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The loop ABCD is moving with velocity ‘ $ϑ$ ’ towards right. The magnetic field is $4 T$ . The loop is connected to a resistor of $8 Ω$ . A 2A induced steady current flows in the loop. Then the value of ‘ $ϑ$ ’ if loop has resistance of $4 Ω$ , is (Given $A B = 30 c m$ , $A D = 30 c m;$ $\sin 37 ° = \frac{3}{5}$ )

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$\frac{50}{3} m s^{- 1}$

$20 m s^{- 1}$

$10 m s^{- 1}$

$\frac{100}{3} m s^{- 1}$

answer is D.

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Detailed Solution

$i = \frac{V}{R} \Rightarrow V = 2 \times 12 = 24 V$

$L = A D \sin 37 ° = 0.18$

$ϑ = \frac{24}{B L} = \frac{24}{4 \times 0.18}$

$= \frac{100}{3} m / s$

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