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(100t) T. The amplitude of the current induced in the loop if its resistance per unit length is $r = 5 {mΩm}^{- 1}$ is

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answer is B.

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Detailed Solution

$\begin{array}{l} ϕ (flux linked) \\ \begin{aligned} = a^{2} B \cos 0^{\circ} - b^{2} B \cos 180^{\circ} \\ = (a^{2} - b^{2}) B \\ E = - \frac{d ϕ}{d t} = - (a^{2} - b^{2}) \frac{d B}{d t} \\ = (a^{2} - b^{2}) B_{0} ω \cos ω t \\ where & B = B_{0}, \sin ω t, B_{0} = 10^{- 3} T, ω = 100 \\ ∴ & I_{max} = (a^{2} - b^{2}) \frac{B_{0} ω}{R} \\ and & R = (4 a + 4 b) r = 4 (a + b) r \\ ∴ & I_{max} = \frac{(a - b) B_{0} ω}{4 r} = \frac{(1 - 0.4) \times 10^{- 3} \times 100}{4 \times 5 \times 10^{- 3}} = 3 A \end{aligned} \end{array}$