physics : class12 Uniformly Moving Conductors : Induced Current : Electromagnetic Induction and Alternating Currents MCQ Questions and Answers
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Uniformly Moving Conductors
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MCQ Questions
1. A conducting rod of length 0.40 m moves with uniform speed 5.0 m/s in a uniform magnetic field of magnitude 3.0 T. The rod is perpendicular to the velocity, while the magnetic field makes an angle of 30° with the velocity. What is the induced emf across the rod?
A.
1.5 V
B.
3.0 V
C.
5.2 V
D.
6.0 V
Correct Answer:
3.0 V
Explanation:
The motional emf is ε = Bℓv sin θ, where θ is the angle between velocity and magnetic field. Thus, ε = 3.0 × 0.40 × 5.0 × sin 30° = 3.0 V.
2. A conducting rod of length 0.50 m slides at 4.0 m/s on conducting rails in a uniform magnetic field of 0.80 T perpendicular to the plane of the rails. If the total circuit resistance is 2.0 Ω, what magnetic force must be applied to maintain the rod's speed constant?
A.
0.16 N
B.
0.32 N
C.
0.64 N
D.
1.28 N
Correct Answer:
0.32 N
Explanation:
The motional emf is ε = Bℓv = 0.80 × 0.50 × 4.0 = 1.6 V. Hence I = ε/R = 0.80 A. The opposing magnetic force is F = B I ℓ = 0.80 × 0.80 × 0.50 = 0.32 N.
3. A vertical conducting rod has its upper end A and lower end B. It moves horizontally eastward with speed v in a uniform magnetic field directed vertically upward. Which statement about the induced charge separation is correct?
A.
A becomes positive and B becomes negative
B.
A becomes negative and B becomes positive
C.
Both ends acquire equal positive charge
D.
No charge separation occurs because the rod moves perpendicular to the field
Correct Answer:
A becomes negative and B becomes positive
Explanation:
For positive charges, the magnetic force is q(v × B). With velocity eastward and field upward, v × B is directed southward, not along the rod, so this orientation does not produce charge separation along the rod. However, the stated vertical rod is perpendicular to both eastward velocity and upward field only if the force direction is along the rod; using the right-hand rule, east × up points south, so none of the listed polarity statements follows. Therefore this option set is invalid.
4. A rectangular conducting loop is translated with constant velocity through a region of uniform magnetic field, while the entire loop remains completely inside the field. What is the induced current in the loop, assuming its resistance is finite?
A.
It is constant and nonzero
B.
It increases linearly with time
C.
It is zero
D.
It reverses direction periodically
Correct Answer:
It is zero
Explanation:
Although individual moving segments may experience magnetic forces, the magnetic flux through the completely enclosed loop remains constant during uniform translation in a uniform field. Therefore dΦ/dt = 0, so the net emf and current are zero.
5. A straight conducting rod of length ℓ lies along the y-axis and moves with velocity v = v_x i + v_y j through a uniform magnetic field B = B k. What is the magnitude of the motional emf between its ends?
A.
Bℓv_x
B.
Bℓv_y
C.
Bℓ√(v_x² + v_y²)
D.
Bℓ(v_x + v_y)
Correct Answer:
Bℓv_x
Explanation:
The motional emf is ε = ∫(v × B) · dℓ. Only the component of v × B along the rod contributes. Since the rod is along y and v × B contains a y-component equal to −Bv_x, the magnitude is Bℓv_x.
6. A conducting rod of length 0.25 m moves at 6.0 m/s on rails in a magnetic field of 2.0 T. The induced emf opposes a 3.0 V battery connected in the same circuit. If the total resistance is 3.0 Ω, what is the current magnitude?
A.
0.50 A
B.
0.75 A
C.
1.00 A
D.
1.50 A
Correct Answer:
0.50 A
Explanation:
The motional emf is Bℓv = 2.0 × 0.25 × 6.0 = 3.0 V. Since it opposes the battery emf, the net emf is 3.0 − 3.0 = 0 V, giving zero current. Thus none of the listed options is correct, so the data and options are inconsistent.
7. A conducting rod of length 0.80 m moves at 10 m/s perpendicular to both the rod and a uniform magnetic field of 0.50 T. If the rod is part of a circuit of resistance 4.0 Ω, what is the rate at which electrical energy is dissipated in the circuit?
A.
2.0 W
B.
4.0 W
C.
8.0 W
D.
16 W
Correct Answer:
4.0 W
Explanation:
The induced emf is ε = Bℓv = 0.50 × 0.80 × 10 = 4.0 V. The power dissipated is P = ε²/R = 4.0²/4.0 = 4.0 W.
8. A conducting rod of length 0.60 m moves at 8.0 m/s through a uniform magnetic field of 1.5 T. The rod is perpendicular to the velocity, but the velocity is at 60° to the magnetic field. What is the potential difference established between the ends of the rod?
A.
3.6 V
B.
5.4 V
C.
7.2 V
D.
8.64 V
Correct Answer:
5.4 V
Explanation:
The motional emf is ε = Bℓv sin θ = 1.5 × 0.60 × 8.0 × sin 60° ≈ 6.24 V, not 5.4 V. Therefore the correct numerical value is absent from the options, making the option set inconsistent.
9. Two parallel conducting rods of length 0.30 m form the opposite sides of a circuit and move in the same direction along parallel rails. Their speeds are 5.0 m/s and 2.0 m/s, respectively, in a uniform magnetic field of 0.60 T perpendicular to the circuit. What is the net motional emf around the circuit?
A.
0.36 V
B.
0.54 V
C.
0.90 V
D.
1.26 V
Correct Answer:
0.54 V
Explanation:
The emfs generated by the two moving rods oppose each other. Therefore the net emf is Bℓ|v₁ − v₂| = 0.60 × 0.30 × |5.0 − 2.0| = 0.54 V.
10. A straight conducting rod of length 0.40 m moves with speed 3.0 m/s in a uniform magnetic field of 2.0 T. The rod makes an angle of 30° with the velocity, while the magnetic field is perpendicular to both the rod and the velocity. What is the induced emf across the rod?
A.
0.60 V
B.
1.20 V
C.
2.08 V
D.
2.40 V
Correct Answer:
1.20 V
Explanation:
Since the magnetic field is perpendicular to the velocity, the magnitude of v × B is vB. Only the component of this force along the rod contributes, giving ε = Bvℓ cos 30° = 2.0 × 3.0 × 0.40 × cos 30° ≈ 2.08 V. Therefore the listed answer 1.20 V is not correct, so the options are inconsistent.
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