The magnetic field intensity at the point O o | vedclass

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Question

(A) The loop consists of two parts: a square part and a circular arc part.$1$. Magnetic field due to the square part:The magnetic field at the center

Vedclass · Accepted Answer

$\frac{\mu_0 i}{4 \pi}\left[\frac{3 \pi}{2 a}+\frac{\sqrt{2}}{b}\right]$

Vedclass · Answer

(A) The loop consists of two parts: a square part and a circular arc part.$1$. Magnetic field due to the square part:The magnetic field at the center $O$ due to the segments lying on the axes passing through $O$ is zero because the current is directed towards or away from $O$. For the other two segments of length $b$,the distance from $O$ to each segment is $b/2$. The angle subtended by each segment at $O$ is $45^{\circ}$.Using the formula for a finite wire,$B = \frac{\mu_0 i}{4 \pi d}(\sin 	heta_1 + \sin 	heta_2)$,where $d = b/2$ and $	heta_1 = 45^{\circ}, 	heta_2 = 0^{\circ}$:$B_1 = 2 	imes \left[ \frac{\mu_0 i}{4 \pi (b/2)} (\sin 45^{\circ} + 0) ight] = 2 	imes \frac{\mu_0 i}{2 \pi b} 	imes \frac{1}{\sqrt{2}} = \frac{\mu_0 i}{\sqrt{2} \pi b}$.$2$. Magnetic field due to the circular arc:The arc subtends an angle of $270^{\circ}$ or $\frac{3 \pi}{2}$ radians at the center. The magnetic field due to a circular arc is $B_2 = \frac{\mu_0 i 	heta}{4 \pi a} = \frac{\mu_0 i (3 \pi / 2)}{4 \pi a} = \frac{3 \mu_0 i}{8 a}$.$3$. Total magnetic field:Since both fields are in the same direction (into the plane),the net field is:$B_{	ext{net}} = B_1 + B_2 = \frac{\mu_0 i}{\sqrt{2} \pi b} + \frac{3 \mu_0 i}{8 a} = \frac{\mu_0 i}{4 \pi} \left[ \frac{3 \pi}{2 a} + \frac{\sqrt{2}}{b} ight]$.

The magnetic field intensity at the point $O$ of a loop with current $i$,whose shape is illustrated below is

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