Two long rails are horizontal and parallel to each other. On one end, the rails are connected by resistance R and on the other end a capacitor of capacitance C is connected as shown in the figure. A connector of mass m and length l can slide on the rails without friction. Uniform magnetic field B, perpendicular to the plane of the figure exist in space. A constant horizontal force F starts acting on the connector. What is the acceleration of connector when its velocity is $\frac{3}{4}$ of its terminal velocity?

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$\frac{3 F}{4 (m + C B^{2} l^{2})}$

$\frac{F}{(m + C B^{2} l^{2})}$

$\frac{F}{4 (m + C B^{2} l^{2})}$

$\frac{4 F}{3 (m + C B^{2} l^{2})}$

answer is C.

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Detailed Solution

Net inertia $= m + B^{2} l^{2} C$
$∴ F - \frac{B^{2} l^{2} V}{R} = (m + B^{2} l^{2} C^{2}) a$
At terminal velocity: $\frac{B^{2} l^{2} V_{0}}{R} = F$
When $V = \frac{3}{4} V_{0} \Rightarrow F - \frac{B^{2} l^{2} V_{0}}{R^{4}} = (m + B^{2} l^{2} C^{2}) a$
$\frac{F}{4} = (m + B^{2} l^{2} C^{2}) a \Rightarrow a = \frac{F}{4 (m + B^{2} l^{2} C^{2})}$

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