JEE MAIN - Physics (2022 - 25th July Morning Shift - No. 13)
A small square loop of wire of side $$l$$ is placed inside a large square loop of wire $$\mathrm{L}(\mathrm{L}>>l)$$. Both loops are coplanar and their centres coincide at point $$\mathrm{O}$$ as shown in figure. The mutual inductance of the system is :
_25th_July_Morning_Shift_en_13_1.png)
$$\frac{2 \sqrt{2} \mu_{0} \mathrm{~L}^{2}}{\pi l}$$
$$\frac{\mu_{0} l^{2}}{2 \sqrt{2} \pi \mathrm{L}}$$
$$\frac{2 \sqrt{2} \mu_{0} l^{2}}{\pi \mathrm{L}}$$
$$
\frac{\mu_{0} \mathrm{~L}^{2}}{2 \sqrt{2} \pi l}
$$
Explanation
We know $$\phi = Mi$$
Let i current be flowing in the larger loop
$$ \Rightarrow \phi \simeq \left[ {4 \times {{{\mu _0}i} \over {4\pi (L/2)}}[\sin 45^\circ + \sin 45^\circ ]} \right] \times $$ Area
$$ = {{2\sqrt 2 {\mu _0}i} \over {\pi L}} \times {I^2}$$
$$ \Rightarrow M = {\phi \over i} = {{2\sqrt 2 {\mu _0}{I^2}} \over {\pi L}}$$
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