Exercise 4.1.4: A straight segment of finite length

As in Problem 4.1.1, every element $Id{z}^{\prime }{a}_z$ gives a contribution along $a_{\varphi }$ with $|a_z\times R|=d$ and $R=\sqrt{d^2+z^{\prime 2}}$. Only the limits change:

With $\sin {\theta }_2=\frac{L/2}{\sqrt{d^2+L^2/4}}$ for the upper end and $\sin {\theta }_1=-\sin {\theta }_2$ for the lower one (angles measured from the perpendicular, positive upwards),

This is the general segment formula: ${\theta }_1\to -\pi /2$, ${\theta }_2\to \pi /2$ gives back ${\mu }_{0}I/(2\pi d)$.

$\sin {\theta }_2=0.10/\sqrt{0.02^2+0.10^2}=0.981$,

against ${\mu }_{0}I/(2\pi d)=100$ μT for the infinite wire: at $d=L/10$ the segment is already $98\%$ "infinite".

In the plane through one end, ${\theta }_1=0$ and $\sin {\theta }_2=L/\sqrt{d^2+L^2}=0.995$:

half the mid-plane value: only one half of the wire is "seen".

Home exercise 4.1.1 (square loop) is this formula with $d=a/2$, ${\theta }_{1,2}=\mp 45^{\circ }$, applied four times. The angles version is the one to remember: $B={\mu }_{0}I(\sin {\theta }_2-\sin {\theta }_1)/4\pi d$, or ${\mu }_{0}I(\cos {\alpha }_1-\cos {\alpha }_2)/4\pi d$ with the angles measured from the wire. Note that no Amperian circle could have given this result: $H$ is not constant on a circle round a finite segment.