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(B) Let the length of the wire be $L$. The circumference of a coil with $N$ turns and radius $R$ is $L = N(2\pi R)$.For coil $P$: $L = N_P(2\pi R_P) =

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$4$

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(B) Let the length of the wire be $L$. The circumference of a coil with $N$ turns and radius $R$ is $L = N(2\pi R)$.For coil $P$: $L = N_P(2\pi R_P) = 4(2\pi R_P) \implies R_P = \frac{L}{8\pi}$.For coil $Q$: $L = N_Q(2\pi R_Q) = 2(2\pi R_Q) \implies R_Q = \frac{L}{4\pi}$.The magnetic field at the center of a circular coil is $B = \frac{\mu_0 N I}{2R}$.Assuming the same current $I$ flows through both coils:$B_P = \frac{\mu_0 N_P I}{2 R_P} = \frac{\mu_0 (4) I}{2 (L/8\pi)} = \frac{16\pi \mu_0 I}{L}$.$B_Q = \frac{\mu_0 N_Q I}{2 R_Q} = \frac{\mu_0 (2) I}{2 (L/4\pi)} = \frac{4\pi \mu_0 I}{L}$.Therefore,the ratio is $\frac{B_P}{B_Q} = \frac{16\pi \mu_0 I / L}{4\pi \mu_0 I / L} = \frac{16}{4} = 4$.

Two circular coils $P$ and $Q$ are made from two identical wires of the same length. The number of turns in coils $P$ and $Q$ are $4$ and $2$,respectively. The magnetic inductions at the centers of $P$ and $Q$ are $B_P$ and $B_Q$,respectively. The ratio $\frac{B_P}{B_Q}$ is

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