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(D) The magnetic field due to a straight wire of finite length at a perpendicular distance $r$ is given by $B = \frac{\mu_0 I}{4 \pi r} (\sin 	heta_1

Vedclass · Accepted Answer

$\frac{2 \sqrt{2} \mu_0 I}{\pi L}$

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(D) The magnetic field due to a straight wire of finite length at a perpendicular distance $r$ is given by $B = \frac{\mu_0 I}{4 \pi r} (\sin 	heta_1 + \sin 	heta_2)$.For a square loop of side $L$, the perpendicular distance from the center to any side is $r = L/2$.The angles subtended by the ends of each side at the center are $	heta_1 = 45^{\circ}$ and $	heta_2 = 45^{\circ}$.Since there are $4$ identical sides, the total magnetic field $B_{total}$ is $4$ times the field due to one side:$B_{total} = 4 	imes \left[ \frac{\mu_0 I}{4 \pi (L/2)} (\sin 45^{\circ} + \sin 45^{\circ}) ight]$$B_{total} = 4 	imes \left[ \frac{\mu_0 I}{2 \pi L} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) ight]$$B_{total} = 4 	imes \left[ \frac{\mu_0 I}{2 \pi L} (\frac{2}{\sqrt{2}}) ight]$$B_{total} = 4 	imes \left[ \frac{\mu_0 I}{\pi L} 	imes \frac{1}{\sqrt{2}} 	imes 2 ight] = \frac{8 \mu_0 I}{2 \sqrt{2} \pi L} = \frac{4 \mu_0 I}{\sqrt{2} \pi L} = \frac{2 \sqrt{2} \mu_0 I}{\pi L}$.

The magnetic field at the point of intersection of diagonals of a square wire loop of side $L$ carrying a current $I$ is

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