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/d² whereas for d << x, it varies as 1/d.

Answer 1

Let AB be the wire of length x with midpoint O.
Given:
Magnitude of current = i
Separation of the point from the wire = d

Now,
The magnetic field on a perpendicular bisector is given by
$B=\frac{{\mathrm{\mu }}_{0}i}{4\mathrm{\pi }d}(\sin \theta +\sin \theta )$
$B=\frac{{\mathrm{\mu }}_{0}i}{4\mathrm{\pi }d}\frac{2x}{\sqrt{x^2+4d^2}}$
So, if d > > x (neglecting x), then
$$\begin{gathered} B=\frac{{\mathrm{\mu }}_{0}i}{4\mathrm{\pi }d}\frac{2x}{2d} \\ \Rightarrow B\propto \frac{1}{d^2} \end{gathered}$$
And, if d < < x (neglecting d), then
$$\begin{gathered} B=\frac{{\mathrm{\mu }}_{0}i}{4\mathrm{\pi }d}\frac{2x}{x} \\ \Rightarrow B\propto \frac{1}{d} \end{gathered}$$