Question

Derive An Expression For Magnetic Field Strength At Any Point On The Axis Of A Circular Current Carrying Loop Using Biot-savart Law.

Answer 1

Biot-Savart’s law: Magnetic field at an axial point P due to a current element dl of the ring.

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

$ordB=\frac{{\mu }_0}{4\pi }\frac{idl\sin \theta }{r^2}$

$ordB=\frac{{\mu }_0}{4\pi }\frac{idl\sin 90}{r^2}(\because \theta =90°)$

From the fig. $r=\sqrt{R^2+X^2}$

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

We resolve dB into vertical and horizontal components.

Now all the vertical components cancel out each other and so only the horizontal components survive, which results in the net magnetic field at P in the horizontal direction.

Net magnetic field $B=\int dB\sin \theta$

$$\begin{gathered} orB=\int \frac{{\mu }_{0}idl}{4\mathrm{\pi }({\mathrm{R}}^2+{\mathrm{x}}^2)}.\frac{R}{{(R^2+x^2)}^{1/2}} \\ orB=\frac{{\mu }_{0}i}{4\mathrm{\pi }(R^2+{\mathrm{x}}^2)}\frac{R}{{(R^2+x^2)}^{1/2}}\int dl \\ orB=\frac{{\mu }_{0}iR}{4\mathrm{\pi }{(R^2+{\mathrm{x}}^2)}^{3/2}}\left(2\mathrm{\pi R}\right) \\ \mathrm{or}\mathrm{B}=\frac{{\mathrm{\mu }}_0{\mathrm{iR}}^2}{2{({\mathrm{R}}^2+{\mathrm{x}}^2)}^{3/2}} \end{gathered}$$