Force between two infinitely long straight parallel conductors carrying currents -
( i ) two infinitely long straight parallel conductors carrying currents in the same direction :
Consider two infinitely long parallel conductors X and Y carrying current $I_{1}$ and $I_{2}$ respectively in the same direction . let d be the distance of separation between these two conductor. The current $I_{1}$ in the conductor X produces a magnetic field around it.
The magnitude of magnetic field at any point on the conductor Y due to current carrying in the conductor X is given by :
$$B_{1} = \frac{\mu_{0}}{4\pi} (\frac{2I_{1}}{d})$$
The direction of $\overrightarrow{B}_{1}$ is perpendicular to the plane of the conductor Y and is directed into the plane as per right hand thumb rule ,
We know a current carrying conductor of length l placed as right angle to the magnetic field (B) experience a force is given by :
$$F =BIl$$
Therefore force experience per unit length of a conductor why in the magnetic field b1 is given by
$$F_{2} = \frac{F}{l} = B_{1}I_{2}$$
Using equation (i) we get
$$F_{2} = \frac{\mu_{0}}{4\pi} (\frac{2I_{1}}{d})I_{2} = \frac{\mu_{0}}{4\pi}(\frac{2I_{1}I_{2}}{d})$$
According to flaming left hand rule the direction of $\overrightarrow{F}_{2}$ is in the plane of the conductor and the directed toward conductor X.
The magnitude of the magnetic field at any point on the conductor X due to current carrying conductor why is given by
$$B_{2} = \frac{\mu_{0}}{4\pi} (\frac{2I_{2}}{d})$$
According to right thumb rule the direction of $B_{2}$ is perpendicular to the plane of the conductor X and directed outward.
The force experience per unit length of a conductor X in the magnetic field $B_{2}$ due to the current carrying conductor why is given by
$$F_{2} = \frac{\mu_{0}}{4\pi} (\frac{2I_{1}}{d})I_{2} = \frac{\mu_{0}}{4\pi}(\frac{2I_{1}I_{2}}{d})$$
According to flaming left hand rule the direction of $\overrightarrow{F}_{1}$ lie in the plane of the conductor and is directed toward conductor Y.
Since $\overrightarrow{F}_{1}$ and $\overrightarrow{F}_{2}$ are equal and opposite so too long parallel conductor carrying current in the same direction attract each other.
(ii) To infinitely long straight parallel conductor carrying current in the opposite direction :
Consider two infinitely long straight parallel conductor X and Y carrying current $I_{1}$ and $I_{2}$ respectively. Let conductor X and Y separated by a distance d.
The magnitude of the magnetic field at any point on the conductor Y due to current carrying conductor X is given by
$$B_{1} = \frac{\mu_{0}}{4\pi} (\frac{2I_{1}}{d})$$
The direction of magnetic field $\overrightarrow{B}_{1}$ is perpendicular to the plane of conductor Y and directed into the plane as per right hand thumb rule.
The force experienced per unit length of conductor why in the magnetic field $\overrightarrow{B}_{1}$ is given by
$$F_{2} = \frac{F}{l} = B_{1}I_{2}$$
$$F_{2} = \frac{\mu_{0}}{4\pi} (\frac{2I_{1}}{d})I_{2} = \frac{\mu_{0}}{4\pi}(\frac{2I_{1}I_{2}}{d})$$
According to flaming left hand rule the direction of $\overrightarrow{F}_{2}$ is away from the conductor X.
The magnitude of magnetic field at any point on the conductor X due to current carrying conductor Y is given by
$$B_{2} = \frac{\mu_{0}}{4\pi} (\frac{2I_{2}}{d})$$
According to right hand thumb rule the direction of $\overrightarrow{B}_{2}$ way to is perpendicular to the plane of the conductor X and directed into the plane.
The force experience per unit length of a conductor X in the magnetic field $\overrightarrow{B}_{2}$ is given by
$$F_{1} = \frac{F}{l} = B_{2}I_{1}$$
$$F_{1} = \frac{\mu_{0}}{4\pi} (\frac{2I_{2}}{d})I_{1} = \frac{\mu_{0}}{4\pi}(\frac{2I_{1}I_{2}}{d})$$
According to flaming left hand rule the direction of $\overrightarrow{F}_{1}$ is away from the conductor Y.
Since $\overrightarrow{F}_{1}$ and $\overrightarrow{F}_{2}$ are equal in magnitude and directed away from each other so too long parallel conductor carrying current in the opposite direction repel each other.