A 1 m long metal rod XY completes the circuit as shown in figure. The plane of the circuit is perpendicular to the magnetic field of flux density 0.15 T. If the resistance of the circuit is 5Ω, the force needed to move the rod in direction, as indicated, with a constant speed of 4 m/s will be ____________ 10⁻³ N.

Solution

Physics sir · 4 steps

At constant speed, your push must exactly cancel the magnetic drag on the current in the rod.

1. Induced EMF. The rod cuts field lines as it moves, so an EMF ε = Bvl = 0.15 × 4 × 1 = 0.6 V appears across XY.
2. Current in circuit. This EMF drives a current through the 5 Ω circuit: I = ε/R = 0.6/5 = 0.12 A.
3. Magnetic drag. The current-carrying rod in the field feels a force BIl opposing its motion (Lenz's law). At constant speed the applied force must equal this drag.

   F = BIl = (B² l² v)/R = ((0.15)² × 1² × 4)/5 = 0.018 N

4. So. F = 0.018 N = 18 × 10⁻³ N, so the answer is 18.

The common slip: 120. Stopping at the current I = Bvl/R = 0.12 A and reporting it gives 120, not 18. Multiply by Bl once more to get the force.

The idea to keep. For a rod sliding on rails, the force to keep constant speed is (B² l² v)/R. Constant speed means zero net force, so the applied force equals the magnetic drag.

Answer from the NTA final answer key. Solution by PrepWiser.