# What is motional emf?

Whenever a conducting rod is moving in a magnetic field, either showing translator motion (or) rotatory motions as a result a P.D is established at the ends of the rod. An electric field appears along with the length of the rod directed from the high potential end to the low potential one. This EMF is called Motional EMF.

The conducting rod MN having length ‘L’ is moving with a velocity ‘v’ in a Magnetic field $\overrightarrow{B}$. The mobile charges present in the conducting rod are also moving in this magnetic field, as a result, a magnetic force (FB) deflects the electron to the end ‘N’ making it –ve.

Due to the P.D an Electric field comes into picture from N to M. This electric field exerts force on the e- opposing the magnetic force. In steady state, this E.F balances the magnetic force.

$\not{e}E=\not{e}VB$

E = VB

$\varepsilon = – \int \overrightarrow{E}\overrightarrow{dl} = VB_{m}^{N}dl$

= VBL

In this case, the three vectors V, B & L are mutually far.

The motional emf established in this case is the maximum i.e VBL.

Although N is at high potential when compared to m, no current flows, because there is no conducting path between ends M & N. A conducting rod moving in a magnetic field, is similar to a source of emf i.e. a battery on the open circuit as shown in the figure. In this case also emf = terminal P.D.

When a π-shaped conductor has used a conducting path then we can see that the current is flowing from N to M in an anticlockwise direction.

$I=\frac{\varepsilon }{R}=\frac{VBL}{R}$

This is the same result which we got when we were using the flux rule

$\varepsilon =\frac{-d{{\phi }_{B}}}{dt}$

According to Einstein’s special theory of relativity, this is not a mere coincidence, it has a lot of significance.

General formula for motional EMF:

Whenever a conducting rod of length ‘L’ is moving in a magnetic field ‘B’, then the motional EMF is given by

$\varepsilon =\left(\overrightarrow{V}\times \overrightarrow{B} \right).,\overrightarrow{L}$ … (4)

But now, we have written a general formula for motional EMF, when the st. rod of length L is moving with a velocity ‘v’ in a magnetic field $\left( \overrightarrow{B} \right)$ in any direction.

This is a scalar triple product whose geometrical interpretation is given by the volume of the parallelepiped, with the 3 vectors V, B & L as coterminous sides.

If any two vectors are parallel & all the vectors are coplanar then the scalar triple product becomes two. i.e. the volume of parallelepiped becomes zero.

That means Motional EMF becomes zero.

If a conductor having some arbitrary shape is moving with a velocity ‘v’ in a magnetic field B in some direction. Then the motional emf is given by

$\varepsilon =2{\left(\overrightarrow{V}\times \overrightarrow{B} \right)} \overrightarrow{dl}$