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

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Four times of its first value

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(C) The magnetic field $B$ at the centre of a circular coil of $N$ turns and radius $r$ carrying current $I$ is given by $B = \frac{\mu_0 N I}{2r}$.Since the length of the wire $L$ is constant,$L = 2\pi r_1 N_1 = 2\pi r_2 N_2$.For the first case,$N_1 = 1$,so $L = 2\pi r_1$,which implies $r_1 = \frac{L}{2\pi}$.For the second case,$N_2 = 2$,so $L = 2\pi r_2 	imes 2$,which implies $r_2 = \frac{L}{4\pi} = \frac{r_1}{2}$.The magnetic field in the first case is $B_1 = \frac{\mu_0 (1) I}{2r_1}$.The magnetic field in the second case is $B_2 = \frac{\mu_0 (2) I}{2r_2} = \frac{\mu_0 (2) I}{2(r_1/2)} = \frac{4 \mu_0 I}{2r_1} = 4 B_1$.Thus,the magnetic field becomes four times its first value.

$A$ length $L$ of wire carries a steady current $I$. It is bent first to form a circular plane coil of one turn. The same length is now bent more sharply to give a double loop of smaller radius. The magnetic field at the centre caused by the same current is

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