# Induced EMF - Methods of producing Induced EMF

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### Induced EMFIntroduction:

Methods of producing Induced EMF can be become more easy to understand , if we first know the concept of Induced EMF. Electromotive force abbreviated as EMF   is the characteristic of any energy source capable of driving electric charge around a circuit. It is actually the work done in moving unit electric charge around the circuit. It is measured in units of  J C-1 or volt.

Some examples of energy source which provide EMF are thermoelectric devices,
electrochemical cells, electrical generators and Solar cells .

If we talk about the concept of Faraday's law of electromagnetic induction , an EMF is induced in a circuit when magnetic flux linked with it changes. This EMF is called Induced EMF.

#### Methods of producing Induced EMF :

Induced EMF from Faraday's law is given as;

É› = −N dÉ¸/dt

Where magnetic flux , É¸ = BAcosÎ¸.

This equation shows that EMF depends upon two factors. They are (1) number of turns N and (2) rate of change of magnetic flux (É¸).

 Methods of producing Induced EMF

#### Magnetic flux can be changed by :

1) Changing the strength of the magnetic field (B).

2) Changing the orientation (Î¸) of the coil with respect to the magnetic field.

3) Changing the area (A) of the coil.

#### A)  Induced EMF by changing the magnetic field :

In one of Faraday's experiment , when magnet is moved towards the coil. The strength of magnetic field at any point of the coil increases. As more and more magnetic field lines pass through the coil. The galvanometer deflection indicates the presence of the current in the coil.

On the other hand, when magnet is moved away from the coil, strength of the magnetic field at any point of the coil decreases. Again, galvanometer shows deflection. It indicates the presence of current in the coil. Thus, EMF current is induced in the coil by changing the strength of magnetic field.

#### B)  Induced EMF due to change in the area (Motional EMF) :

Induced EMF produced as a result of change in the area of a closed circuit, due to the circuit or part of it through a uniform magnetic field is know as motional EMF.

Consider a closed loop or circuit ACDE placed in a uniform magnetic field B directed into the page. Let the loop move with a uniform velocity V.

Change in the magnetic flux linked with the closed circuit in the time interval ∆t,  when coil moves out of magnetic field through a small distance ∆x  Is given as :

∆É¸B = −B × l × ∆x

or                        ∆É¸B = −Bl∆x

Negative sign shows that area of the closed circuit inside the magnetic field decreases.

According to Faraday's law ,  Induced EMF in the circuit is as  :

É›  =  − ∆É¸B/∆t = Bl∆x/∆t

since       ∆x/∆t = v

∴         É›  =  Blv      ....................(7)

Equation (7) gives the expression for motional EMF.

If R be the resistance of the closed circuit , then the induced current flowing in the circuit is as:

I = É›/R = Blv/R

#### Direction of the Induced EMF divided by current is given with the help of :

i) Fleming's right hand rule

ii) Lenz's Law.

#### Alternative Method :

Motional EMF in a conductor moving in magnetic field obtained from Lorentz Force

 Alternative method

Consider a conductor PQ of length L moving freely in a uniform magnetic field B acting perpendicular to the plane of paper with uniform velocity on a rectangular conductor ABCD of negligible resistance (Figure).  Let any arbitrary charge q in the conductor also move in the field with same velocity.

Magnitude of Lorentz Force on this charge;

Fm = qvB

Direction of this force on the charge as per Fleming's  left hand rule comes out to be along PQ towards Q.

Work done in moving the charge from P to Q is given by;

W = Fm × PQ  = (qvB) l

Electromotive force (EMF)  is defined as the work done per unit charge,  so EMF across conductor PQ is given by :

É› = w/q  = qvBl/q  = vBl

i.e.,              É› = Bvl

Points To Remember:

#### Note that all charges experience the same Lorentz Force  irrespective of their position in the conductor PQ.

Example:

A 10 m long metallic bar moves in a direction at right angle to the magnetic field with a speed of 5 m S  ̄¹ . e. m. f of 25 v is induced in it,  find the magnetic field intensity.

Solution :

Motional e. m. f induced in a metallic bar is given by;

É›  = Blv

or      B = É›/lv

Here ;                   É› = 25 v , l = 10 m  and v = 5 m s ̄¹

∴                      B = 25/ 10×5     =    0.5 T

Numerical : A spoke of length 50 cm is rotated in a uniform magnetic field of 0.4 × 10  ̄⁴ T with a speed of 2 r. p. s

i)  Find the induced e. m. f in it.

ii)  If there are 16 spokes such that starting ends (S)  of all spokes as well as finishing ends (F)  of all spokes are joined separately, what will be the total e. m. f.

Solution;

i)  Induced e. m. f, É› = 1/2 BÏ‰l²

= 1/2 B (2Ï€Ï…)l²

1/2 × 0.4 × 10  ̄⁴ × 2Ï€ × 2 × ( 50 × 10 ̄²)

=  6.3 × 10  ̄⁵ v

ii) Total EMF will be 6.3 × 10 ̄ V only because all spokes are connected in parallel.

#### Conclusion

This article has discussed about "methods of producing Induced EMF". The concept of Induced EMF has much broader importance in physics. Especially, this concept is necessary to know before we switch to Electromagnetism. There are different articles related to Induction discussed so far in this blog.

However, this article provides you complete and a brief information. It also helps you through examples and numerical. Alternative methods have also been discussed to help you in getting all the information.

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