Question

Use Biot-Savart law to derive the expression for the magnetic field on the axis of a current carrying circular loop of radius R.

Draw the magnetic field lines due to a circular wire carrying current I.

Answer 1

Imagine a circular coil of radius R with centre O.Let the current flowing through the circular loop be I. suppose P is any point on the axis at a distance of r from the centre 0. Let the circular coil be made up of a large number of small elements of current, each having a length of dl.
According to Biot-Savart's law, the magnetic field at Point P will be

$dB=\frac{{\mu }_{0}I}{4\pi }\times \frac{|dl\times r|}{r^3}$

where, $r^2=x^2+R^2$

$|dl\times r|=rdl$ [ $\because$ Both are perpendicular]

Here, r is the position vector of point O from the current element.


dB has two components i.e.,$d{B}_{x}andd{B}_y$.

$d{B}_y$ is cancelled out and only the x-component remains.

$\therefore d{B}_x=dBcos\theta$

$cos\theta =\frac{R}{\sqrt{x^2+R^2}}$

$d{B}_x=\frac{{\mu }_{0}Idl}{4\pi }.\frac{R}{{(x^2+R^2)}^{\frac{3}{2}}}$

But, $\int dl=2\pi R$

So, $B=\frac{{\mu }_{0}IR\times 2\pi R}{4\pi {(x^2+R^2)}^{\frac{3}{2}}}$

For n turns in the circular loop,

$\overrightarrow{B}=\frac{{\mu }_{0}nI{R}^2}{2{(x^2+R^2)}^{\frac{3}{2}}}.\hat{i}$