Electromagnetic Induction – Motional EMF and Eddy Current: NEET MCQs with Answers

15 free questions on Electromagnetic Induction – Motional EMF and Eddy Current for NEET (UG). Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Motional EMF and Eddy Current: Eddy currents.

Question 1

A straight rod of length l moves without rotating, at speed v, in a plane perpendicular to a uniform magnetic field B. In case (a) the velocity makes an angle θ with the rod. In case (b) the velocity makes an angle θ with the normal to the rod, drawn in the same plane. In case (c) the velocity is directed along the rod. What are the emfs induced across the rod in cases (a), (b) and (c), in that order?
  1. Bvl sin θ, Bvl cos θ and zero
  2. Bvl, Bvl and zero
  3. Bvl sin θ, Bvl cos θ and Bvl
  4. Bvl cos θ, Bvl sin θ and zero
Show answer
Answer: (A) Bvl sin θ, Bvl cos θ and zero
Explanation: Only the velocity component perpendicular to the rod produces an emf, so ε = Blv⊥. This component is v sin θ in case (a) and v cos θ in case (b). In case (c) the rod moves along its own length, cuts no field lines, and the emf is zero.

Question 2

A rod of length l slides at a constant speed v on two smooth parallel conducting rails joined at one end through a resistance R, in a uniform magnetic field B perpendicular to the plane of the rails. All other resistance is negligible. If different steady speeds are tried, how do the induced emf ε, the applied force F and the mechanical power P supplied depend on v?
  1. ε ∝ v², F ∝ v, P ∝ v²
  2. ε ∝ v, F ∝ v², P ∝ v³
  3. ε ∝ v, F ∝ v, P ∝ v²
  4. ε ∝ v, F is constant, P ∝ v
Show answer
Answer: (C) ε ∝ v, F ∝ v, P ∝ v²
Explanation: The motional emf is ε = Blv, which is proportional to v. The current is Blv/R, so the magnetic force on the rod is B²l²v/R, and at constant speed the applied force must equal it. The power supplied is P = Fv = B²l²v²/R, which varies as v².

Question 3

A metal rod 25 cm long lies across two parallel conducting rails in a uniform 0.40 T magnetic field perpendicular to the plane of the rails. An external agent keeps the rod moving along the rails at a steady 10 cm/s, and the closed circuit has a total resistance of 5.0 mΩ. At what rate is heat produced in the circuit?
  1. 0.010 W
  2. 0.005 W
  3. 0.020 W
  4. 0.040 W
Show answer
Answer: (C) 0.020 W
Explanation: The emf is Blv = 0.40 × 0.25 × 0.10 = 0.010 V, so the current is 0.010/0.0050 = 2 A. The heating rate is I²R = 4 × 0.0050 = 0.020 W. This equals the mechanical power supplied by the external agent, as energy conservation requires.

Question 4

A metal rod 0.5 m long lies north-south and slides east at a steady 4 m/s through a uniform 0.25 T magnetic field directed vertically downward. The rod is not part of any circuit. Once the free electrons have stopped drifting along the rod, which description is correct?
  1. There is an electric field of 1 V/m in the rod, and its north end is positive
  2. There is an electric field of 0.5 V/m in the rod, and its north end is positive
  3. There is an electric field of 1 V/m in the rod, and its south end is positive
  4. There is no electric field in the rod, because no current flows
Show answer
Answer: (A) There is an electric field of 1 V/m in the rod, and its north end is positive
Explanation: The magnetic force on a moving positive charge, along v × B, points north, so the free electrons are driven towards the south end. Charge builds up until the electric force balances the magnetic force, qE = qvB. This gives E = vB = 4 × 0.25 = 1 V/m, and the north end is left positive. The potential difference across the ends is El = 0.5 V, equal to the motional emf Blv.

Question 5

A straight metal rod 40 cm long lies along the y-axis and moves at 5 m/s along +x in a uniform magnetic field of 0.2 T along +z. The rod is not part of a closed circuit. Once charges have stopped shifting along the rod, what is the electric field inside it, and which end is positive?
  1. 1 V/m along +y; the −y end is positive
  2. Zero, since a conductor has no electric field inside it; neither end is positive
  3. 1 V/m along −y; the +y end is positive
  4. 0.4 V/m along +y; the −y end is positive
Show answer
Answer: (A) 1 V/m along +y; the −y end is positive
Explanation: The magnetic force on a positive charge is along v × B, which is along −y, so free electrons are pushed towards the +y end and that end becomes negative. The field set up by these charges points from the positive −y end to the negative +y end. Charges stop shifting when eE = evB, so E = vB = 5 × 0.2 = 1 V/m. The potential difference between the ends is then 1 × 0.4 = 0.4 V.

