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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.

what is the retarding force on the rod when k is closed ?

$(A) \; 7.5 \times 10^{-2} N$
$(B)\; 7.5 \times 10^2 N$
$(C)\; 75 \times 10^2 N$
$(D)\; 75 \times 10^{-2} N$

1 Answer

Given :
$l= 15 cm = 15 \times 10^{-2} m$
$B = 0.50 T$
$R = 9m \Omega = 9 \times 10^{-3} \Omega$
$v = 12 cm/s = 12 \times 10^{-2} m/s$
$\begin{align*} \text{The regarding force in the magnetic field is given by} F = BI l & = B . \frac{e}{R} . l \\ & = \frac{0.5 \times 9 \times 10^{-3} \times 15 \times 10^{-2}}{9 \times 10^{-3} } \\ & = 7.5 \times 10^{-2} N\end{align*}$
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