The ratio of the magnetic field at the centre | vedclass

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(B) Let the total length of the wire be $L$ and the current flowing through it be $i$.For a circular coil of radius $r$,the circumference is $L = 2\pi

Vedclass · Accepted Answer

$\frac{\pi^2}{8\sqrt{2}}$

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(B) Let the total length of the wire be $L$ and the current flowing through it be $i$.For a circular coil of radius $r$,the circumference is $L = 2\pi r$,so $r = \frac{L}{2\pi}$.The magnetic field at the centre of the circular coil is $B_{	ext{Circular}} = \frac{\mu_0 i}{2r} = \frac{\mu_0 i}{2(L/2\pi)} = \frac{\mu_0 \pi i}{L}$.For a square coil of side $a$,the perimeter is $L = 4a$,so $a = \frac{L}{4}$.The magnetic field at the centre of a square coil is $B_{	ext{Square}} = 4 	imes \left( \frac{\mu_0 i}{4\pi (a/2)} 	imes 2 \sin(45^\circ) ight) = \frac{2\sqrt{2}\mu_0 i}{\pi a}$.Substituting $a = \frac{L}{4}$,we get $B_{	ext{Square}} = \frac{2\sqrt{2}\mu_0 i}{\pi (L/4)} = \frac{8\sqrt{2}\mu_0 i}{\pi L}$.Now,the ratio is $\frac{B_{	ext{Circular}}}{B_{	ext{Square}}} = \frac{\mu_0 \pi i / L}{8\sqrt{2}\mu_0 i / \pi L} = \frac{\pi}{1} 	imes \frac{\pi}{8\sqrt{2}} = \frac{\pi^2}{8\sqrt{2}}$.

The ratio of the magnetic field at the centre of a current-carrying circular wire and the magnetic field at the centre of a square coil made from the same length of wire will be

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