ElectrostaticsMedium
Question
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)
Options
A.$\frac{2mgR}{B^{2}l^{2}}$
B.$\frac{8mgR}{B^{2}l^{2}}$
C.$\frac{2mgR}{B^{2}{\text{ }L}^{2}}$
D.$\frac{mgR}{B^{2}l^{2}}$
Solution
at equilibrium (Or for terminal velocity)
$${mg = iB\mathcal{l} \Rightarrow mg = \left( \frac{Bv\mathcal{l}}{R} \right)B\mathcal{l} }{V = \frac{mgR}{B^{2}\mathcal{l}^{2}}}$$
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