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

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$4 B$

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(D) The magnetic field at the centre of a circular coil with $N$ turns and radius $r$ carrying current $I$ is given by $B = \frac{\mu_0 N I}{2r}$.Initially,$N_1 = 2$ and $B_1 = B = \frac{\mu_0 (2) I}{2r_1} = \frac{\mu_0 I}{r_1}$.When the coil is rewound to have $N_2 = 4$ turns,the total length of the wire remains constant. Since the length $L = N(2\pi r)$,we have $N_1 r_1 = N_2 r_2$.Substituting the values: $2 r_1 = 4 r_2$,which gives $r_2 = r_1 / 2$.The new magnetic field $B_2$ is $B_2 = \frac{\mu_0 N_2 I}{2r_2} = \frac{\mu_0 (4) I}{2(r_1 / 2)} = \frac{4 \mu_0 I}{r_1}$.Comparing $B_2$ with $B_1$: $B_2 = 4 	imes (\frac{\mu_0 I}{r_1}) = 4 B_1 = 4 B$.

$A$ current passing through a circular coil of two turns produces a magnetic field $B$ at its centre. The coil is then rewound so as to have four turns and the same current is passed through it. The magnetic field at its centre now is

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