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Two metal bars are fixed vertically and are connected on the top by a capacitor $C$ . A sliding conductor of length $l$ and mass $m$ slides with its ends in contact with the bars. The arrangement is placed in a uniform horizontal magnetic field directed normal to the plane of the figure. The conductor is released from rest. Find the displacement $x (t)$ of the conductor as a function of time $t$ .

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$x = \frac{9 m g t^{2}}{2 (m + C B^{2} l^{2})}$

$x = \frac{5 m g t^{2}}{2 (m + C B^{2} l^{2})}$

$x = \frac{9 m g t^{2}}{2 (m - C B^{2} l^{2})}$

$x = \frac{m g t^{2}}{2 (m + C B^{2} l^{2})}$

answer is A.

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Detailed Solution

Let $v$ be the velocity at some instant. Then, motional emf, $V = B v l$

Charge stored in capacitor $q = C V = (C B l) v$

current in the wire $= \frac{d q}{d t} = (C B l) \frac{d v}{d t}$

Magnetic force, $F_{m} = i l B = (C B^{2} l^{2}) \frac{d v}{d t}$ (upwards)

$∴$ Net force, $F_{n e t} = m g - F_{m}$

or $m \frac{d v}{d t} = m g - (C B^{2} l^{2}) \frac{d v}{d t}$

$∴ \frac{d v}{d t} =$ acceleration, $a = \frac{m g}{m + C B^{2} l^{2}}$

Since, $a =$ constant

$∴ x = \frac{1}{2} a t^{2} = \frac{m g t^{2}}{2 (m + C B^{2} l^{2})}$

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