Question

Apply Biot-Savert law to a point on the axis of a current carrying circular loop.

Answer 1

Consider a circular coil of radius R carrying a current I. Let P be a point on its axis at a distance x from the centre O.

According to the Biot-Savart law, the magnetic field due to a small current element \[I d\vec l\] is

\[dB=\frac{\mu_0}{4\pi}\frac{I\,dl\sin\theta}{r^2}\]

For a point on the axis of the circular loop,

θ = 90°

Hence,

dB = \[\frac{\mu_0}{4\pi}\frac{I\,dl}{r^2}\]

From geometry,

r = \[\sqrt{R^2+x^2}\]

The magnetic field dB can be resolved into two components. The components perpendicular to the axis cancel due to the diametrically opposite current elements, while the components along the axis add.

If α is the angle such that

sin⁡ α = \[\frac{R}{r}\]

then the axial component is

\[dB_x=dB\sin\alpha\]

Therefore,

\[dB_x= \frac{\mu_0}{4\pi} \frac{I\,dl}{r^2} \frac{R}{r}\]

\[dB_x= \frac{\mu_0}{4\pi} \frac{IR\,dl}{r^3}\]

For the complete circular loop,

B = \[\frac{\mu_0 IR}{4\pi r^3} \int dl\]

Since

∫dl =2 π R

we get

B = \[\frac{\mu_0 I R^2}{2r^3}\]

Substituting

r² = R² + x²

we get

B = \[\frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}\]

This is the magnetic field at a point on the axis of a current-carrying circular loop.

For a coil of N turns,

B = \[\frac{\mu_0 N I R^2} {2(R^2+x^2)^{3/2}}\]

The direction of the magnetic field is along the axis of the loop and is given by the right-hand thumb rule.