Question

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

• A. $nB$
• B. $n^2 B$
• C. $2nB$
• D. $2n^2 B$

Answer 1

Answer: (B)

The circumference of the first loop is the length of the wire $=2 \pi R$
The magnetic field at the center of this wire $=B=\frac{\mu_{0} I}{2 R}$
The same wire is bent into $n$ circular coils.
Thus new radius is found out by
$n \times 2 \pi r =2 \pi R$
$\Longrightarrow r=\frac{R}{n}$
Thus magnetic field at the center of these loops $=n \frac{\mu_{0} I}{2 r}$
$=\frac{n^{2} \mu_{0} I}{2 R}=n^{2} B $