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Consider a conducting wire of length $\mathrm{L}$ bent in the form of a circle of radius $\mathrm{R}$ and another conductor of length 'a' (a < < R) is bent in the form of a square. The two loops are then placed in same plane such that the square loop is exactly at the centre of the circular loop. What will be the mutual inductance between the two loops?
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Verified Answer
The correct answer is:
$\mu_{0} \frac{\pi \mathrm{a}^{2}}{16 \mathrm{~L}}$
Hint:
$2 \pi R=L_{\cdots} \ldots \ldots \ldots .(1)$
$4 s=a \ldots \ldots \ldots . .(2)$
$\frac{\mu_{0} I \times S^{2}}{2 R}=\phi, M=\frac{\mu_{0} S^{2}}{2 R}$
$=\frac{\mu_{0}}{2} \times \frac{a^{2} \times 2 \pi}{16 \times L}=\frac{\mu_{0} a^{2} \pi}{16 L}$
$2 \pi R=L_{\cdots} \ldots \ldots \ldots .(1)$
$4 s=a \ldots \ldots \ldots . .(2)$
$\frac{\mu_{0} I \times S^{2}}{2 R}=\phi, M=\frac{\mu_{0} S^{2}}{2 R}$
$=\frac{\mu_{0}}{2} \times \frac{a^{2} \times 2 \pi}{16 \times L}=\frac{\mu_{0} a^{2} \pi}{16 L}$
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