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QJEEM24Ph8242: JEE Main 2024 Physics - Class 12 JEEM24Ph8242 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 of side L(L=l2)L\left(L=l^2\right) . The loops are coplanar and their centers coincide. The value of the mutual inductance of the system is x×10−7H\sqrt{x} \times 10^{-7} \mathrm{H} , where x=x= _________.

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JEE Main 2024 (Online) 31st January Morning Shift Physics - Electromagnetic Induction Question 17 English Explanation

Flux linkage for inner loop.

amp;ϕ=Bcenter ⋅ℓ2amp;=4×μ0i4πL2(sin⁡45+sin⁡45)ℓ2amp;ϕ=22μ0iπLℓ2amp;M=ϕi=22μ0ℓ2πL=22μ0πamp;=224ππ×10−7amp;=82×10−7Hamp;=128×10−7Hamp;x=128\begin{aligned}& \phi=\mathrm{B}_{\text {center }} \cdot \ell^2 \\& =4 \times \frac{\mu_0 \mathrm{i}}{4 \pi \frac{\mathrm{L}}{2}}(\sin 45+\sin 45) \ell^2 \\& \phi=2 \sqrt{2} \frac{\mu_0 \mathrm{i}}{\pi \mathrm{L}} \ell^2 \\& \mathrm{M}=\frac{\phi}{\mathrm{i}}=\frac{2 \sqrt{2} \mu_0 \ell^2}{\pi \mathrm{L}}=2 \sqrt{2} \frac{\mu_0}{\pi} \\& =2 \sqrt{2} \frac{4 \pi}{\pi} \times 10^{-7} \\& =8 \sqrt{2} \times 10^{-7} \mathrm{H} \\& =\sqrt{128} \times 10^{-7} \mathrm{H} \\& \mathrm{x}=128\end{aligned}

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