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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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The correct answer is:
$9 B_{0}$
The magnetic field produced at the centre of the circular coil carrying current is given by
$$
B V=\frac{\mu_{0} N I}{2 r}
$$
For one turn
$$
\begin{aligned}
&N=1 \\
&B_{0}=\frac{\mu_{0} I}{2 r}
\end{aligned}
$$
As the coil is rewound
$$
\begin{aligned}
r^{\prime} &=\frac{r}{3}, N^{\prime}=3 \\
\therefore \quad B^{\prime} &=\frac{\mu_{0} I \times 3}{2 \times\left(\frac{r}{3}\right)}=\frac{9 \mu_{0} I}{2 r}=9 B_{0}
\end{aligned}
$$
$$
B V=\frac{\mu_{0} N I}{2 r}
$$
For one turn
$$
\begin{aligned}
&N=1 \\
&B_{0}=\frac{\mu_{0} I}{2 r}
\end{aligned}
$$
As the coil is rewound
$$
\begin{aligned}
r^{\prime} &=\frac{r}{3}, N^{\prime}=3 \\
\therefore \quad B^{\prime} &=\frac{\mu_{0} I \times 3}{2 \times\left(\frac{r}{3}\right)}=\frac{9 \mu_{0} I}{2 r}=9 B_{0}
\end{aligned}
$$
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