Question

A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be:

Answer 1

Magnetic field Q centre of a coil with n turns is given by
$B=\frac{{\mu }_{0}nI}{2a}$
For a coil of radius R and 1 turn,
$B=\frac{{\mu }_{0}I}{2R}$
When coil is made of n turns, its radius becomes ‘r’
Total length of wire is constant
$\therefore$ (Perimeter of) = (Perimeter of n loops)
$\therefore 2vR=n\times 2\pi r$
$\therefore r=\frac{R}{n}$
The magnetic field Q centre will be
$B_1=\frac{{\mu }_{0}nI}{2r}=\frac{{\mu }_{0}nI}{2\left(\frac{R}{n}\right)}=n^2\left(\frac{{\mu }_{0}I}{2R}\right)$
$\therefore B_1=n^{2}B$