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Question

(B) Let the length of the wire be $L$. For a circle of radius $r$,$L = 2\pi r$,so $r = \frac{L}{2\pi}$. The magnetic field at the center is $B_c = \fr

Vedclass · Accepted Answer

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

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

What is the ratio of the magnetic field at the center of a circle and a square made from a wire of the same length?

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