Question

When a long wire carrying a steady current is bent into a circular coil of one turn, the magnetic induction at its centre is B. When the same wire carrying the same current is bent to form a circular coil of n turns of a smaller radius, the magnetic induction at the centre will be

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

Answer 1

The correct option is D

$n^{2}B$

$B=\frac{{\mu }_{0}I}{2r}$. For a coil of n turns, $2\pi r=n\left(2\pi r^{\prime }\right)$
or $r^{\prime }=\frac{r}{n}$, where r' is the radius of the coil of n turns.
$\therefore B^{\prime }=\frac{n{\mu }_{0}I}{2r^{\prime }}=\frac{n{\mu }_{0}I}{\frac{2r}{n}}=n^{2}B$
Hence the correct choice is (d)