Figure shows a metal rod PQ resting on the rails AB and positioned between the poles of a permanent magnet. The rails, the rod and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = 180.0 $m\Omega$. Assume the field to be uniform.

The power required (by an external agent) to keep the rod moving at the same speed (= 12 $cms^{-1}$) when K is closed will:

$4.5 \times 10^{-4} W$

The correct answer is Option (3) → $4.5 \times 10^{-4} W$

The rod moves with velocity

v = 12 cm s⁻¹ = 0.12 m s⁻¹

Induced emf

ε = B l v

ε = 0.50 × 0.15 × 0.12

ε = 0.009 V

Current in the circuit

R = 180 mΩ = 0.18 Ω

I = ε / R

I = 0.009 / 0.18

I = 0.05 A

Power dissipated

P = I²R

P = (0.05)² × 0.18

P = 0.0025 × 0.18

P = 4.5 × 10⁻⁴ W