JEE MAIN - Physics (2023 - 29th January Morning Shift - No. 2)
Find the mutual inductance in the arrangement, when a small circular loop of wire of radius '$$R$$' is placed inside a large square loop of wire of side $$L$$ $$(L \gg R)$$. The loops are coplanar and their centres coincide :
_29th_January_Morning_Shift_en_2_1.png)
$$M=\frac{\sqrt{2} \mu_{0} R}{L^{2}}$$
$$M=\frac{2 \sqrt{2} \mu_{0} R}{L^{2}}$$
$$M=\frac{2 \sqrt{2} \mu_{0} R^{2}}{L}$$
$$M=\frac{\sqrt{2} \mu_{0} R^{2}}{L}$$
Explanation
_29th_January_Morning_Shift_en_2_2.png)
$$ \begin{aligned} & B \text { at centre }=\frac{\mu_{0} i}{4 \pi\left(\frac{L}{2}\right)}\left(\frac{2}{\sqrt{2}}\right) \times 4 \\\\ &= \frac{\sqrt{2} \mu_{0} i}{2 \pi L} \times 4 \\\\ &=\left(\frac{2 \sqrt{2} \mu_{0} i}{\pi L}\right) \end{aligned} $$
Mutual inductance $=\frac{B \cdot A}{i}$
$$ =\frac{2 \sqrt{2} \mu_{0} i}{\pi L} \times \frac{\pi R^{2}}{i} $$
$$ =\left(\frac{2 \sqrt{2} \mu_{0} R^{2}}{L}\right) $$
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