Obtain the expression for the force between two parallel current-carrying wires.

(N/A) Consider two long parallel wires $a$ and $b$ separated by a distance $d$,carrying currents $I_{a}$ and $I_{b}$ respectively.
The conductor $a$ produces a magnetic field $\overrightarrow{B}_{a}$ at all points along the conductor $b$. According to the right-hand rule,the direction of this field is perpendicular to the plane containing the wires.
The magnitude of the magnetic field produced by wire $a$ at distance $d$ is given by:
$B_{a} = \frac{\mu_{0} I_{a}}{2 \pi d}$ ... $(1)$
The conductor $b$ carrying current $I_{b}$ experiences a magnetic force due to the field $B_{a}$. The force on a segment of length $L$ of wire $b$ is given by:
$\overrightarrow{F} = I \vec{L} \times \overrightarrow{B}$
Since the current and magnetic field are perpendicular,the magnitude of the force $F_{ba}$ on wire $b$ due to wire $a$ is:
$F_{ba} = I_{b} L B_{a} = \frac{\mu_{0} I_{a} I_{b} L}{2 \pi d}$ ... $(2)$
Similarly,the force $F_{ab}$ on a segment of length $L$ of wire $a$ due to the current in wire $b$ is equal in magnitude to $F_{ba}$ but acts in the opposite direction:
$\overrightarrow{F}_{ba} = -\overrightarrow{F}_{ab}$ ... $(3)$
This result is consistent with Newton's third law of motion.
It is observed that currents flowing in the same direction attract each other,while currents flowing in opposite directions repel each other.