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Shows a square conducting loop of side 8cm being taken out of a uniform magnetic field of $B = 1 .5 T$ which is directed perpendicular and into the plane of the loop. The loop is connected to a circuit as shown in the figure. Resistance of loop is $2 Ω$ and that of the wires which connect the loop with the network of resistances is negligible. Speed of the loop such that current through it has a steady value $2 mA,$ is

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18 cm/s

12 cm/s

10 cm/s

8 cm/s

answer is C.

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Detailed Solution

Resistance of network is $\frac{1}{R_{P}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} = \frac{1}{12} + \frac{1}{12} + \frac{1}{12} = \frac{1}{4}$
$R_{P} = 4 Ω$

$∴$ Total resistance, $R = 2 + 4 = 6 Ω$

$∵ i = \frac{e}{R} = \frac{B l v}{R}$
$v = \frac{iR}{Bl} = \frac{2 \times 10^{- 3} \times 6}{1 .5 \times 8 \times 10^{- 2}} = 0 .1 m / s = 10 cm / s$

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