Question

(a) Write using Biot-Savart law, the expression for the magnetic field $\overrightarrow{B}$ due to an element $\overrightarrow{dl}$ carrying current I at a distance $\overrightarrow{r}$ from it in vector form.
Hence derive the expression for the magnetic field due to a current carrying loop of radius R at a point P distant X from its centre along the axis of the loop.

(b) Explain how Biot-Savart law enables one to express the Ampere's circuital law in the integral form, viz.,
$\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_{0}I$
Where I is the total current passing through the surface.

Answer 1

(a) Suppose we have a conductor of length l in which current i is flowing. We need to calculate the magnetic field at a point P in vacuum. If $i\overrightarrow{dl}$ is one of the infinitely small current element, the magnetic field $\overrightarrow{dB}$ at point P is given by



$\overrightarrow{dB}\propto \frac{i\overrightarrow{dl}\times \overrightarrow{r}}{r^3}$

$\overrightarrow{dB}=\frac{{\mu }_0}{4\pi }\frac{i\overrightarrow{dl}\times \overrightarrow{r}}{r^3}$

Where $\frac{{\mu }_0}{4\pi }$ is a proportionality constant.

Suppose there is a circular coil of radius R, carrying a current i. Let P be a point at the axis of the coil at a distance x from the centre, at which the field in required.



Consider a conducting element dl of the loop. The magnetic field due to dl is given by the Biot-Savart law,

$dB=\frac{{\mu }_0}{4\pi }\frac{i|\overrightarrow{dl}\times \overrightarrow{r}|}{r^3}$

$dB=\frac{{\mu }_0}{4\pi }\frac{idl}{(R^2+x^2)}$

The direction of dB is perpendicular to the plane formed by dl and r. It has an X-component $d{B}_x$, and a component perpendicular to X-axis, $d{B}_{\perp }$. When the components perpendicular to the X-axis are summed over, they cancel out and we obtain null result. Thus only the X-component survives.

So, the resultant field $\overrightarrow{B}$ at P given by

$B=\int dBsin\theta$

$B=\frac{{\mu }_0}{4\pi }\frac{i}{r^2}\int dlsin\theta$

$B=\frac{{\mu }_0}{4\pi }\frac{iR}{r^3}\int dl[\because sin\theta =\frac{R}{r}]$

But $\int dl=2\pi R$ and $r=(R^2+x^2{)}^{1/2}$

$\therefore B=\frac{{\mu }_0}{4\pi }\frac{2\pi i{R}^2}{(R^2+x^2{)}^{3/2}}$

If the coil has N turns, then each turn will contribute equally to B. Then,

$B=\frac{{\mu }_{0}Ni{R}^2}{2(x^2+R^2{)}^{3/2}}$

(b) According to Biot-Savart law the line integral of the magnetic field $\overrightarrow{B}$ around any 'closed' path is equal to ${\mu }_0$ times the net current I threading through the area enclosed by the path.

i.e. $\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_{0}I$

Where ${\mu }_0$ in the permeability of free space. Ampere's circuital law in electromagnetism is analogous to Gauss' law in electrostatics.