Electromagnetic Induction Question 8
Question 8 - 2024 (31 Jan Shift 1)
A small square loop of wire of side $\ell$ is placed inside a large square loop of wire of side $\mathrm{L}\left(\mathrm{L}=\ell^{2}\right)$. The loops are coplanar and their centers coinside. The value of the mutual inductance of the system is $\sqrt{\mathrm{x}} \times 10^{-7} \mathrm{H}$, where $\mathrm{x}=$
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Answer: (128)
Solution:
Flux linkage for inner loop.
$$ \begin{aligned} & \phi=\mathrm{B}{\text {center }} \cdot \ell^{2} \ & =4 \times \frac{\mu{0} \mathrm{i}}{4 \pi \frac{\mathrm{L}}{2}}(\sin 45+\sin 45) \ell^{2} \end{aligned} $$
$\phi=2 \sqrt{2} \frac{\mu_{0} \mathrm{i}}{\pi \mathrm{L}} \ell^{2}$
$\mathrm{M}=\frac{\phi}{\mathrm{i}}=\frac{2 \sqrt{2} \mu_{0} \ell^{2}}{\pi \mathrm{L}}=2 \sqrt{2} \frac{\mu_{0}}{\pi}$
$=2 \sqrt{2} \frac{4 \pi}{\pi} \times 10^{-7}$
$=8 \sqrt{2} \times 10^{-7} \mathrm{H}$
$=\sqrt{128} \times 10^{-7} \mathrm{H}$
$x=128$
