A coil of inductance $L$ connects the upper end of two copper bars separated by a distance $l$ . A horizontal conducting connector of mass $m$ starts falling with zero initial velocity along the bars without losing contact. The whole system is located in uniform magnetic field $B$ perpendicular to the plane of bars.
During the motion

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Maximum displacement of connector is $\frac{m g L}{B^{2} l^{2}}$

Maximum displacement of connector is $\frac{m g L}{2 B^{2} l^{2}}$

Maximum velocity of connector is $\frac{g}{l B} \sqrt{m L}$

Maximum velocity of connector is $\frac{2 g}{l B} \sqrt{m L}$

answer is B.

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Detailed Solution

$m g - i l B = m \frac{d v}{d t} B v l = - \frac{L d i}{d t} i = \frac{B x l}{L} m g - \frac{B^{2} l^{2} x}{L} = m \frac{d^{2} x}{d t^{2}} m \frac{d^{2} x}{d t^{2}} + \frac{B^{2} l^{2} x}{L} = m g m \frac{d^{2} v}{d t^{2}} + \frac{B^{2} l^{2} v}{L} = B^{2} l^{2} x = - \frac{m g L}{B^{2} l^{2}} \cos (ω t) + \frac{m g L}{B^{2} l^{2}} ω = \frac{B l}{\sqrt{m L}}$