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A small circular loop of wire of radius a is located at the Centre of a much larger circular wire loop of radius b. The two loops are in the same plane. The outer loop of radius b carries an alternating current ${I = I}_{0} cos (ω t)$ . The emf induced in the smaller inner loop is nearly

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$\frac{π μ_{0} I_{0}}{2} . \frac{a^{2}}{b} ω \sin (ω t)$

$π μ_{0} I_{0} \frac{a^{2}}{b} ω \sin (ω t)$

$\frac{π μ_{0} I_{0}}{2} . \frac{a^{2}}{b} ω \cos (ω t)$

$\frac{π μ_{0} I_{0} b^{2}}{a} ω \cos (ω t)$

answer is A.

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Detailed Solution

For two concentric circular coil,

Mutual inductance, $M = \frac{μ_{0} π N_{1} N_{2} a^{2}}{2 b}$

here, N₁ = N₂ = 1

Hence, $M = \frac{μ_{0} π a^{2}}{2 b}$ ……. (i)

and given $I = I_{0} \cos ω t$ ……. (ii)

Now according to Faraday’s second law induced emf

$e = - M \frac{d I}{d t}$

From equation (ii),

$e = \frac{- μ_{0} π a^{2}}{2 b} \frac{d}{d t} (I_{0} \cos ω t)$

$e = \frac{μ_{0} π a^{2}}{2 b} I_{0} ω \sin ω t$

$e = \frac{π μ_{0} I_{0}}{2} . \frac{a^{2}}{b} ω \sin ω t$

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