Question

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

• A. $1/r$
• B. $1/r^{3/2}$
• C. $1/r^2$
• D. $1/r^3$

Answer 1

Answer: (D)

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,
$B=\int dB\sin \beta$
$=\frac{{\mu }_0}{\mu \pi }\int \frac{idl\sin \theta }{r^2}\sin \beta$
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 $=\pi R^2$
$\therefore B=\frac{{\mu }_0}{2\pi }\frac{Ai}{r^3}$
$\Rightarrow B\propto \frac{1}{r^3}$