Question

A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is $B$. It is then bent into a circular coil of $n$ turns. The magnetic field at the centre of the coil will be

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

Answer 1

The correct option is B $n^{2}B$

The magnetic field at the centre of a coil of n turns and radius $R$ carrying a current $I$ is

$B=\frac{{\mu }_{0}nI}{2R}$
For $n=1,B=\frac{{\mu }_{0}I}{2R}$

When the wire is bent into a circular coil of $n$ turns, the radius of coil becomes

$R^{\prime }=\frac{L}{2\pi n}=\frac{R}{n}$

The magnetic field at the centre of coil will be

$B^{\prime }=\frac{{\mu }_{0}nI}{2R^{\prime }}$

$=\frac{{\mu }_{0}nI}{2R/n}$

$=\frac{{\mu }_{0}I{n}^2}{2R}=n^{2}B$