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

Explanation:

To show: for d >> x, the magnetic field at P varies as 1/d2 whereas for d << x, it varies as 1/d

Step 1:

Let AB be the length x wire with midpoint O.

Current Magnitude = i

Divide the point from wire = d

The magnetic field given on a perpendicular bisector is

$B=\frac{\mu_{0} i}{4 \pi d}(\sin \theta+\sin \theta)$

$B=\frac{\mu_{0} i}{4 \pi d} \frac{2 x}{\sqrt{x^{2}+4 d^{2}}}$

Step 2:

Case i) So,if  d > > x ( neglecting x), then  

$B=\frac{\mu_{0} i}{4 \pi d} \frac{2 x}{2 d}$

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

Case ii) And,if d < < x ( neglecting d ),then  

$B=\frac{\mu_{0} i}{4 \pi d} \frac{2 x}{x}$

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

Hence showed.

Answer 2

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