The magnitude of the magnetic field (in SI un | vedclass

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(C) The magnetic field $B$ due to a single straight wire of length $a$ at a perpendicular distance $r$ is given by $B = \frac{\mu_0 I}{4 \pi r} (\sin

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$500$

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(C) The magnetic field $B$ due to a single straight wire of length $a$ at a perpendicular distance $r$ is given by $B = \frac{\mu_0 I}{4 \pi r} (\sin 	heta_1 + \sin 	heta_2)$.For a hexagonal coil, the distance from the center to the side is $r = a \cos 30^{\circ} = a \frac{\sqrt{3}}{2}$.The angles subtended at the center by the ends of the side are $	heta_1 = 	heta_2 = 30^{\circ}$.For one side, $B_1 = \frac{\mu_0 I}{4 \pi (a \sqrt{3} / 2)} (\sin 30^{\circ} + \sin 30^{\circ}) = \frac{\mu_0 I}{2 \pi a \sqrt{3}} (1) = \frac{\mu_0 I}{2 \pi a \sqrt{3}}$.For a hexagon with $N=50$ turns, the total magnetic field is $B = 6 	imes N 	imes B_1 = 6 	imes 50 	imes \frac{\mu_0 I}{2 \pi a \sqrt{3}} = \frac{150 \mu_0 I}{\pi a \sqrt{3}}$.Given $a = 10 \, cm = 0.1 \, m$, we have $B = \frac{150 \mu_0 I}{\pi (0.1) \sqrt{3}} = \frac{1500}{\sqrt{3}} \frac{\mu_0 I}{\pi} = 500 \sqrt{3} \frac{\mu_0 I}{\pi}$.

The magnitude of the magnetic field (in $SI$ units) at the centre of a hexagonal coil of side $10 \, cm$, having $50$ turns and carrying a current $I$ (Ampere), in units of $\frac{\mu_{0} I}{\pi}$ is: (in $\sqrt{3}$)

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