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 circle loop of n turns. The magnetic field at the centre of the coil will be?

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

Answer 1

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

$B=\frac{{\mu }_{0}I}{2r}$
Total new magnetic field:
$B^{\prime \prime }=n\frac{{\mu }_{0}I}{2r^{\prime }}$
$Newradius\left(r^{\prime }\right)\Rightarrow 2\pi r=n\times 2\pi r^{\prime }$
$r^{\prime }=\frac{r}{n}$
Magnetic field at the centre of one of the new loops:
$B^{\prime }=\frac{r_{0}I}{2r^{\prime }}=\frac{n{r}_{0}I}{2r}$
Total magnetic field at the centre of all the loops:
$B^{\prime \prime }=n{B}^{\prime }$
$=n^2\frac{{\mu }_{0}I}{2r}$
$B^{\prime \prime }=n^{2}B$