Question 6

A square wire loop of side 0.4 m and resistance 2 Ω lies wholly inside a uniform magnetic field of 0.5 T, with its plane perpendicular to the field. It is pulled out at a steady speed through the straight edge of the field, with one side of the loop parallel to that edge. What is the work done by the pulling force if the loop is taken out completely in (i) 0.1 s and (ii) 0.4 s?
  1. (i) 64 mJ, (ii) 16 mJ
  2. (i) 32 mJ, (ii) 8 mJ
  3. (i) 32 mJ, (ii) 2 mJ
  4. (i) 32 mJ, (ii) 32 mJ
Show answer
Answer: (B) (i) 32 mJ, (ii) 8 mJ
Explanation: At speed v the emf is Blv, the current is Blv/R and the magnetic force to be balanced is B²l²v/R. The pull acts over a distance l, and v = l/t, so W = B²l⁴/(Rt). This gives 0.0064/(2 × 0.1) = 0.032 J for 0.1 s and 0.0064/(2 × 0.4) = 0.008 J for 0.4 s. Slower removal needs less work because less energy is dissipated as heat.

Question 7

A conducting ring of radius 5 cm translates in its own plane at 2 m/s, without rotating, in a uniform 0.4 T magnetic field perpendicular to that plane. A is the foremost point of the ring in the direction of motion, and C is a point on the ring 90° around from A. What are the magnitude of the potential difference between A and C and the net emf around the whole ring?
  1. 0.057 V and zero
  2. 0.04 V and zero
  3. 0.04 V and 0.08 V
  4. 0.08 V and zero
Show answer
Answer: (B) 0.04 V and zero
Explanation: Only the separation of the two points perpendicular to the velocity counts. A and C are separated by one radius in that direction, so the potential difference is Bvr = 0.4 × 2 × 0.05 = 0.04 V. The flux through the ring stays constant during pure translation in a uniform field, so the net emf around it is zero.
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Question 8

A straight rod slides on two smooth parallel conducting rails 0.4 m apart, which are joined at one end through a 0.2 Ω resistor. All other resistance is negligible, and a 0.5 T uniform magnetic field is perpendicular to the plane of the rails. What mechanical power must an external agent supply to keep the rod moving at a steady 10 m/s?
  1. 20 W
  2. 5 W
  3. 40 W
  4. 10 W
Show answer
Answer: (A) 20 W
Explanation: The induced emf is Blv = 0.5 × 0.4 × 10 = 2 V, so the current is 2/0.2 = 10 A. The magnetic braking force is BIl = 0.5 × 10 × 0.4 = 2 N. At steady speed the external force equals this, so the power is Fv = 2 × 10 = 20 W, which matches the heating I²R = 20 W. Emf and force grow in proportion to v, while power grows as v².

Question 9

Two parallel conducting rails lie along the x-direction in the xy-plane, and a rod of length l lying along y slides on them at speed v along +x. A uniform magnetic field of magnitude B is applied in turn along (a) the x-axis, (b) the y-axis and (c) the z-axis. The emf induced in the rod in the three cases, in that order, is
  1. 0, Blv, 0
  2. Blv, 0, 0
  3. 0, Blv, Blv
  4. 0, 0, Blv
Show answer
Answer: (D) 0, 0, Blv
Explanation: An emf appears in the rod only if the field has a component perpendicular to both the velocity and the rod, so that the rod cuts field lines. With the field along x it is parallel to the velocity, and with the field along y it is parallel to the rod, so the emf is zero in both cases. With the field along z the rod, the velocity and the field are mutually perpendicular and the emf is Blv.

Question 10

A metal rod moves on frictionless parallel rails in a uniform magnetic field perpendicular to the plane of the rails. The circuit joining the rails contains a switch, which is kept open. Statement I: An emf exists across the rod while it moves. Statement II: No external force is needed to keep the rod moving at constant velocity.
  1. Both statements are incorrect
  2. Statement I is correct but Statement II is incorrect
  3. Both statements are correct
  4. Statement I is incorrect but Statement II is correct
Show answer
Answer: (C) Both statements are correct
Explanation: The motional emf Blv depends only on the rod's motion through the field, so it exists even in an open circuit. With the switch open no steady current flows, so there is no magnetic force BIl on the rod. With no friction either, the rod needs no external force to continue at constant velocity.

