A long wire carrying a steady current is bent into a circular loop ...

Question

A long wire carrying a steady current is bent into a circular loop of 1 turn. The magnetic field at the centre is $B$. It is then bent into a circular coil of $n$ turns. What will be the magnetic field at the centre of this coil?

Accepted Answer

For 1 turn: radius $R_1$, $B = \frac{\mu_0 I}{2R_1}$

For $n$ turns: same wire length, so $n \cdot 2\pi R_n = 2\pi R_1 \Rightarrow R_n = R_1/n$
$$B_n = \frac{\mu_0 n I}{2R_n} = \frac{\mu_0 n I}{2 \cdot R_1/n} = \frac{\mu_0 n^2 I}{2R_1} = n^2 B$$

The magnetic field becomes $n^2 B$.

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