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(B) The magnetic field at the centre due to a straight wire of length $L$ carrying current $i$ at a perpendicular distance $r$ is given by $B = \frac{

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$\frac{{{\mu _0}ni}}{{2\pi a}}	an \frac{\pi }{n}$

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(B) The magnetic field at the centre due to a straight wire of length $L$ carrying current $i$ at a perpendicular distance $r$ is given by $B = \frac{{{\mu _0}i}}{{4\pi r}}(\sin \phi_1 + \sin \phi_2)$.For one side of the regular polygon,the angle subtended by half the side at the centre is $	heta = \frac{\pi}{n}$.The perpendicular distance from the centre to the side is $r = a \cos 	heta = a \cos(\frac{\pi}{n})$.The angles at the ends of the side are $\phi_1 = \phi_2 = 	heta = \frac{\pi}{n}$.Thus,the magnetic field due to one side is $B_1 = \frac{{{\mu _0}i}}{{4\pi (a \cos 	heta)}} (2 \sin 	heta) = \frac{{{\mu _0}i}}{{2\pi a}} 	an 	heta = \frac{{{\mu _0}i}}{{2\pi a}} 	an(\frac{\pi}{n})$.Since there are $n$ such sides,the net magnetic field at the centre is $B_{net} = n 	imes B_1 = \frac{{n{\mu _0}i}}{{2\pi a}} 	an(\frac{\pi}{n})$.

In the following figure,a wire bent in the form of a regular polygon of $n$ sides is inscribed in a circle of radius $a$. The net magnetic field at the centre will be

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