Question

A current carrying wire produces a magnetic field in its surrounding space. With the help of a diagram, derive an expression for the magnetic field at a point on the axis of a circular current loop.

Answer 1

We need to find out the magnetic field at a point P due to the current carrying loop.

Magnetic field at P due to current element $dl$, $dB=\frac{{\mu }_{o}Idl}{4\pi r^2}$

From figure, $r=\sqrt{R^2+x^2}$

$⟹$ $dB=\frac{{\mu }_{o}Idl}{4\pi (R^2+x^2)}$

Now, the vertical components of $dB$ cancel out each other and only the horizontal components survive.

Thus net magnetic field at P must be in horizontal direction.

Net magnetic field at P, $B_p=\int dB\cos \theta$

From figure, $\cos \theta =\frac{R}{\sqrt{R^2+x^2}}$

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

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

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