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(A) The magnetic field due to a straight wire segment of length $L$ at a perpendicular distance $d$ is given by $B = \frac{\mu_0 I}{4 \pi d} (\sin 	h

Vedclass · Accepted Answer

$n \frac{\mu_0 I}{2 \pi R} 	an \frac{\pi}{n}$

Vedclass · Answer

(A) The magnetic field due to a straight wire segment of length $L$ at a perpendicular distance $d$ is given by $B = \frac{\mu_0 I}{4 \pi d} (\sin 	heta_1 + \sin 	heta_2)$.For a regular polygon of $n$ sides,the angle subtended by each side at the centre is $2\pi/n$. Thus,the angles $	heta_1$ and $	heta_2$ at the centre for each side are $\pi/n$.The perpendicular distance $d$ from the centre to the side is $d = R \cos(\pi/n)$,where $R$ is the distance to the vertex.The magnetic field due to one side is $B_1 = \frac{\mu_0 I}{4 \pi (R \cos(\pi/n))} (\sin(\pi/n) + \sin(\pi/n)) = \frac{\mu_0 I}{4 \pi R \cos(\pi/n)} \cdot 2 \sin(\pi/n) = \frac{\mu_0 I}{2 \pi R} 	an(\pi/n)$.Since there are $n$ such sides,the total magnetic field at the centre is $B = n \cdot B_1 = n \frac{\mu_0 I}{2 \pi R} 	an \frac{\pi}{n}$.

$A$ loop carrying current $I$ has the shape of a regular polygon of $n$ sides. If $R$ is the distance from the centre to any vertex,then the magnitude of the magnetic induction vector $B$ at the centre of the loop is

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