Question

Derive the expression for magnetic field at a point on the axis of a circular current carrying loop.

Answer 1

In the above diagram,

Let O be the centre of the ring, R be the radius of the ring. The point P lie outside the ring on the axis, let the distance between the P and o be $x$, where we have to calculate the magnetic field.

Let $\overrightarrow{dB}$ be the magnetic field due to a small length $dl$ of the ring,

As per bio-severt rule,

$\overrightarrow{dB}=\frac{{\mu }_{o}Idl}{4\pi r^2}$

We know that $r=\sqrt{(x^2+R^2)}$

So $\overrightarrow{db}=\frac{{\mu }_{o}Idl}{4\pi \sqrt{x^2+R^2}}$

This field have vertical${\overrightarrow{dB}}_y=\overrightarrow{dB}sin\theta$ and horizontal component${\overrightarrow{dB}}_x=\overrightarrow{dB}cos\theta$,

Vertical component will cancel out each other, only horizontal components are responsible,

So, $\overrightarrow{B}={\overrightarrow{dB}}_1+{\overrightarrow{dB}}_2+.....$

$\overrightarrow{dB}=\frac{{\mu }_{o}Idlsin\theta }{4\pi \sqrt{R^2+x^2}}$

$\overrightarrow{dB}=\frac{{\mu }_{o}Ix}{4\pi (R^2+x^2{)}^{\frac{3}{2}}}$

The magnetic field due to the circular current loop of radius a at a point which is a distance R away, and is on its axis,

So $\overrightarrow{B}=\frac{{\mu }_{o}I{x}^2}{2(R^2+x^2{)}^{\frac{3}{2}}}$