The ratio of the magnetic field at the centre | vedclass

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the total length of the wire be $L$. For a circular coil of radius $r$,$L = 2\pi r$,so $r = \frac{L}{2\pi}$. The magnetic field at the centre of the circular coil is $B_{circular} = \frac{\mu_0 i}{2r} = \frac{\mu_0 i}{2(L/2\pi)} = \frac{\mu_0 \pi i}{L}$. For a square coil of side $a$,$L = 4a$,so $a = \frac{L}{4}$. The magnetic field at the centre of a square coil due to one side is $B_1 = \frac{\mu_0 i}{4\pi (a/2)} (\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 i}{2\pi a} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{\mu_0 i}{\sqrt{2} \pi a}$. Since there are four sides,the total magnetic field is $B_{square} = 4 	imes B_1 = \frac{4 \mu_0 i}{\sqrt{2} \pi a} = \frac{4 \mu_0 i}{\sqrt{2} \pi (L/4)} = \frac{16 \mu_0 i}{\sqrt{2} \pi L}$. The ratio is $\frac{B_{circular}}{B_{square}} = \frac{\mu_0 \pi i / L}{16 \mu_0 i / (\sqrt{2} \pi L)} = \frac{\pi}{1} 	imes \frac{\sqrt{2} \pi}{16} = \frac{\sqrt{2} \pi^2}{16} = \frac{\pi^2}{8\sqrt{2}}$.

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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