JEE MAIN - Physics (2021 - 31st August Morning Shift - No. 8)
A small square loop of side 'a' and one turn is placed inside a larger square loop of side b and one turn (b >> a). The two loops are coplanar with their centres coinciding. If a current I is passed in the square loop of side 'b', then the coefficient of mutual inductance between the two loops is :
$${{{\mu _0}} \over {4\pi }}8\sqrt 2 {{{a^2}} \over b}$$
$${{{\mu _0}} \over {4\pi }}{{8\sqrt 2 } \over a}$$
$${{{\mu _0}} \over {4\pi }}8\sqrt 2 {{{b^2}} \over a}$$
$${{{\mu _0}} \over {4\pi }}{{8\sqrt 2 } \over b}$$
Explanation
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$$B = \left[ {{{{\mu _0}} \over {4\pi }}{I \over {b/2}} \times 2\sin 45} \right] \times 4$$
$$\phi = 2\sqrt 2 {{{\mu _0}} \over \pi }{I \over b} \times {a^2}$$
$$\therefore$$ $$M = {\phi \over I} = {{2\sqrt 2 {\mu _0}{a^2}} \over {\pi b}} = {{{\mu _0}} \over {4\pi }}8\sqrt 2 {{{a^2}} \over b}$$
Option (a)
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