Mar 23, 2017

Two electric current carrying units would exert a magnetic force on one another.

#### Explanation:

We know that an electric current produces a magnetic field in its surroundings.

The magnitude of the magnetic field may be given using Biot-Savart law or Ampere's law.

From Lorentz force law, one finds that a magnetic field is capable of exerting a force on a moving charged particle.

So consider two wires in each other's vicinity, both carrying a certain current and hence producing magnetic fields of their own.
Now, electric currents are constituted by the flow of electric charges. So, the magnetic field produced by wire 1 interacts with the moving charges inside the wire 2 and exert a Lorentz force.

That is what precisely happens.

By Newton's third law, the field of wire 2 inturn exerts a similar force in opposite direction on the charges moving inside wire 1.

Quantitatively analysing the simplest case where two parallel long wires carry currents ${I}_{1}$ and ${I}_{2}$.

Using Biot-Savart law or by Ampere's law that the magnetic field by first wire at the location of the second is,

${\vec{B}}_{1} = {\mu}_{0} / \left(2 \pi\right) \cdot {I}_{1} / a$ pointing in the circumferential direction where $a$ distance between the two parallel wires.

Also, by Lorentz force law, one obtains that for wire 2 carrying current ${I}_{2}$ of length $l$ in Magnetic field ${B}_{1}$, it experiences a net Lorentz force,

$\vec{F} = {I}_{2} \left(\vec{l} X {\vec{B}}_{1}\right)$

Since in this case the magnetic field is in circumferential direction and the direction of current perpendicular to it, the magnitude of force is,

$F = {I}_{2} \cdot l \cdot {B}_{1}$

Thus force per unit length of the second wire would be,

$f = {I}_{2} \cdot {B}_{1}$

$\implies f = {\mu}_{0} / \left(2 \pi\right) \cdot \frac{{I}_{1} {I}_{2}}{a}$