The ratio of the magnetic field at the centre | vedclass

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $L$ be the total length of the wire. For the circular coil of radius $R$,$L = 2\pi R$,so $R = \frac{L}{2\pi}$. The magnetic field at the center is $B_{C_1} = \frac{\mu_0 i}{2R} = \frac{\mu_0 i}{2(L/2\pi)} = \frac{\mu_0 i \pi}{L}$.For the square coil of side $a$,$L = 4a$,so $a = \frac{L}{4}$. The magnetic field at the center due to one side of length $a$ is $B_{side} = \frac{\mu_0 i}{4\pi(a/2)} (\sin 45^{\circ} + \sin 45^{\circ}) = \frac{\mu_0 i}{2\pi a} (2 \cdot \frac{1}{\sqrt{2}}) = \frac{\sqrt{2} \mu_0 i}{\pi a}$.Since there are $4$ sides,the total magnetic field at the center is $B_{C_2} = 4 \cdot B_{side} = \frac{4\sqrt{2} \mu_0 i}{\pi a} = \frac{4\sqrt{2} \mu_0 i}{\pi (L/4)} = \frac{16\sqrt{2} \mu_0 i}{\pi L}$.The ratio is $\frac{B_{C_1}}{B_{C_2}} = \frac{\mu_0 i \pi / L}{16\sqrt{2} \mu_0 i / \pi L} = \frac{\pi^2}{16\sqrt{2}}$.Wait,re-evaluating the provided solution steps: $B_{C_2} = 4 \cdot \frac{\mu_0 i}{4\pi(a/2)} [\sin 45^{\circ} + \sin 45^{\circ}] = \frac{\mu_0 i}{\pi a/2} [\sqrt{2}] = \frac{2\sqrt{2} \mu_0 i}{\pi a}$. Substituting $a = \frac{\pi R}{2}$,we get $B_{C_2} = \frac{2\sqrt{2} \mu_0 i}{\pi (\pi R / 2)} = \frac{4\sqrt{2} \mu_0 i}{\pi^2 R}$.Then $\frac{B_{C_1}}{B_{C_2}} = \frac{\mu_0 i / 2R}{4\sqrt{2} \mu_0 i / \pi^2 R} = \frac{\pi^2}{8\sqrt{2}}$.

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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