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 centre 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 centre 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 centre of a square coil is given by $B_2 = 4 	imes \frac{\mu_0 i}{4\pi(a/2)} [\sin 45^{\circ} + \sin 45^{\circ}] = \frac{\mu_0 i}{\pi (a/2)} 	imes 2 \sin 45^{\circ} = \frac{2\mu_0 i}{\pi a} 	imes \sqrt{2} = \frac{2\sqrt{2}\mu_0 i}{\pi a}$.Substituting $a = L/4$,we get $B_2 = \frac{2\sqrt{2}\mu_0 i}{\pi (L/4)} = \frac{8\sqrt{2}\mu_0 i}{\pi L}$.The ratio is $\frac{B_1}{B_2} = \frac{\mu_0 i \pi / L}{8\sqrt{2}\mu_0 i / \pi L} = \frac{\pi}{1} 	imes \frac{\pi}{8\sqrt{2}} = \frac{\pi^2}{8\sqrt{2}}$.

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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