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Question

(A) The magnetic field $B$ at the center due to a square loop of side $b$ carrying current $I$ is given by the sum of the fields due to its four sides

Vedclass · Accepted Answer

$\frac{\mu_{0}}{4 \pi} 8 \sqrt{2} \frac{a^{2}}{b}$

Vedclass · Answer

(A) The magnetic field $B$ at the center due to a square loop of side $b$ 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 $B_{1} = \frac{\mu_{0} I}{4 \pi (b/2)} (\sin 45^{\circ} + \sin 45^{\circ}) = \frac{\mu_{0} I}{2 \pi b} (2 	imes \frac{1}{\sqrt{2}}) = \frac{\sqrt{2} \mu_{0} I}{\pi b}$.Since there are four sides,the total magnetic field at the center is $B = 4 	imes B_{1} = \frac{4 \sqrt{2} \mu_{0} I}{\pi b}$.Since $b \gg a$,we assume the magnetic field is uniform over the area of the smaller loop.The magnetic flux $\phi$ linked with the smaller loop of area $A = a^{2}$ is $\phi = B 	imes A = \frac{4 \sqrt{2} \mu_{0} I}{\pi b} 	imes a^{2}$.The coefficient of mutual inductance $M$ is defined as $M = \frac{\phi}{I} = \frac{4 \sqrt{2} \mu_{0} a^{2}}{\pi b}$.To match the given options,we multiply and divide by $4$:$M = \frac{\mu_{0}}{4 \pi} 	imes 16 \sqrt{2} \frac{a^{2}}{b}$.Wait,re-evaluating the field calculation: $B_{side} = \frac{\mu_{0} I}{4 \pi d} (\sin 	heta_1 + \sin 	heta_2)$. Here $d = b/2$,$	heta_1 = 	heta_2 = 45^{\circ}$.$B_{side} = \frac{\mu_{0} I}{4 \pi (b/2)} (2 \sin 45^{\circ}) = \frac{\mu_{0} I}{2 \pi b} \sqrt{2} = \frac{\mu_{0} I}{\sqrt{2} \pi b}$.Total $B = 4 	imes \frac{\mu_{0} I}{\sqrt{2} \pi b} = \frac{2 \sqrt{2} \mu_{0} I}{\pi b}$.Flux $\phi = B a^{2} = \frac{2 \sqrt{2} \mu_{0} I a^{2}}{\pi b}$.$M = \frac{\phi}{I} = \frac{2 \sqrt{2} \mu_{0} a^{2}}{\pi b} = \frac{\mu_{0}}{4 \pi} \frac{8 \sqrt{2} a^{2}}{b}$.

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