Example

Question:

Figure 4.13 shows a long straight wire of a circular cross-section (radius \(a\)) carrying steady current \(I\). The current \(I\) is uniformly distributed across this cross-section. Calculate the magnetic field in the region \(r < a\) and \(r > a\).

Solution:

(a) For \(r > a\):
The Amperian loop is concentric with the cross-section. The current enclosed by the loop is \(I\).
The magnetic field at distance \(r\) (outside the wire) is \[ B = \frac{\mu_0 I}{2\pi r} \] so \( B \propto \frac{1}{r} \) for \( r > a \). (b) For \( r < a \) (inside the wire):
The current enclosed is only a fraction of total current: \[ I_e = I \frac{\pi r^2}{\pi a^2} = I\frac{r^2}{a^2} \] By Ampère’s law, \[ B(2\pi r) = \mu_0 I \frac{r^2}{a^2} \implies B = \frac{\mu_0 I}{2\pi a^2} r \] so \( B \propto r \) for \( r < a \). Figure 4.14 shows the plot: \(B\) rises linearly for \(r < a\), then decreases as \(1/r\) for \(r > a\).