Question

State Biot-Savart law. Derive an expression for the intensity of magnetic field at the centre of a current carrying circular loop on its basis.

Answer 1

Biot Savart law states that If there is a current carrying wire and a point $P$ close to it, then the intensity of the magnetic field produced at that point, due to a very small part of the wire is given by

$dB=\frac{{\mu }_{{}_0}}{4\pi }\times \frac{Idlsin\theta }{r^2}$

Here ${\mu }_{{}_0}$ is the magnetic permeability of air or vacuum, $I$ is the current, $dl$ is the small part, $r$ is the line joining the $dl$ and the $P$, and $\theta$ is the smaller angle between $dl$ and $r.$

Using this law we can derive an expression for the intensity of magnetic field at the centre of a current carrying circular loop on its basis.

$B=\int dB=\frac{{\mu }_{{}_0}}{4\pi }\times \int \frac{Idlsin\theta }{r^2}$

Since all elements on the parts (like dl) of the loop are perpendicular to $r$, $\theta$ is ${90}^O$ everywhere. $\therefore sin\theta =1$ everywhere.

The $r$ and $I$ being constants, the come out of the integral.

$\therefore B=\frac{{\mu }_{{}_0}I}{4\pi r^2}\times \int dl$

Here$\int dl$ is $=2\pi r$

Hence $\therefore B=\frac{{\mu }_{{}_0}I\times 2\pi r}{4\pi r^2}$

This is the final equation.

Please note that cancelling $r$, $2$ and $\pi$ is not recommended, because that will make situations complicated.