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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 $\frac{B_{0} a}{4 k λ}$ where $k =$ ____

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answer is B.

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Detailed Solution

E = \frac{d ϕ}{dt}

$ϕ = BA = B_{0} {ta}^{2} \to (1)$

$E = \frac{d ϕ}{dt} = B_{0} a^{2}$

Total resistance $R = λ 12 (a) \to (2)$

$i = \frac{E}{R} = \frac{B_{0} a^{2}}{λ 12 (a)} = \frac{B_{0} a}{12 λ}$

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