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:

$4.5 \times 10^{-3} W$

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

The power required by external agent is,

$P_{ext}=F_{ext}V$

and,

The induced current in the circuit,

$I=\frac{Blv}{R}=\frac{(0.5)(0.15)(0.12)}{0.180}$

$=0.05A$

$F_{mag}=BlI$

$=(0.5)(0.15)(0.05)$

$=0.00375N$

$∴P_{ext}=F_{mag}V$

$=(0.00375)(0.12)$   $[F_{ext}=F_{mag}]$

$=0.45mW$