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A steady current flows in a long wire. It is bent into a circular loop of one turn and the magnetic field at the centre of the coil is $B$. If the same wire is bent into a circular loop of $n$ turns, the magnetic field at the centre of the coil is
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Verified Answer
The correct answer is:
$n^2 B$
The magnetic field a circular coil (for $n$ number of loops)
$$
\begin{gathered}
B=\frac{\mu_0 n i}{2 r} \\
n_1=1, n_2=2 \\
2 \pi r_1=n \times 2 \pi r_2
\end{gathered}
$$
When,
So $\quad \begin{aligned} \frac{r_1}{r_2} & =\frac{n}{1} \\ \frac{B_1}{B_2} & =\frac{\mu_0 n_1 i / 2 r_1}{\mu_0 n_2 i / 2 r_2} \\ \frac{B}{B_2} & =\frac{n_1}{n_2} \cdot \frac{r_1}{r_2} \\ \frac{B}{B_2} & =\frac{1}{n} \cdot \frac{1}{n} \\ \Rightarrow \quad B_2 & =n^2 B\end{aligned}$
$$
\begin{gathered}
B=\frac{\mu_0 n i}{2 r} \\
n_1=1, n_2=2 \\
2 \pi r_1=n \times 2 \pi r_2
\end{gathered}
$$
When,
So $\quad \begin{aligned} \frac{r_1}{r_2} & =\frac{n}{1} \\ \frac{B_1}{B_2} & =\frac{\mu_0 n_1 i / 2 r_1}{\mu_0 n_2 i / 2 r_2} \\ \frac{B}{B_2} & =\frac{n_1}{n_2} \cdot \frac{r_1}{r_2} \\ \frac{B}{B_2} & =\frac{1}{n} \cdot \frac{1}{n} \\ \Rightarrow \quad B_2 & =n^2 B\end{aligned}$
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