Current I varies according to the relation $i = 3.36 (1 + 2 t) \times 10^{- 2}$ A. A small circular loop of radius $10^{- 3} m,$ with its plane parallel to the wire, is placed at a distance of 1m from the wire. The magnitude of the current in the coil is (resistance of the coil is $8 .4 \times 10^{- 4} Ω$ )

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$16 π \times 10^{- 12} A$

$8 π \times 10^{- 12} A$

$4 π \times 10^{- 12} A$

$12 π \times 10^{- 12} A$

answer is A.

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Detailed Solution

Flux induced in the coil = BA
$\begin{array}{l} ϕ = \frac{μ_{0}}{2 π} \frac{i}{d} {πr}^{2} [d = 1 m] e = \frac{dϕ}{dt} = [\frac{μ_{0} r^{2}}{2 d}] \frac{di}{dt} \\ = 2 π \times 10^{- 7} \times 10^{- 6} \times 3 .36 \times 2 \times 10^{- 2} [∵ \frac{di}{dt} = 3 .36 \times 2 \times 10^{- 2}] \\ = 13 .44 \times 10^{- 15} \times πV ∴ i = \frac{e}{R} = \frac{13 .44 \times 10^{- 15} \times π}{8 .4 \times 10^{- 4}} = 16 π \times 10^{- 12} A \end{array}$