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(B) The magnetic field $B$ at the center of a circular coil of $N$ turns and radius $r$ carrying current $I$ is given by $B = \frac{\mu_0 N I}{2r}$.Le

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$1 : 4$

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(B) The magnetic field $B$ at the center of a circular coil of $N$ turns and radius $r$ carrying current $I$ is given by $B = \frac{\mu_0 N I}{2r}$.Let $L$ be the total length of the wire.For the first coil $(N_1 = 1)$: The circumference $2\pi r_1 = L$,so $r_1 = \frac{L}{2\pi}$. The magnetic field is $B_1 = \frac{\mu_0 (1) I}{2(L/2\pi)} = \frac{\mu_0 \pi I}{L}$.For the second coil $(N_2 = 2)$: The total length $2(2\pi r_2) = L$,so $r_2 = \frac{L}{4\pi}$. The magnetic field is $B_2 = \frac{\mu_0 (2) I}{2(L/4\pi)} = \frac{4\mu_0 \pi I}{L}$.Taking the ratio: $\frac{B_1}{B_2} = \frac{\mu_0 \pi I / L}{4\mu_0 \pi I / L} = \frac{1}{4}$.Thus,the ratio is $1 : 4$.

$A$ coil of one turn is made of a wire of a certain length,and then from the same length of wire,a coil of two turns is made. If the same current is passed in both cases,what is the ratio of the magnetic inductions at their centers?

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