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(B) The magnetic field at the center of a circular coil of $n$ turns,radius $r$,and current $I$ is given by $B = \frac{\mu_0 n I}{2r}$.Let the total l

Vedclass · Accepted Answer

$9\,B_1$

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(B) The magnetic field at the center of a circular coil of $n$ turns,radius $r$,and current $I$ is given by $B = \frac{\mu_0 n I}{2r}$.Let the total length of the wire be $L$. For $n=1$ loop,$L = 2\pi r_1$,so $r_1 = \frac{L}{2\pi}$. The magnetic field is $B_1 = \frac{\mu_0 (1) I}{2r_1} = \frac{\mu_0 I}{2(L/2\pi)} = \frac{\mu_0 I \pi}{L}$.For $n=3$ loops,the new radius $r_2$ satisfies $L = 3(2\pi r_2)$,so $r_2 = \frac{L}{6\pi} = \frac{r_1}{3}$.The new magnetic field $B_2$ is $B_2 = \frac{\mu_0 (3) I}{2r_2} = \frac{3\mu_0 I}{2(r_1/3)} = 9 \left( \frac{\mu_0 I}{2r_1} \right) = 9 B_1$.

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