XPQY is a vertical smooth long loop having a total resistance R where PX is parallel to QY and separation between them is \(l\). A constant magnetic field B perpendicular to the plane of the loop exists in the entire space. A rod CD of length L (L > \(l\)) and mass m is made to slide down from rest under the gravity as shown in figure. The terminal speed acquired by the rod is __________ m/s.
(g = acceleration due to gravity)

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Answer & Solution

Correct Answer

(D) \( \frac{mgR}{B^{2}l^{2}} \)

Step-by-step Solution

Detailed explanation

at equilibrium (Or for terminal velocity)
\( mg=iBl\Rightarrow mg=(\frac{Bvl}{R})Bl \)
\( V=\frac{mgR}{B^{2}l^{2}} \)

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