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(A) The magnetic field at the center of a circular coil of $N$ turns, radius $R$, and current $I$ is given by $B = \frac{\mu_0 NI}{2R}$.For a single t

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$4B$

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(A) The magnetic field at the center of a circular coil of $N$ turns, radius $R$, and current $I$ is given by $B = \frac{\mu_0 NI}{2R}$.For a single turn $(N=1)$ of wire of length $L$, the circumference is $2\pi R = L$, so $R = \frac{L}{2\pi}$.Thus, $B = \frac{\mu_0 (1) I}{2(L/2\pi)} = \frac{\mu_0 \pi I}{L}$.When the same wire is bent into $N'=2$ turns, the new radius $R'$ satisfies $2\pi R' = L/2$, so $R' = \frac{L}{4\pi} = \frac{R}{2}$.The new magnetic field $B'$ is $B' = \frac{\mu_0 N' I}{2R'} = \frac{\mu_0 (2) I}{2(R/2)} = 4 \left( \frac{\mu_0 I}{2R} \right) = 4B$.

The magnetic field at the center of a circular coil carrying current '$I$' for a single turn of a given length of wire is '$B$'. The same wire is bent into a circular coil having two turns. When the same current '$I$' passes through it, the value of the magnetic field becomes:

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