A plane loop, shaped as two squares of sides a = 1 m and b = 0.4 m is introduced into a uniform magnetic field $⊥$ to the plane of loop (figure). The magnetic field varies as $B = 10^{- 3} \sin (100 t) T .$ The amplitude of the current induced in the loop if its resistance per unit length is $r = 5 m Ω m^{- 1}$ is

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answer is B.

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Detailed Solution

$ϕ (f l u x l i n k e d) = a^{2} B \cos 0^{0} - b^{2} B \cos 180^{0}$

$= (a^{2} - b^{2}) B$

$E = - \frac{d ϕ}{d t} = - (a^{2} - b^{2}) \frac{d B}{d t}$

$= (a^{2} - b^{2}) B_{0} ω \cos ω t$

$W h e r e B = B_{0}, \sin ω t, B_{0} = 10^{- 3} T, ω = 100$

$∴ I_{m a x} = (a^{2} - b^{2}) \frac{B_{0} ω}{R}$

$a n d R = (4 a + 4 b) r = 4 (a + b) r$

$∴ I_{m a x} = \frac{(a - b) B_{0} ω}{4 r} = \frac{(1 - 0.4) \times 10^{- 3} \times 100}{4 \times 5 \times 10^{- 3}} = 3 A$

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