JEE Main 2026 · 21 January · morning · Single correct · +4 / −1
A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T . If the resistance of the total circuit is 2Ω then the force needed to move the rod towards right with constant speed (v) of 1.5 m/s is ____ N.
- (a) 5.7× 10⁻²
- (b) 7.5× 10⁻³
- (c) 5.7× 10⁻³
- (d) 7.5× 10⁻²
Answer and solution
Answer (b) 7.5× 10⁻³
Solution
Physics sir · 4 steps
Constant speed means your push exactly cancels the magnetic drag on the rod, so find that drag.
- Motional EMF. The rod cuts the field lines as it moves, so an EMF ε = Blv = 0.10 × 1 × 1.5 = 0.15 V appears across AB.
- Current in circuit. This EMF drives a current through the 2 Ω circuit: I = ε/R = 0.15/2 = 0.075 A.
- Magnetic drag. The current-carrying rod sits in the field, so it feels a force BIl. By Lenz's law this force points left and opposes the motion. At constant speed the applied force must be equal and opposite to it.
F = BIl = (B² l² v)/R = ((0.10)² × 1² × 1.5)/2
- So. F = 0.015/2 = 7.5 × 10⁻³ N, which is option (b).
The idea to keep. For a rod sliding on rails, the drag force is (B² l² v)/R. B appears twice: once to make the EMF and once to push on the current.
Answer from the NTA final answer key. Solution by PrepWiser.