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:
Answer & explanation
Correct answer: option 3
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