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Offline Centres

A small square loop of wire of side “ $l$ ” is placed inside a large square loop of wire $L (L ≫ l)$ . Both loops are coplanar and their centres coincide at point “O” as shown in figure. The mutual inductance of the system is

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$\frac{2 \sqrt{2} μ_{0} L^{2}}{π l}$

$\frac{μ_{0} L^{2}}{2 \sqrt{2} π l}$

$\frac{μ_{0} l^{2}}{2 \sqrt{2} π l}$

$\frac{2 \sqrt{2} μ_{0} l^{2}}{π L}$

answer is C.

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Detailed Solution

$ϕ_{2} α I_{1}$

$ϕ_{2} = M_{21} I_{1}$

$M_{21} = \frac{ϕ_{2}}{I_{1}} = \frac{B_{1} A_{2}}{I_{1}}$

Now $B_{1} = \frac{μ_{0} I_{1}}{4 π L / 2} [\sin 45 + \sin 45] \times 4$

$= \frac{μ_{0} I_{1}}{2 π L} \times \frac{8}{\sqrt{2}} = \frac{2 \sqrt{2} μ_{0} I_{1}}{π L}$

So $M_{21} = \frac{2 \sqrt{2} μ_{0} I_{1}}{π L I_{1}} l^{2} = \frac{μ_{0} l^{2}}{π L} \times 2 \sqrt{2}$

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