JEE Physics 2026 Question with Solution

Given: rod length $L = 1 \, \text{m}$, magnetic field $B = 0.10 \, \text{T}$, total resistance $R = 2 \, \Omega$, speed $v = 1.5 \, \text{m/s}$.

Find: the external force required to move the rod with constant speed.

A conducting rod moving in a magnetic field produces motional emf, which drives current in the closed circuit. The current-carrying rod then experiences a magnetic force opposing the motion. For constant velocity, the applied force must balance this magnetic force.

Use:

$$\mathcal{E} = BLv$$ $$I = \frac{\mathcal{E}}{R}$$ $$F_m = ILB$$

Substitute the values:

$$\mathcal{E} = (0.10)(1)(1.5) = 0.15 \, \text{V}$$ $$I = \frac{0.15}{2} = 0.075 \, \text{A}$$ $$F_m = (0.075)(1)(0.10) = 0.0075 \, \text{N}$$

For constant speed,

$$F_{\text{app}} = F_m = 0.0075 \, \text{N} = 7.5 \times 10^{-3} \, \text{N}$$

Therefore, the force needed is $7.5 \times 10^{-3} \, \text{N}$, so the correct option is B.

An alternative approach uses energy conservation:

$$P_{\text{mech}} = F_{\text{app}} v$$ $$P_{\text{elec}} = I^2R = \frac{\mathcal{E}^2}{R} = \frac{(BLv)^2}{R}$$

Equating mechanical and electrical power,

$$F_{\text{app}} v = \frac{B^2L^2v^2}{R}$$ $$F_{\text{app}} = \frac{B^2L^2v}{R}$$ $$F_{\text{app}} = \frac{(0.10)^2(1)^2(1.5)}{2} = 7.5 \times 10^{-3} \, \text{N}$$