A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the cent

Question

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

Accepted Answer

1. **Identify the initial magnetic field:** The magnetic field $B$ at the centre of a circular loop of radius $R$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2R}$ .
2. **Relate radius to the number of turns:** Let the total length of the wire be $L$. For one turn, $L = 2\pi R$. For $n$ turns of radius $r$, the same wire length is used, so $L = n(2\pi r)$. Equating the two, we get $2\pi R = n(2\pi r)$, which implies $r = \frac{R}{n}$.
3. **Calculate the new magnetic field ($B'$):** The magnetic field at the centre of a coil with $n$ turns is $B' = \frac{\mu_0 n I}{2r}$ .
4. **Substitute the new radius:**
   $$B' = \frac{\mu_0 n I}{2(R/n)} = n^2 \left( \frac{\mu_0 I}{2R} \right)$$
5. **Conclusion:** Since $B = \frac{\mu_0 I}{2R}$, we find that $B' = n^2 B$.

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