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A rectangular loop with a slide wire of length l is kept in a uniform magnetic field as shown in figure. The resistance of slider is R. neglecting self inductance of the loop find the current in the connector during its motion with a velocity v.

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$\frac{BIV}{R_{1} + R_{2} + R}$

$\frac{BIv (R_{1} + R_{2})}{R + (R_{1} + R_{2})}$

$\frac{BIv (R_{1} + R_{2})}{{RR}_{1} + {RR}_{2} + R_{1} R_{2}}$

$Blv (\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}})$

answer is C.

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Detailed Solution

Inductor behaves like a cell of emf $ε^{'} =$ Blv and equivalent ckt is
$BIV = ε^{'}$
Hence, i $\begin{array}{l} = \frac{BIv}{R + \frac{R_{1} R_{2}}{R_{1} + R_{2}}} \\ = \frac{BIv (R_{1} + R_{2})}{{RR}_{1} + {RR}_{2} + R_{1} R_{2}} \end{array}$

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