A wire bent in the shape of a regular n-polyg | vedclass

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(A) For an $n$-sided polygon,the magnetic field due to one side at the centre of the loop using the Biot-Savart law is given by:$B_1 = \frac{\mu_0 I}{

Vedclass · Accepted Answer

$\frac{n \mu_0 I}{2 \pi l} \sin (\pi / n)$

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(A) For an $n$-sided polygon,the magnetic field due to one side at the centre of the loop using the Biot-Savart law is given by:$B_1 = \frac{\mu_0 I}{4 \pi l} (\sin 	heta_1 + \sin 	heta_2)$As there are $n$ sides,the angle subtended by one side at the centre is $\alpha = \frac{2 \pi}{n}$.Therefore,$	heta_1 = 	heta_2 = \frac{1}{2} 	imes \frac{2 \pi}{n} = \frac{\pi}{n}$.Substituting this into the expression for $B_1$:$B_1 = \frac{\mu_0 I}{4 \pi l} (\sin \frac{\pi}{n} + \sin \frac{\pi}{n}) = \frac{\mu_0 I}{2 \pi l} \sin \frac{\pi}{n}$.The magnetic field at the centre due to all $n$ segments is the sum of the fields due to each segment:$B = n 	imes B_1 = \frac{n \mu_0 I}{2 \pi l} \sin \left(\frac{\pi}{n}ight)$.

$A$ wire bent in the shape of a regular $n$-polygonal loop carries a steady current $I$. Let $l$ be the perpendicular distance of a given segment and $R$ be the distance of a vertex from the centre of the loop. The magnitude of the magnetic field at the centre of the loop is given by

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