Q.Two wires shown in the figure are connected in a series circuit and the same amount of current, 10 A, passes through both, but in opposite directions. The separation between the two wires is 8 mm. The length AB is $S=22$ cm. Obtain the direction and magnitude of the magnetic field due to the current in wire 2 on the section AB of wire 1. Also obtain the magnitude and direction of the force on wire 1. [$\mu_0=4\pi\times10^{-7}$ T.m/A]

[!TLDR] Field due to wire 2 at wire 1: $B=\mu_0I/(2\pi d)=2.5\times10^{-4}$ T. Force on the 22 cm section of wire 1: $F=BIS=5.5\times10^{-4}$ N, repulsive (currents are antiparall …

The two wires carry the same current $I=10$ A but in OPPOSITE directions, separated by $d=8\ \text{mm}=0.008$ m. The field due to wire 2, at the location of wire 1, is $B=\dfrac{\mu_0 I}{2\pi d}=\dfrac{(4\pi\times10^{-7})(10)}{2\pi(0.008)}=\dfrac{(2\times10^{-7})(10)}{0.008}=2.5\times10^{-4}$ T. The force on the length $S=22\ \text{cm}=0.22$ m of wire 1 (section AB), due to this field, is $F=BIS=(2.5\times10^{-4})(10)(0.22)=5.5\times10^{-4}$ N (equivalently, directly from the parallel-wire force-per-length formula, $F/L=\mu_0I_1I_2/(2\pi d)$, times $L=S$). Since the two currents flow in OPPOSITE (antiparallel) directions, the force is REPULSIVE -- the section AB of wire 1 is pushed directly AWAY from wire 2, perpendicular to both wires (the book's own printed answer for this item confirms 'Repulsive, $5.5\times10^{ …

Find the field of one wire at the other's location using $B=\mu_0I/(2\pi d)$, then the force on the given len …

Assuming antiparallel currents attract (it is parallel currents that attract, antipa …