Home exercise 4.1.3: Force on a wire segment

Put the segment in the half-plane $\varphi =0$, i.e. at $x=d$, $y=0$, parallel to $z$. From Problem 4.1.1,

the same at every point of the segment, since all of it is at the distance $d$.

With $I$ flowing along $+z$, $dl=dz{a}_z$ and

Reversing $I$ reverses $dl$ and hence $F$: antiparallel currents repel with the same magnitude. Numerically,

Units: ${\mu }_{0}I{I}^{\prime }l/d$ is (H/m)(A${}^2$) $=$ (V s/m)(A) $=$ J/m $=$ N. This is Worked example 4.1.1 for a segment: the force per unit length is ${\mu }_{0}I{I}^{\prime }/(2\pi d)$ whatever $l$, because every element of the segment sees the same $B$. The direction follows from the right-hand rule twice: $B$ of the wire circles it, and $Idl\times B$ then points across the gap. Half a millinewton on a metre is the order of magnitude felt by the leads of any $10$ A appliance, unnoticeable until the current is a short-circuit current.