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

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$1$:$9$

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(A) The magnetic field $B$ at the centre of a circular coil with $N$ turns,radius $r$,and current $I$ is given by $B = \frac{\mu_0 N I}{2r}$.Let the total length of the wire be $L$.For a single turn $(N_1 = 1)$,the circumference is $2 \pi r_1 = L$,so $r_1 = \frac{L}{2 \pi}$.The magnetic field is $B_1 = \frac{\mu_0 (1) I}{2 (L / 2 \pi)} = \frac{\mu_0 I \pi}{L}$.For three turns $(N_2 = 3)$,the total length is $3(2 \pi r_2) = L$,so $r_2 = \frac{L}{6 \pi}$.The magnetic field is $B_2 = \frac{\mu_0 (3) I}{2 (L / 6 \pi)} = \frac{9 \mu_0 I \pi}{L}$.The ratio of the magnetic inductions is $\frac{B_1}{B_2} = \frac{\mu_0 I \pi / L}{9 \mu_0 I \pi / L} = \frac{1}{9}$.

$A$ wire of certain length carries a steady current. It is first bent to form a circular coil of one turn. The same wire is then bent to form a circular coil of three turns. The ratio of magnetic inductions at the centre of the coil in the two cases is

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