JEE MAIN - Physics (2024 - 31st January Morning Shift - No. 28)
A small square loop of wire of side $$l$$ is placed inside a large square loop of wire of side $$L\left(L=l^2\right)$$. The loops are coplanar and their centers coincide. The value of the mutual inductance of the system is $$\sqrt{x} \times 10^{-7} \mathrm{H}$$, where $$x=$$ _________.
Answer
128
Explanation
_31st_January_Morning_Shift_en_28_1.png)
Flux linkage for inner loop.
$$\begin{aligned} & \phi=\mathrm{B}_{\text {center }} \cdot \ell^2 \\ & =4 \times \frac{\mu_0 \mathrm{i}}{4 \pi \frac{\mathrm{L}}{2}}(\sin 45+\sin 45) \ell^2 \\ & \phi=2 \sqrt{2} \frac{\mu_0 \mathrm{i}}{\pi \mathrm{L}} \ell^2 \\ & \mathrm{M}=\frac{\phi}{\mathrm{i}}=\frac{2 \sqrt{2} \mu_0 \ell^2}{\pi \mathrm{L}}=2 \sqrt{2} \frac{\mu_0}{\pi} \\ & =2 \sqrt{2} \frac{4 \pi}{\pi} \times 10^{-7} \\ & =8 \sqrt{2} \times 10^{-7} \mathrm{H} \\ & =\sqrt{128} \times 10^{-7} \mathrm{H} \\ & \mathrm{x}=128 \end{aligned}$$
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