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)

$1/r^3$

Answer 2

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 $=\pi R^{2}$ $\therefore B=\frac{\mu_{0}}{2 \pi} \frac{A i}{r^{3}}$ $\Rightarrow B \propto \frac{1}{r^{3}}$