Question 11

A metal rod of length 20 cm slides without friction on two parallel conducting rails, staying perpendicular to them, in a uniform magnetic field of 0.40 T perpendicular to the plane of the rails. The closed circuit has a total resistance of 5.0 mΩ. What mechanical power must an external agent supply to keep the rod moving at a steady 25 cm/s?
  1. 0.16 W
  2. 0.32 W
  3. 0.08 W
  4. 0.04 W
Show answer
Answer: (C) 0.08 W
Explanation: The emf is Blv = 0.40 × 0.20 × 0.25 = 0.020 V, so the current is 0.020/0.005 = 4 A. The magnetic force on the rod is BIl = 0.40 × 4 × 0.20 = 0.32 N. The external power is Fv = 0.32 × 0.25 = 0.08 W, which equals the heat I²R produced in the circuit.

Question 12

A metal rod of length 0.5 m and resistance 2 Ω rotates at 100 rad/s about one end, at the centre of a circular metal ring of radius 0.5 m, with its free end sliding on the ring. A uniform field of 0.4 T acts perpendicular to the plane of the ring. The centre and the ring are joined by a wire of negligible resistance, and the ring also has negligible resistance. The power dissipated is:
  1. 2.5 W
  2. 12.5 W
  3. 50 W
  4. 25 W
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Answer: (B) 12.5 W
Explanation: The emf across a rod rotating about one end is ε = ½Bωl² = ½ × 0.4 × 100 × 0.25 = 5 V. The only resistance in the circuit is that of the rod. The power is ε²/R = 25/2 = 12.5 W.

Question 13

A square conducting loop of side 20 cm lies in the xy-plane with its sides parallel to the axes. It moves at a constant 5 cm/s along +x through a magnetic field that points along +z everywhere. The field does not depend on time or on y, but it increases steadily by 2 × 10⁻³ T per cm as one moves along −x. The magnitude of the emf induced in the loop is:
  1. 4 × 10⁻² V
  2. 8 × 10⁻⁵ V
  3. 2 × 10⁻³ V
  4. 4 × 10⁻⁴ V
Show answer
Answer: (D) 4 × 10⁻⁴ V
Explanation: The field changes by 0.2 T per metre along x, so at a speed of 0.05 m/s the field at any point of the moving loop changes at 0.2 × 0.05 = 0.01 T/s. The rate of change of flux is this value times the area, 0.01 × 0.04 = 4 × 10⁻⁴ V. The same result follows from the difference of Blv across the two sides perpendicular to the motion.

Question 14

A rod of length 0.5 m slides at a constant 5 m/s along two parallel frictionless conducting rails joined at one end. The plane of the rails is perpendicular to a uniform field of 0.4 T, and the total circuit resistance is 2 Ω, which stays constant. What external force keeps the speed constant, and at what rate does this force do work?
  1. 0.1 N and 0.25 W
  2. 0.1 N and 0.5 W
  3. 0.5 N and 2.5 W
  4. 0.2 N and 1.0 W
Show answer
Answer: (B) 0.1 N and 0.5 W
Explanation: The emf is Blv = 0.4 × 0.5 × 5 = 1 V, so the current is 1/2 = 0.5 A. The magnetic force on the rod is BIl = 0.4 × 0.5 × 0.5 = 0.1 N, and the external force must balance it. Its mechanical power is Fv = 0.1 × 5 = 0.5 W. This equals the Joule heating I²R = 0.25 × 2 = 0.5 W, as energy conservation requires.

Question 15

A straight metal rod 1.2 m long rotates in a horizontal plane at 3 revolutions per second about a vertical axis through its midpoint. A uniform vertical magnetic field of 5 × 10⁻⁵ T is present. What are the magnitudes of the emf between the midpoint and one end, and of the emf between the two ends? (Take π = 3.14)
  1. 6.8 × 10⁻⁴ V and zero
  2. Zero and 1.7 × 10⁻⁴ V
  3. 1.7 × 10⁻⁴ V and 3.4 × 10⁻⁴ V
  4. 1.7 × 10⁻⁴ V and zero
Show answer
Answer: (D) 1.7 × 10⁻⁴ V and zero
Explanation: Each half of the rod has length r = 0.6 m and ω = 2π × 3 = 18.84 rad/s. The emf of a half is ½Bωr² = 0.5 × 5 × 10⁻⁵ × 18.84 × 0.36 ≈ 1.7 × 10⁻⁴ V. Both ends are at the same potential relative to the midpoint because both halves push charge the same way, outward or inward. The emf between the two ends is therefore zero.

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