Verified Solution • 2025–26 Session moderate

QJEEM22Ph28c8: JEE Main 2022 Physics - Class 12 JEEM22Ph28c8 Solution: A small square loop of wire of side is placed inside a large…

A small square loop of wire of side ll is placed inside a large square loop of wire \mathrm{L}(\mathrm{L}>>l) . Both loops are coplanar and their centres coincide at point O\mathrm{O} as shown in figure. The mutual inductance of the system is :

JEE Main 2022 (Online) 25th July Morning Shift Physics - Electromagnetic Induction Question 54 English

Multiple Choice Options

  • A 22μ0 L2πl\frac{2 \sqrt{2} \mu_{0} \mathrm{~L}^{2}}{\pi l}
  • B μ0l222πL\frac{\mu_{0} l^{2}}{2 \sqrt{2} \pi \mathrm{L}}
  • C 22μ0l2πL\frac{2 \sqrt{2} \mu_{0} l^{2}}{\pi \mathrm{L}}
  • D μ0 L222πl\frac{\mu_{0} \mathrm{~L}^{2}}{2 \sqrt{2} \pi l}

Step-by-Step Verified Solution

We know ϕ=Mi\phi = Mi

Let i current be flowing in the larger loop

⇒ϕ≃[4×μ0i4π(L/2)[sin⁡45∘+sin⁡45∘]]×\Rightarrow \phi \simeq \left[ {4 \times {{{\mu _0}i} \over {4\pi (L/2)}}[\sin 45^\circ + \sin 45^\circ ]} \right] \times Area

=22μ0iπL×I2= {{2\sqrt 2 {\mu _0}i} \over {\pi L}} \times {I^2}

⇒M=ϕi=22μ0I2πL\Rightarrow M = {\phi \over i} = {{2\sqrt 2 {\mu _0}{I^2}} \over {\pi L}}

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