Mechanism Of The Induced EMF
Electromagnetic Induction of Class 12
MECHANISM OF THE INDUCED EMF
When a conducting rod of length l moves at constant velocity v perpendicular to a uniform magnetic field (as shown in Fig. 4.10) an electron experiences a magnetic force ) directed downward along the rod. As a result, electron migrates to the lower end of the rod and leave unbalanced positive charge at the top. This redistribution of charge sets up an electrostatic field E directed downward.
In the absence of any external circuit an equilibrium is established between the electric and magnetic forces.
Thus,B = 0
or = 0
The action of the magnetic force may be considered similar to the action of a certain electric field (opposing the Coulomb field) of magnitude (4.5) Since the field is not due to the separation of charges, but to nonelectrostatic forces (in this case magnetic), therefore, it is called an extraneous field. Unlike coulomb field it is directed from negative to positive charges. 


When the moving rod is connected to an external resistor (shown in Fig. 4.11), the following points are observed. (i)The potential difference between the ends of the conductor is less than when the circuit was open. (ii)The strength of the extraneous field E* remain unchanged because it depends only on V and B. 


The direction of induced current coincides with the direction of motion of positive charges i.e. with the resultant electric field. In the external part of the circuit, positive charges flow from A to B. Thus, the work done by the net external field is Wext = q (ϕA  ϕB) In the external part of the circuit, positive charges flow from B to A. The work done in the internal part of the circuit is 

Wint = ∫ q
orWint = q ∫  q (ϕA  ϕB)
Hence the total work done is
W = Wext + Wint = q ∫ . d(4.6)
The net work done in moving a charge along a closed circuit is equal to the work done by nonelectrostatic forces. Being a conservative force an electrostatic force does no work along a closed path.
The electromotive force is defined as the ratio of the work done in moving a charge along a closed circuit, to the magnitude of charge.
E = W/q = ∫(4.7)
How does an induced emf is produced in a stationary conductor ?
Let us consider two cases as shown in Fig. 4.13.

Fig. (4.13)(a)A loop moves in the magnetic field produced by the magnet. Charges flow under the action of the magnetic force. (b)The magnet moves toward a stationary loop. The induced current is produced in the same direction as if the loop were moving with respect to the magnet. 
According to the principle of relativity, all inertial frames are equivalent and phenomena are determined only by the relative velocity with which the loop and magnet approach each other. But what forces in second case (shown in Fig. 4.13 b) cause circulation of charges around the loop.
Let the reference frame xyz be fixed to a conducting loop, and frame x'y'z' to the source of a magnetic field (bar magnet). In the reference frame xyz, linked to the loop, the charges are stationary with respect to the frame, and therefore, not subject to the magnetic force. Hence the current induced in the loop must be produced by an induced electric field which is not in reference frame x'y'z'.
E = (4.8)
There is an induced electric field in any closed path, whether in matter or in empty space, through which the magnetic field is changing.
Characteristic of Induced Electric Field
1.It is not a Coulomb field.
2.Unlike Coulomb field the lines of induced field form closed loop. Therefore, it is called a circuital field or vortex field.
3.It is not conservative.
Consequently, the energy characteristic of an induced field is the induced electromotive force and not the potential.
Conclusion
In general emf is defined as the work done by a nonelectrostatic agent in carrying a unit charge around a closed loop.
E =
The total force acting on a charged particle is given by Lorentz force
F = q
Thus, E =
The first term is associated with an induced electric field and arise when the magnetic field varies with time. The second term arises when a conductor moves relative to a magnetic field. This motional emf arises from the magnetic force on a moving charge.
Hence, the total induced emf may be written as
E = (4.9)
Example: 4.5 Figure 4.14 shows a conducting disc of radius R rotating at constant angular velocity ω. Its plane is perpendicular to a uniform and constant magnetic field B. Find the emf generated between the centre and the rim.
Solution A disc may be treated as a collection of radially oriented rods. Consider a small segment of length dr at a distance r from the centre. 

The speed of the segment is v = ωr. The electrons in the segment are subjected to the magnetic force directed radially inward.
Since v is perpendicular to B, therefore,   = vB
The length element is dl = (dr) .
Thus, ( = vBdr = ωBrdr
Using equation (4.9), we get
E = ∫
with the given directions of , the centre is at a lower potential than the rim.
Example 4.6 The current in an ideal solenoid of a radius R varies as a function of time. Find the induced electric field at points. (a) inside, and (b) outside the solenoid. Express the result in terms of dB/dt Solution From the condition of symmetry the value of the induced electric field will be the same at all points on any circular loop concentric with the solenoid. (a)Consider a loop of radius r < R 

E.dl= Edl
since E is parallel is dl
Now
Using equation (4.8)
orE(2πr) = πr2 dB/dt
orE = r/2 dB/dt (r < R)..(i)
The induced electric field intensity increases linearly with distance from the centre.
(b)When r > R, the flux through the loop is ΦB = (πR2) B
Thus, E(2πr) = πR2 dB/dt orE = R^{2}/2r dB/dt (r > R) (ii) Outside the solenoid, the induced electric field strength is inversely proportional to the distance from the centre. The variation of induced electric field E is plotted with distance from the centre of a solenoid as shown in (Fig. 4.16). 
