A current carrying conductor produces its own magnetic field. Such a conductor is placed in a magnetic field. The magnetic field of the conductor will interact with the external magnetic field. As a result of which the conductor may experience a force. It can be observed by the following experiment.

Consider a copper rod which can move on a pair of copper rails. The whole arrangement is placed in a uniform magnetic field. The magnetic field on the rod is directed vertically upwards as shown in fig.

When a current is passed through a copper rod from a batter, a rod moves on the rails. The relative direction of current, magnetic field and motion of conductor (force) are shown in the figure.

It can be observed that the force on a conductor is always at right angle to the plane containing the rod and the direction of the magnetic field B and is predicted by Fleming’s Left-Hand rule. The magnitude of the force acting on the rod depends on the following factors.

- The force F is directly proportional to Sin𝛂 where 𝛂 is the angle between conductor L and magnetic field B.

$F\propto Sin\alpha $

It means that the force is zero if the rod is placed parallel to the magnetic field and force is maximum when the conductor is placed at right angle to a field.

- The force F is directly proportional to the current I flowing through the conductor.

$F\propto ~I$

- The force F is directly proportional to the length of the conductor L inside the magnetic field B.

$F\propto ~L$

- The force F is directly proportional to the strength of magnetic field applied.

$F\propto B$

Combining all four factors, we can write a simple following equation

$F\propto BIL~Sin\alpha $

Or

$F=KBIL~Sin\alpha $

Where K is the constant of proportionality. The SI units of force, current, conductor length and magnetic field are given below:

Force is measured in Newton (N)

Current is measured in ampere (A)

Conductor length is in Meter (m)

Magnetic field strength is measured in Tesla Whereas;

$Tesla={}^{N}/{}_{Am}$

By putting all units is above formula, we can observe that K has no units.

**Example**

A 20 cm wire carrying a current of 10 A is placed in a uniform magnetic field of 0.3 Tesla. If the wire makes an angle of 40^{o} with the direction of magnetic field, find the magnitude of the force acting on the wire?

**Solution**

Length of the wire = L= 20 cm = 0.2 m

Current in the wire = I= 10 A

Strength of magnetic field = B = 0.30 T

Angle along magnetic field =α= 40^{o}

In order to calculate Force, we have following formula:

$F=BIL~Sin\alpha $

So,

$F=0.30*10*0.2*Sin~({{40}^{o}})$

$F=0.386~N$