Question

Write the formula of Biot-Sevart law in vector form. Obtain an expression of magnetic field on the axis of a current carrying circular loop. Draw necessary diagram.

Answer 1

According to biot-savart law, magnetic field caused due to a current carrying element is given by:

$\overrightarrow{dB}=\frac{\mu I\overrightarrow{dl}\times \hat{r}}{4\pi r^2}$

Consider the loop given in the attached figure.

At first, lets focus a small element $dl$ along the loop at $A$. For this, element, applying Biot-Savart Law,

${\overrightarrow{dB}}_A=\frac{{\mu }_{o}I\times \left(dl\hat{k}\right)\times (x\hat{i}-a\hat{j})}{(a^2+x^2{)}^{3/2}}$

Similarly, magnetic field due to a current element of same small element at $B$ is:

${\overrightarrow{dB}}_B=\frac{{\mu }_{o}I\times (-dl\hat{k})\times (x\hat{i}+a\hat{j})}{(a^2+x^2{)}^{3/2}}$

If we add these two components, then first term of each cross product will be cancelled. Hence, we need to integrate only the second term.

So, net magnetic field is:

$\overrightarrow{B}={\int }_0^{2\pi a}\frac{{\mu }_{o}Iadl}{4\pi (a^2+x^2{)}^{3/2}}\hat{i}$

$\overrightarrow{B}=\frac{{\mu }_{o}I{a}^2}{2(a^2+x^2{)}^{3/2}}\hat{i}$

From the figure,

$x=a\cot \varphi$

$\overrightarrow{B}=\frac{{\mu }_{o}I{\sin }^3\varphi }{2a}\hat{i}$