Question

(a) Write using Biot - Savart Law, the expression for magnetic field $\overrightarrow{B}$ due to an element $\overrightarrow{dl}$ carrying current $I$ at a distance $\overrightarrow{r}$ from it in a 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.,
$\overrightarrow{B}.\overrightarrow{dl}={\mu }_{o}I$
where $I$ is the total current passing through the surface.

Answer 1

(a)

According to Biot-Savart Law, magnetic field due to a current carrying element dl is:

$\overrightarrow{dB}=\frac{{\mu }_{o}I}{4\pi r^3}(\overrightarrow{dl}\times \overrightarrow{r})$

The magnetic field due to a current carrying circular loop is shown in the figure. Vertical components of the magnetic field cancel out by the diametrically opposite end on the wire.

Horizontal component due to an element dl is given by:

$dB=\frac{{\mu }_{o}I}{4\pi r^2}sin\theta dl$

$dB=\frac{{\mu }_{o}IR}{4\pi r^3}dl$

$\int dB=\frac{{\mu }_{o}IR}{4\pi r^3}\int dl$

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

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

(b)

From Biot-Savart Law, it can be concluded that the magnetic field around an infinitely long current carrying conductor is

$B=\frac{{\mu }_{o}I}{2\pi r}$

or $B\times 2\pi r={\mu }_{o}I$

$\int B.dl={\mu }_{o}I$