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Question

(A) The magnetic field due to a straight wire of length $a$ at a perpendicular distance $r$ from its centre is given by $B = \frac{\mu_0 i}{4 \pi r} (

Vedclass · Accepted Answer

$\frac{9 \mu_0 i}{4 \pi L}$

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(A) The magnetic field due to a straight wire of length $a$ at a perpendicular distance $r$ from its centre is given by $B = \frac{\mu_0 i}{4 \pi r} (\sin 	heta_1 + \sin 	heta_2)$.For an equilateral triangle of side $a = 2L$,the distance $r$ from the centre to any side is $r = \frac{a}{2 	an 60^{\circ}} = \frac{2L}{2 \sqrt{3}} = \frac{L}{\sqrt{3}}$.The angles subtended by the ends of the side at the centre are $	heta_1 = 	heta_2 = 60^{\circ}$.The magnetic field due to one side is $B_1 = \frac{\mu_0 i}{4 \pi (L/\sqrt{3})} (\sin 60^{\circ} + \sin 60^{\circ}) = \frac{\mu_0 i \sqrt{3}}{4 \pi L} (\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}) = \frac{\mu_0 i \sqrt{3}}{4 \pi L} (\sqrt{3}) = \frac{3 \mu_0 i}{4 \pi L}$.Since there are $3$ identical sides,the total magnetic field at the centre is $B = 3 	imes B_1 = 3 	imes \frac{3 \mu_0 i}{4 \pi L} = \frac{9 \mu_0 i}{4 \pi L}$.

The magnetic field at the centre of an equilateral triangular loop of side $2L$ and carrying a current $i$ is

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