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(B) $1$. For a straight wire of length $L = \pi^2 \, m$ carrying current $i = 2 \, A$,the magnetic field $B_1$ at a distance $d = 1 \, cm = 10^{-2} \,

Vedclass · Accepted Answer

$1:50$

Vedclass · Answer

(B) $1$. For a straight wire of length $L = \pi^2 \, m$ carrying current $i = 2 \, A$,the magnetic field $B_1$ at a distance $d = 1 \, cm = 10^{-2} \, m$ is given by the Biot-Savart law for an infinitely long wire (assuming the point is near the center of the wire): $B_1 = \frac{\mu_0 i}{2\pi d} = \frac{\mu_0 	imes 2}{2\pi 	imes 10^{-2}} = \frac{\mu_0}{\pi 	imes 10^{-2}}$.$2$. When the wire is bent into a circle of radius $R$,its circumference is $2\pi R = L = \pi^2$. Thus,$R = \frac{\pi^2}{2\pi} = \frac{\pi}{2} \, m$.$3$. The magnetic field at the center of this circular loop is $B_2 = \frac{\mu_0 i}{2R} = \frac{\mu_0 	imes 2}{2 	imes (\pi/2)} = \frac{\mu_0}{\pi/2} = \frac{2\mu_0}{\pi}$.$4$. The ratio of the magnetic field at the center to the first case is $\frac{B_2}{B_1} = \frac{2\mu_0 / \pi}{\mu_0 / (\pi 	imes 10^{-2})} = 2 	imes 10^{-2} = \frac{2}{100} = \frac{1}{50}$.$5$. Therefore,the ratio is $1:50$.

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