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(B) The magnetic field $B$ at the centre of a circular coil with $N$ turns,radius $r$,and current $I$ is given by $B = \frac{\mu_{0} N I}{2 r}$.Initia

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$9 B_{0}$

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(B) The magnetic field $B$ at the centre of a circular coil with $N$ turns,radius $r$,and current $I$ is given by $B = \frac{\mu_{0} N I}{2 r}$.Initially,for $N=1$ turn,the magnetic field is $B_{0} = \frac{\mu_{0} I}{2 r}$.When the coil is rewound to have $N' = 3$ turns using the same length of wire,the new radius $r'$ becomes $r' = \frac{r}{3}$.The new magnetic field $B'$ at the centre is $B' = \frac{\mu_{0} N' I}{2 r'} = \frac{\mu_{0} (3) I}{2 (r/3)}$.Simplifying this,we get $B' = \frac{9 \mu_{0} I}{2 r} = 9 B_{0}$.

$A$ circular coil carrying a certain current produces a magnetic field $B_{0}$ at its centre. The coil is now rewound so as to have $3$ turns and the same current is passed through it. The new magnetic field at the centre is

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