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Three identical square loops are made of a single wire as shown. The magnetic field depending on time as $B = B_{0} t$ is present in the region as shown. The side of the square is a and resistance per unit length of the wire is $λ$ . The induced current in the loop is $\frac{B_{0} a}{4 kλ}$ where k =________

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answer is 3.

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Detailed Solution

$E = \frac{dϕ}{dt}; ϕ = BA = B_{0} {ta}^{2} \dots . . (1) E = \frac{dϕ}{dt} = B_{0} a^{2}$ Total resistance
$\begin{array}{l} R = λ . 12 (a) \dots \dots (2) \\ i = \frac{E}{R} = \frac{B_{0} a^{2}}{λ 12 a} = \frac{B_{0} a}{4 (3) λ} \end{array}$

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