The induced current in the second loop is observed to be clockwise when viewed from the same side as the first loop. From that viewing side, what is the direction of the increasing current in the first loop?
A.
Clockwise
B.
Anticlockwise
C.
It cannot be determined without the loop resistance
D.
It must alternate periodically
Correct Answer:
Clockwise
Explanation:
If the first-loop current is clockwise as viewed from that side, its magnetic field through the second loop points away from the viewer and is increasing. The induced current in the second loop must oppose this increase, producing a field toward the viewer, which requires an anticlockwise current. Therefore, observing an anticlockwise induced current means the increasing current in the first loop is clockwise.
10. A conducting loop of area A and resistance R rotates in a uniform magnetic field B with angular speed omega. The angle between the field and the normal to the loop is theta(t) = omega t. What is the average magnitude of the induced current over one-quarter of a rotation, from t = 0 to t = pi/(2 omega)?
A.
BA/(R)
B.
2BA/(pi R)
C.
BA omega/(R)
D.
pi BA/(2R)
Correct Answer:
2BA/(pi R)
Explanation:
The flux is BA cos(omega t), so the emf is BA omega sin(omega t). The average emf over the interval is [BA]/[pi/(2 omega)] = 2BA omega/pi, because the flux changes from BA to zero. Dividing by R gives the average current magnitude 2BA/(pi R).
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