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 dissipated as heat in the closed circuit is (assume v= 2 m s⁻¹):
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $4.5 \times 10^{-3} W$
Induced emf in the rod
ε = B l v
ε = 0.50 × 0.15 × 2
ε = 0.15 V
Current in the circuit
I = ε / R
R = 180 mΩ = 0.18 Ω
I = 0.15 / 0.18 ≈ 0.83 A
Power dissipated
P = I²R
P = (0.83)² × 0.18
P ≈ 4.5 × 10⁻³ W
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