Thank you for reporting, we will resolve it shortly
Q.
A circular coil of radius $R$ carries an electric current. The magnetic field due to the coil at a point on the axis of the coil located at a distance $r$ from the center of the coil, such that $r > > R$, varies as
For a circular coil, the component of the field $B$ perpendicular to the axis at $P$ cancel each other while along the axis add up.
The resultant magnetic field at point $P$ will be due to the components along the axis. Hence,
$B=\int d B \sin \beta$
$=\frac{\mu_{0}}{\mu \pi} \int \frac{i d l \sin \theta}{r^{2}} \sin \beta$
and as here angle $\theta$ between the element $\overrightarrow{dl}$ and $\vec{ r }$ is $\frac{\pi}{2}$ every where and $r$ is same for all elements while $\sin \beta=\frac{R}{r}$, so
Hence, we have
$B =\frac{\mu_{0}}{4 \pi} \frac{2 \pi i R^{2}}{x^{3}}$
where $ x=\left(R^{2}+r^{2}\right)^{1 / 2}$
$B=\frac{\mu_{0}}{4 \pi} \frac{2 \pi i R^{2}}{\left(R^{2}+r^{2}\right)^{3 / 2}}$
Given, $r > > R$ then we have, neglecting $R$,
$B=\frac{\mu_{0}}{4 \pi} \frac{2 \pi i R^{2}}{r^{3}}$
Also area $=\pi R^{2}$
$\therefore B=\frac{\mu_{0}}{2 \pi} \frac{A i}{r^{3}} $
$ \Rightarrow B \propto \frac{1}{r^{3}}$
