Exercise 4.1.8: Force on a rectangular loop near a long wire

The near side, at $\rho =d$, carries $I_2$ parallel to $I_1$: it is attracted (Home exercise 4.1.3),

The far side, at $\rho =d+w$, carries $I_2$ antiparallel to $I_1$: it is repelled,

On the two short sides $dl=\pm d\rho a_{\rho }$ and $B\propto a_{\varphi }$, so $dl\times B\propto \pm a_z$: the forces are along the wire, equal and opposite on the two sides (the same $B(\rho )$ along each, opposite currents), and cancel. Their magnitude is $\frac{{\mu }_0{I}_1{I}_2}{2\pi }\ln \frac{d+w}{d}=0.32$ mN each.

Net force: $F=F_1-F_2$ towards the wire,

Reversing $I_2$ reverses every force: the loop is repelled with $3.2$ mN. The loop is attracted when its near side runs with $I_1$, repelled otherwise, the rule "parallel currents attract" applied to the side that sits in the stronger field.

A loop in a uniform field feels no net force (the two long sides cancel exactly, as in the DC motor of Worked example 4.1.4). The net force here exists only because $B$ falls off as $1/\rho$. Written as $F=-I_2\partial \Phi /\partial d$ (the pull towards the wire, $\partial \Phi /\partial d<0$) with the flux of Exercise 4.1.7, $F=I_2\frac{{\mu }_0{I}_{1}l}{2\pi }\left(\frac{1}{d}-\frac{1}{d+w}\right)$, the same value: the loop moves so as to increase the flux linking it when the currents are parallel.