The ratio of the magnetic field at the centre | vedclass

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(B) Let the length of the wire be $L$. For a circular loop of radius $R$,$L = 2\pi R$,so $R = \frac{L}{2\pi}$.The magnetic field at the center of the

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

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(B) Let the length of the wire be $L$. For a circular loop of radius $R$,$L = 2\pi R$,so $R = \frac{L}{2\pi}$.The magnetic field at the center of the circular loop is $B_1 = \frac{\mu_0 i}{2R} = \frac{\mu_0 i}{2(L/2\pi)} = \frac{\mu_0 i \pi}{L}$.For a square coil of side $a$,$L = 4a$,so $a = \frac{L}{4}$.The magnetic field at the center of a square coil is the sum of the fields due to its four sides. For one side,the field is $B_{side} = \frac{\mu_0 i}{4\pi(a/2)} (\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 i}{2\pi a} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{\mu_0 i}{2\pi a} \sqrt{2} = \frac{\mu_0 i}{\sqrt{2}\pi a}$.Total field $B_2 = 4 	imes B_{side} = \frac{4\mu_0 i}{\sqrt{2}\pi a} = \frac{2\sqrt{2}\mu_0 i}{\pi a}$.Substituting $a = L/4$,$B_2 = \frac{2\sqrt{2}\mu_0 i}{\pi (L/4)} = \frac{8\sqrt{2}\mu_0 i}{\pi L}$.The ratio is $\frac{B_1}{B_2} = \frac{\mu_0 i \pi / L}{8\sqrt{2}\mu_0 i / \pi L} = \frac{\pi^2}{8\sqrt{2}}$.

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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