Question

Derive an expression for magnetic field strength at any point on the axis of a circular current carrying loop using Biot-Savart's law.

Answer 1

Biot-Savart's law : Magnetic field at an axial point P due to a current element of the ring $dB=\frac{{\mu }_o}{4\pi }\frac{i(\overrightarrow{dl}\times \hat{r})}{r^2}$

where $r=\sqrt{R^2+x^2}$

$\therefore$ We get $B=\frac{{\mu }_o}{4\pi }\frac{idl}{(R^2+x^2)}$

We resolve $dB$ into vertical and horizontal components. Now all the vertical components cancel out each other and so only the horizontal components survive which results in the net magnetic field at P in the horizontal direction.

Net magnetic field $B=\int dB\sin \theta$

$B=\int \frac{{\mu }_{o}idl}{4\pi (R^2+x^2)}\frac{R}{(R^2+x^2{)}^{1/2}}$

Or $B=\frac{{\mu }_{o}iR}{4\pi (R^2+x^2{)}^{3/2}}\int dl$

Or $B=\frac{{\mu }_{o}iR}{4\pi (R^2+x^2{)}^{3/2}}\left(2\pi R\right)$

$⟹$ $B=\frac{{\mu }_{o}i{R}^2}{2(R^2+x^2{)}^{3/2}}$