Question

A steady current flows in a long wire. It is bent into a circular loop of one turn and the magnetic field at the centre of the coil is $B$. If the same wire is bent into a circular loop of $n$ turns, the magnetic field at the centre of the coil is:

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

Answer 1

Answer: (D)

$n^{2}B$

Let the length of the wire be $L$ and radius of the coil be $r$

$\therefore$ $L=2\pi r$

Magnetic field at the centre of coil $B=\frac{{\mu }_{o}I}{2r}$ ........(1)

Now the same wire is bent into circular loops of n turns.

$\therefore$ $L=n2\pi r^{\prime }$

$2\pi r=2n\pi r^{\prime }$ $⟹r^{\prime }=\frac{r}{n}$ ........(2)

Magnetic field at the centre of circular loops of n turns $B^{\prime }=\frac{{\mu }_{o}I}{4\pi r^{\prime }}\times 2\pi n=\frac{{\mu }_{o}I}{2r^{\prime }}n$

Using equation (2), we get $B^{\prime }=\frac{{\mu }_{o}I}{2\frac{r}{n}}n=\frac{{\mu }_{o}I}{2r}{n}^2$

$⟹$ $B^{\prime }=n^{2}B$