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

\[B = \frac{\mu_0 i}{4\pi d}(\sin\theta + \sin\theta)\]

\[B = \frac{\mu_0 i}{4\pi d}\frac{2x}{\sqrt{x^2 + 4 d^2}}\]

So, if  d > > x (neglecting x), then

 

\[B = \frac{\mu_0 i}{4\pi d}\frac{2x}{2d}\]
\[ \Rightarrow B \propto \frac{1}{d^2}\]

And, if d < < x (neglecting d), then

 

\[B = \frac{\mu_0 i}{4\pi d}\frac{2x}{x}\]
\[ \Rightarrow B \propto \frac{1}{d}\]