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JEE Mains · Physics · STD 12 -Electromagnetic induction
  • JEE Mains 2024
  • Hard
  • Numerical

A small square loop of wire of side \(\ell\) is placed inside a large square loop of wire of side \(L\) \(\left(\mathrm{L}=\ell^2\right)\). The loops are coplanar and therr centers coinside. The value of the mutual inductance of the system is \(\sqrt{\mathrm{x}} \times 10^{-7} \mathrm{H}\), where X = _______.

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Answer & Solution

Correct Answer

(C) \(128\)

Step-by-step Solution

Detailed explanation

Flux linkage for inner loop.

\(\phi=\mathrm{B}_{\text {crater }} \ell^2\)

\(=4 \times \frac{\mu_0 \mathrm{i}}{4 \pi \frac{\mathrm{L}}{2}}(\sin 45+\sin 45) \ell^2\)

\(\phi=2 \sqrt{2} \frac{\mu_0 \mathrm{i}}{\pi \mathrm{L}} \ell^2\)

\(\mathrm{M}=\frac{\phi}{\mathrm{i}}=\frac{2 \sqrt{2} \mu_0 \ell^2}{\pi \mathrm{L}}=2 \sqrt{2} \frac{\mu_0}{\pi}\)

\(=2 \sqrt{2} \frac{4 \pi}{\pi} \times 10^{-7}\)

\(=8 \sqrt{2} \times 10^{-7} \mathrm{H}\)

\(=\sqrt{128} \times 10^{-7} \mathrm{H}\)

\(x=128\)

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