Question

Derive an expression for the magnetic induction at a point on the axis of a current carrying circular coil using Biot-Savart law.

Answer 1

Consider a circular loop with radius R and center O. assuming the plane of the coil to be perpendicular to the plane of

the paper with current I flowing in the direction shown.

Let A be any point on the axis at direction x from the center O.

From figure we see that BA$\perp d{B}^{{}^{\prime }}$ and CA$\perp d{B}^{{}^{\prime }}$----(1)

From we BA=CA=$\sqrt { { r }^{ 2 }+{ x }^{ 2 } }$ {from Pythagoras theorem}

Now according to Biot-Sevart law, magnetic field at A due to the element at B

$dB=\frac{{\mu }_0}{4}\frac{Id{B}^{{}^{\prime }}sin{90}^0}{r^2+x^2}$

where r=radius of the loop and x is the distance from point O to A on the central axis. The net

magnetic field $B=\oint dBsin\varphi$

$=\oint {\mu }_{0}4\pi \left(Id{B}^{{}^{\prime }}{r}^2+x^2\right)\left(r\sqrt{r^2+x^2}\right)$

$={\mu }_{0}4\pi \left(Ir{{(r}^2+x^2)}^{\frac{3}{2}}\right)\oint d{B}^{{}^{\prime }}$

$={\mu }_{0}4\pi \left(Ir{{(r}^2+x^2)}^{\frac{3}{2}}\right)\left(2\pi r\right)$

$B=\frac{{\mu }_{0}I{r}^2}{2{{(r}^2+x^2)}^{\frac{3}{2}}}$

For n number of turns

$B=n{\mu }_{0}I{r}^{2}2{{(r}^2+x^2)}^{\frac{3}{2}}$