A current I flows around a closed path in the | vedclass

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(C) The path consists of $8$ arcs in total. Since they form a closed loop and subtend equal angles at the center $P$,each arc subtends an angle $	het

Vedclass · Accepted Answer

$\frac{1}{8} \frac{\mu_0 I}{r}$; perpendicular to the plane of the paper and directed inward.

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(C) The path consists of $8$ arcs in total. Since they form a closed loop and subtend equal angles at the center $P$,each arc subtends an angle $	heta = \frac{2\pi}{8} = \frac{\pi}{4}$ at the center.There are $4$ arcs of radius $r$ and $4$ arcs of radius $2r$.The magnetic field due to an arc of radius $R$ subtending angle $	heta$ is $B = \frac{\mu_0 I 	heta}{4\pi R}$.For the $4$ arcs of radius $r$: $B_1 = 4 	imes \left( \frac{\mu_0 I (\pi/4)}{4\pi r} ight) = \frac{\mu_0 I}{4r}$. The direction is inward (using the right-hand rule).For the $4$ arcs of radius $2r$: $B_2 = 4 	imes \left( \frac{\mu_0 I (\pi/4)}{4\pi (2r)} ight) = \frac{\mu_0 I}{8r}$. The direction is outward (using the right-hand rule).The net magnetic field is $B_{net} = B_1 - B_2 = \frac{\mu_0 I}{4r} - \frac{\mu_0 I}{8r} = \frac{\mu_0 I}{8r}$.Since $B_1 > B_2$,the net field is directed inward.

$A$ current $I$ flows around a closed path in the horizontal plane as shown in the figure. The path consists of eight arcs with alternating radii $r$ and $2r$. Each segment of arc subtends an equal angle at the common centre $P$. The magnetic field produced by the current path at point $P$ is

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