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A long wire carries a steady current. It is bent into a coil of one turn such that magnetic induction at centre is $B$, then same wire is bent to from a coil of smaller radius of $n$ turns such that magnetic induction at centre is $B$, then
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Verified Answer
The correct answer is:
$B^{\prime}=n^2 B$
Initially, $B=\frac{\mu_0 I}{2 r}$
For coil with smaller radius $r^{\prime}=\frac{r}{n}$, such that $2 \pi r=n\left(2 \pi r^{\prime}\right)$ :
$\begin{aligned}
& \Rightarrow B^{\prime}=n\left(\frac{\mu_0 I}{2 r^{\prime}}\right)=\frac{\mu_0 n I}{2(r / n)} \\
& \Rightarrow B^{\prime}=n^2 B
\end{aligned}$
For coil with smaller radius $r^{\prime}=\frac{r}{n}$, such that $2 \pi r=n\left(2 \pi r^{\prime}\right)$ :
$\begin{aligned}
& \Rightarrow B^{\prime}=n\left(\frac{\mu_0 I}{2 r^{\prime}}\right)=\frac{\mu_0 n I}{2(r / n)} \\
& \Rightarrow B^{\prime}=n^2 B
\end{aligned}$
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