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Two parallel long smooth conducting rails separated by a distance $l$ are connected by a movable conducting connector of mass $m$ . Terminals of the rails are connected by the resistor $R$ and the capacitor $C$ as shown in figure. A uniform magnetic field $B$ perpendicular to the plane of the rail is switched on. The conductor is dragged by a constant force $F$ . Find the speed of the connector as a function of time if the force $F$ is applied at $t = 0$ . Also, find the terminal velocity of the connector.

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$v = \frac{2 F R}{B^{2} l^{2}} [1 + e^{- (\frac{B^{2} l^{2}}{m R - R B^{2} l^{2} C}) t}]; v_{T} = \frac{9 F R}{B^{2} l^{2}}$

$v = \frac{F^{2} R}{9 B^{2} l^{2}} [1 - e^{- (\frac{B^{2} l^{2}}{m R + R B^{2} l^{2} C}) t}]; v_{T} = \frac{F R}{9 B^{2} l^{2}}$

$v = \frac{F R}{B^{2} l^{2}} [1 - e^{- (\frac{B^{2} l^{2}}{m R + R B^{2} l^{2} C}) t}]; v_{T} = \frac{F R}{B^{2} l^{2}}$

$v = \frac{4 F R}{B^{2} l^{2}} [1 - e^{- (\frac{B^{2} l^{2}}{m R + R B^{2} l^{2} C}) t}]; v_{T} = \frac{8 F R}{B^{2} l^{2}}$

answer is C.

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Detailed Solution

Let $v$ be the velocity of connector at some instant of time. Then,

$V_{a b} = B v l, i_{1} = \frac{B v l}{R}, q = C (B v l) ∴ i_{2} = \frac{d q}{d t} = C B l \frac{d v}{d t}$

Now, $i = i_{1} + i_{2} = \frac{B v l}{R} + C B l \frac{d v}{d t}$

Magnetic force, $F_{m} = i l B = \frac{B^{2} l^{2}}{R} . v + B^{2} l^{2} C . \frac{d v}{d t}$

Further, $F_{n e t} = F - F_{m}$

or $m \frac{d v}{d t} = F - \frac{B^{2} l^{2}}{R} v - B^{2} l^{2} C \frac{d v}{d t}$

$∴ \int_{0}^{v} \frac{d v}{F - \frac{B^{2} l^{2}}{R} v} = \int_{0}^{t} \frac{d t}{m + B^{2} l^{2} C}$

Integrating we get,

$v = \frac{F R}{B^{2} l^{2}} [1 - e^{- (\frac{B^{2} l^{2}}{m R + R B^{2} l^{2} C}) t}]$

Terminal velocity in this case is: $v_{T} = \frac{F R}{B^{2} l^{2}}$

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