Question

State Biot – Savart law. Deduce the expression for the magnetic field at a point on the axis of a current carrying circular loop of radius ‘R’ at a distance ‘x’ from the centre. Hence, write the magnetic field at the centre of a loop.

Answer 1

Biot–Savart’s law

According to Biot–Savart’s law, the magnitude of the magnetic field dB is proportional to the current I and the element length dl, and inversely proportional to the square of the distance r.

$dB\propto \frac{ldl\sin \theta }{r^2}$

$dB=\left(\frac{{\mu }_{\circ }}{4\mathrm{\pi }}\right)\left(\frac{ldl\sin \theta }{r^2}\right)$

Here, $\left(\frac{\mu \circ }{4\pi }\right)$ is a constant of proportionality

In the vector form, it is given by,

$\overrightarrow{dB}=\left(\frac{{\mu }_{\circ }}{4\mathrm{\pi }}\right)\left(\frac{l\overrightarrow{dl}\times \overrightarrow{r}}{r^3}\right)$

Deducing the expression for the magnetic field

Consider a circular loop carrying a steady current I. The loop is placed in the y–z plane with its center at origin O and has a radius R.

Let x be the distance of point P from the centre of the loop where the magnetic field is to be calculated. Consider a conducting element dl of the loop. The magnitude dB of the magnetic field due to dl is given by the Biot–Savart’s law as

$dB=\left(\frac{{\mu }_{\circ }}{4\mathrm{\pi }}\right)\left(\frac{ldl\sin \theta }{r^2}\right)$

From the figure, we see that $r^2=x^2+R^2$

Any element of the loop will be perpendicular to the displacement vector from the element to the axial point. Hence, we have θ=900 or sinθ=1.

Thus, we have

$dB=\left(\frac{{\mu }_{\circ }}{4\mathrm{\pi }}\right)\left(\frac{ldl}{r^2}\right)=\left(\frac{{\mu }_{\circ }}{4\mathrm{\pi }}\right)\left(\frac{ldl}{x^2+R^2}\right).......\left(1\right)$

The direction of dB is perpendicular to the plane formed by dl and r. It has an x-component dBx and a component perpendicular to x-axis dB_⊥

The perpendicular components cancel each other when summed over. Therefore, only the x-component contributes. The net contribution is obtained by integrating dB_x=dBcosθ

dB_x=dBcosθ= $\left(\frac{{\mu }_{\circ }}{4\mathrm{\pi }}\right)\left(\frac{ldl}{x^2+R^2}\times \frac{R}{\sqrt{\left(x^2+R^2\right)}}\right)=\left(\frac{{\mathrm{\mu }}_{\circ }\mathrm{ldl}}{4\mathrm{\pi }}\right)\times \left(\frac{R}{{\left(x^2+R^2\right)}^{{\displaystyle \frac{3}{2}}}}\right)$

From the figure, we see that the summation of dl yields the circumference of the loop 2πR. Hence, the magnetic field at point P caused by the entire loop is

$B=B_x\hat{i}=\frac{{\mu }_{\circ }I\left(2\mathrm{\pi R}\right)}{4\mathrm{\pi }}\times \frac{R}{\left({\left(x^2+R^2\right)}^{{\displaystyle \frac{3}{2}}}\right)}\hat{i}$
$B=\frac{{\mu }_{\circ }I{R}^2}{\left(2{\left(x^2+R^2\right)}^{{\displaystyle \frac{3}{2}}}\right)}\hat{i}$

Case: At the centre x = 0, so we have

$B=\frac{{\mu }_{\circ }I{R}^2}{\left(2{\left(R^2\right)}^{{\displaystyle \frac{3}{2}}}\right)}\hat{i}=\frac{{\mu }_{\circ }I{R}^2}{2R^3}\hat{i}=\frac{{\mu }_{\circ }I}{2R}\hat{i}.$