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Figure shows a metal rod PQ resting on the smooth rail AB and positioned between the poles of a permanent magnet. The rail, 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 = $9.0 m \Omega$. Assume the field to be uniform.

How much power is required at the same speed, to keep the rod moving at the same speed. (=12 cm/s) when k is closed ?

$(a)\; 90 \times 10^{-3} W$
$(b)\; 0.9 \times 10^{-3} W$
$(c)\; 900 \times 10^{-3} W$
$(d)\; 9 \times 10^{-3} W$

1 Answer

$\begin{align*}\text{To keep the rod moving at the same speed the required power } &= Retarding \;Force \times velocity \\ & = 7.5 \times 10^{-2} \times 12 \times 10^{-2} \\ & =9 \times 10^{-3} w \end{align*}$
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