Find the mutual inductance in the arrangement | vedclass

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(C) The magnetic field $B$ at the center of a square loop of side $L$ carrying current $i$ is given by the sum of the fields due to its four sides.For

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$M = \frac{2 \sqrt{2} \mu_0 R^2}{L}$

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(C) The magnetic field $B$ at the center of a square loop of side $L$ carrying current $i$ is given by the sum of the fields due to its four sides.For one side,the field at the center is $B_1 = \frac{\mu_0 i}{4 \pi (L/2)} (\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 i}{2 \pi L} (2 \cdot \frac{1}{\sqrt{2}}) = \frac{\sqrt{2} \mu_0 i}{2 \pi L} = \frac{\mu_0 i}{\sqrt{2} \pi L}$.Since there are four sides,the total magnetic field at the center is $B = 4 \cdot \frac{\mu_0 i}{\sqrt{2} \pi L} = \frac{2 \sqrt{2} \mu_0 i}{\pi L}$.Since $L \gg R$,we assume the magnetic field is uniform over the area of the small circular loop.The magnetic flux $\phi$ through the circular loop of area $A = \pi R^2$ is $\phi = B \cdot A = \left( \frac{2 \sqrt{2} \mu_0 i}{\pi L} \right) (\pi R^2) = \frac{2 \sqrt{2} \mu_0 R^2 i}{L}$.By definition,$\phi = Mi$,therefore $M = \frac{2 \sqrt{2} \mu_0 R^2}{L}$.

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:

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