# A small square loop of wire of side l is placed inside a large square loop of side L. The loops are coplanar and their centres coincide. The mutul inductance of the system is proportional to (a) l /L. (b) l^2/L

Dear Student,

Magnetic Field at the centre of outer square current(I) carrying loop is

${B}_{1}=\frac{2\sqrt{2}{\mu }_{0}I}{\pi L}$

Since L>>l we approximate the magnetic field through entire inner coil by B1
Thus the flux through inner coil

${\phi }_{21}={B}_{1}{A}_{2}\phantom{\rule{0ex}{0ex}}=\left(\frac{2\sqrt{2}{\mu }_{0}I}{\pi }\right)×\frac{{l}^{2}}{L}$

Mutual Inductance = $\frac{{\phi }_{21}}{{I}_{1}}$

option(2) CORRECT ANSWER

Regards.

• 67
Hello Pradeep dear, the option l^2 / L is the right one
• -29
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