Question

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

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

Answer 1

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

$B=\frac{n{\mu }_{0}i}{2r}$
$I^{st}Case:B_1=\frac{{\mu }_{0}i}{2r}$
$I{I}^{nd}Case:2\pi r=n\cdot 2\times R$
$R=\frac{r}{n}$
$B_2=\frac{n{\mu }_{0}i}{2\frac{r}{n}}$
$=\frac{n^2{\mu }_{0}i}{2r}$
$\frac{B_2}{B_1}=n^2$