A small square loop of side ‘a’ and one turn is placed inside a larger square loop of side b and one turn (b >>a). The two loops are coplanar with their centres coinciding. If a current I is passed in the square loop of side ‘b’, then the coefficient of mutual inductance between the two loops is:

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$\frac{μ_{0}}{4 π} \frac{8 \sqrt{2}}{b}$

$\frac{μ_{0}}{4 π} \frac{8 \sqrt{2}}{a}$

$\frac{μ_{0}}{4 π} 8 \sqrt{2} \frac{a^{2}}{b}$

$\frac{μ_{0}}{4 π} 8 \sqrt{2} \frac{b^{2}}{a}$

answer is A.

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Detailed Solution

Magnetic field at centre due to each side
Question Image

$= \frac{μ_{0} I}{4 π \frac{b}{2}} (\sin 45^{\circ} + \sin 45^{\circ})$
Field due to all 4 sides
$= B = \frac{μ_{0}}{4 π} \frac{I}{\frac{b}{2}} (\sin 45^{\circ} + \sin 45^{\circ}) 4 = B = [\frac{μ_{0}}{4 π} \frac{I}{\frac{b}{2}} \times 2 \sin 45] \times 4 = 2 \sqrt{2} \frac{μ_{0}}{π} \frac{I}{b}$

Flux through small square will be $\Rightarrow ϕ = 2 \sqrt{2} \frac{μ_{0}}{π} \frac{I}{b} \times a^{2}$
Coefficient of mutual induction, $M i_{b} = ϕ_{a}$ is given as $M = \frac{ϕ}{I} = \frac{2 \sqrt{2} μ_{0} a^{2}}{π b} = \frac{μ_{0}}{4 π} 8 \sqrt{2} \frac{a^{2}}{b}$