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QJEEM21Ph0711: JEE Main 2021 Physics - Class 12 JEEM21Ph0711 Solution: A small square loop of side 'a' and one turn is placed inside a …

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 :

Multiple Choice Options

  • A μ04π82a2b{{{\mu _0}} \over {4\pi }}8\sqrt 2 {{{a^2}} \over b}
  • B μ04π82a{{{\mu _0}} \over {4\pi }}{{8\sqrt 2 } \over a}
  • C μ04π82b2a{{{\mu _0}} \over {4\pi }}8\sqrt 2 {{{b^2}} \over a}
  • D μ04π82b{{{\mu _0}} \over {4\pi }}{{8\sqrt 2 } \over b}

Step-by-Step Verified Solution

JEE Main 2021 (Online) 31st August Morning Shift Physics - Electromagnetic Induction Question 65 English Explanation
B=[μ04πIb/2×2sin⁡45]×4B = \left[ {{{{\mu _0}} \over {4\pi }}{I \over {b/2}} \times 2\sin 45} \right] \times 4

ϕ=22μ0πIb×a2\phi = 2\sqrt 2 {{{\mu _0}} \over \pi }{I \over b} \times {a^2}

∴\therefore M=ϕI=22μ0a2πb=μ04π82a2bM = {\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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