Question

A long straight wire, of circular cross-section (radius = a) carries a current I which is uniformly distributed across the cross-section of the wire. 

Use Ampere's circuital law to calculate magnetic field B(r), due to this wire, at a Point distance r < a and r > a from its axis. Draw a graph Showing the dependence of B (r) on r.

Answer 1

As per Ampere's circuital law:

\(\oint \vec B.\vec {dl} = \mu_0I\)

(i) For r < a

I_e = current enclosed by Amperian circuit loop of radius r.

\(= I. \frac{\pi r^2}{\pi a^2}\)

\(= \frac{Ir^2}{ a^2}\)

\(\therefore B\oint \vec {dl} = \mu _0. I_e\)

\(= \frac{\mu_0Ir^2}{a^2}\)

Or \(B = \frac{\mu_0 Ir^2}{a^2} . (\frac 1{2\pi r})\)

\(= \left(\frac{\mu_0}{2\pi} \frac I{a^2}\right)\)

(ii) For r > a

\(\oint \vec B.\vec {dl} = \mu_0I\)

\(\therefore B.2\pi r = \mu_0I\)

Or \(B = \frac{\mu _0I}{2\pi r}\)

The graph of, B(r) vs r, is as shown

Answer 2

As per Ampere's circuital law:

(i) For r < a

I_e = current enclosed by Amperian circuit loop of radius r.

The graph of, B(r) vs r, is as shown