JEE MAIN - Physics (2017 - 8th April Morning Slot - No. 6)
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 = Io cos ($$\omega $$t). The emf induced in the smaller inner loop is nearly :
$${{\pi {\mu _o}{I_o}} \over 2}.{{{a^2}} \over b}\,\omega \sin \left( {\omega t} \right)$$
$${{\pi {\mu _o}{I_o}} \over 2}.{{{a^2}} \over b}\,\omega \cos \left( {\omega t} \right)$$
$$\pi {\mu _o}{I_o}\,{{{a^2}} \over b}\omega \sin \left( {\omega t} \right)$$
$${{\pi {\mu _o}{I_o}\,{b^2}} \over a}\omega \cos \left( {\omega t} \right)$$
Explanation
Mutual inductance,
M = $${{{\mu _0}\pi {N_1}{N_2}\,{a^2}} \over {2b}}$$
here $${{N_1}}$$ = N2 = 1
$$\therefore\,\,\,$$ M = $${{{\mu _0}\pi {a^2}} \over {2b}}$$
Current I = I0 cos ($$\omega $$t)
According to Faraday's law,
e = $$-$$ M $${{dI} \over {dt}}$$
= $$-$$ $${{{\mu _0}\pi {a^2}} \over {2b}}$$ $${d \over {dt}}$$ (I0 cos $$\omega $$t)
= + $${{{\mu _0}\pi {a^2}} \over {2b}}$$ I0 $$\omega $$ sin $$\omega $$t
= $${{\pi {\mu _0}{I_0}} \over 2}$$ . $${{{a^2}} \over b}$$ $$\omega $$ sin $$\omega $$ t
M = $${{{\mu _0}\pi {N_1}{N_2}\,{a^2}} \over {2b}}$$
here $${{N_1}}$$ = N2 = 1
$$\therefore\,\,\,$$ M = $${{{\mu _0}\pi {a^2}} \over {2b}}$$
Current I = I0 cos ($$\omega $$t)
According to Faraday's law,
e = $$-$$ M $${{dI} \over {dt}}$$
= $$-$$ $${{{\mu _0}\pi {a^2}} \over {2b}}$$ $${d \over {dt}}$$ (I0 cos $$\omega $$t)
= + $${{{\mu _0}\pi {a^2}} \over {2b}}$$ I0 $$\omega $$ sin $$\omega $$t
= $${{\pi {\mu _0}{I_0}} \over 2}$$ . $${{{a^2}} \over b}$$ $$\omega $$ sin $$\omega $$ t
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