Question

Consider a straight piece of length $x$ of a wire carrying a current $i$. Let $P$ be a point on the perpendicular bisector of the piece, situated at a distance $d$ from its middle point. Show that for $d>>x$, the magnetic field at $P$ varies as $1/d^2$ whereas for $d<<x$, it varies as $1/d$.

Answer 1

$B=\frac{{\pi }_{0}i}{4\pi d}2\sin \theta$

$=\frac{{\pi }_{0}i}{4\pi d}\frac{2\times x}{2\times \sqrt{d^2+\frac{x^2}{4}}}=\frac{{\mu }_{0}ix}{4\pi d\sqrt{d^2+\frac{x^2}{4}}}$

a) When $d>>x$

Neglecting x w.r.t. d

$B=\frac{{\mu }_{0}ix}{\mu xd\sqrt{d^2}}=\frac{{\mu }_{0}ix}{\mu \pi d^2}$

$\therefore B\propto \frac{1}{d^2}$

b) When $x>>d$, neglecting d w.r.t. x

$B=\frac{{\mu }_{0}ix}{4\pi dx/2}=\frac{2{\mu }_{0}i}{4\pi d}$

$\therefore B\propto \frac{1}{d}$