## What is Induced EMF?

*Induced EMF is the generated electromotive force which drives a current in a closed conducting loop when it is moved in a magnetic field.*

- Induced EMF is generated in a conducting loop due to variation in
*magnetic flux*associated with the loop. - Phenomenon of induced EMF is a result of
*electromagnetic induction.*

### Production of Induced EMF

From *Faraday’s second law*, magnitude of induced EMF in a loop is given by –

\epsilon = - N \left ( \frac {d\phi}{dt} \right )

Here, magnetic flux is given by –

\phi = BA \cos \theta

And ( \theta ) is the angle made by the normal of the plane of loop with direction of magnetic field ( B ) .

This shows that, induced emf depends upon –

- Number of turns ( N ) .
- Rate of change in magnetic flux \left ( \frac {d\phi}{dt} \right ) .

Magnetic flux can be changed by –

- Varying the strength of the
*magnetic field*( B ) . - Changing the area ( A ) of the coil.
- Altering the orientation of the coil i.e. changing the angle ( \theta ) with respect to the magnetic field.

#### 1. EMF by changing the Magnetic Field

In Faraday’s experiment, when magnet is moved towards the coil, the strength of magnetic field at any point of the coil increases because more and more magnetic field lines pass through the coil. Hence, the galvanometer shows deflection.

When magnet is moved away from the coil, strength of the magnetic field linking with the coil decreases. Again, galvanometer shows deflection.

Thus, EMF is induced in the coil by changing the strength of the magnetic field.

#### 2. EMF by changing the Area of loop

Induced EMF produced by changing the area of a closed circuit or loop by the movement of the circuit or part of it through a uniform magnetic field is known as motional EMF.

This can be done by moving the loop into the region of magnetic field or away from it.

Consider a closed loop ACDE of length ( l ) moving with uniform velocity ( v ) placed in a uniform magnetic field ( B ) as shown in figure.

The magnetic field is perpendicularly directed into the plane of the figure.

When the loop moves out of the magnetic field through a small distance ( \Delta x ) , then change in magnetic flux will be –

\Delta \phi = - B \times Area of loop.

= - B ( l \Delta x )

= - B l \Delta x

Negative sign indicates that, area of closed loop inside the magnetic field is decreasing. Therefore, from Faraday’s laws, the induced EMF in the loop will be –

\epsilon = \left ( \frac {\Delta \phi}{\Delta t} \right )

= \left ( \frac {B l \Delta x}{\Delta t} \right )

= B l \left ( \frac {\Delta x}{\Delta t} \right )

But, \quad \left ( \frac {\Delta x}{\Delta t} \right ) = v

Therefore, induced EMF is given by –

\epsilon = B l v

#### 3. Emf by changing the Orientation of loop

If a loop is pivoted and continuously rotated in the region of a magnetic field, magnetic flux linked with the loop will be changing continuously and induced EMF will produce in the loop. This principle is used in electric generators.

#### 4. Motional EMF by a rotating rod

Motional induced EMF can be produced by rotating a rod in a magnetic field.

Let, a rod PQ of length ( l ) rotates with angular velocity ( \omega ) in a uniform magnetic field of strength ( B ) . The direction of field is directed into the plane of paper and perpendicular to it.

Let, the rod turns by an angle ( \theta ) in time ( t ) . Then, swept area by the rotating rod will be –

A = \left ( \frac {1}{2} \right ) \times l \times QQ'

= \left ( \frac {1}{2} \right ) \times l \times l \theta

= \left ( \frac {1}{2} \right ) l^2 \theta

Flux linked with swept area will be –

\phi = BA \cos 0 \degree = BA = B \left ( \frac {1}{2} \right ) l^2 \theta

Because, the angle between the normal to the swept area ( A ) and direction of magnetic field ( B ) is ( 0 \degree ) . Magnitude of induced EMF in the rod will be –

\epsilon = \left ( \frac {d \phi}{d t} \right )

= \left ( \frac {d }{d t} \right ) \phi

= \left ( \frac {d}{dt} \right ) \left ( \frac {1}{2} \right ) B l^2 \theta

= \left ( \frac {1}{2} \right ) Bl^2 \left ( \frac {d \theta}{dt} \right )

= \left ( \frac {1}{2} \right ) Bl^2 \omega

Since, \quad \left ( \frac {d \theta}{dt} \right ) = \omega

Therefore, \quad \epsilon = \left ( \frac {1}{2} \right ) Bl^2 \omega