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A rectangular loop with a sliding conductor of length $l$ is located in a uniform magnetic field perpendicular to the plane of loop. The magnetic induction perpendicular to the plane of loop. The magnetic induction perpendicular to the plane equal to $B$ . The part $a d$ and $b c$ has electric resistance $R_{1}$ and $R_{2}$ , respectively. The conductor starts moving with constant acceleration $a_{0}$ at time $t = 0$ . Neglecting the self-inductance of the loop and resistance of conductor. Find
external force required to move the conductor with the given acceleration.

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$F = \frac{B^{2} l^{2} a_{0} t}{R_{1} R_{2}} (R_{1} + R_{2}) + m a_{0}$

$F = m a_{0}$

$F = \frac{B^{2} l^{2}}{R_{1} R_{2}} + m a_{0}$

$F = \frac{B L a_{0}}{R_{1}}$

answer is B.

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Detailed Solution

At time $t$ , $v = a_{0} t$

Motional emf, $V = B v l = B a_{0} l t$

Total resistance $= \frac{R_{1} R_{2}}{R_{1} + R_{2}}$

$∴ i = \frac{(B a_{0} l t) (R_{1} + R_{2})}{R_{1} R_{2}}$

From right hand rule, we can see that points $a$ and $b$ will be at higher potential and $c$ and $d$ at lower potentials.

$F_{m} = i l B = \frac{B^{2} l^{2} a_{0} t}{R_{1} R_{2}} (R_{1} + R_{2})$

Let $F$ be the external force applied, then,

$F - F_{m} = m a_{0}$

$∴ F = F_{m} + m a_{0} = \frac{B^{2} l^{2} a_{0} t}{R_{1} R_{2}} (R_{1} + R_{2}) + m a_{0}$

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