Electromotive Force

We know that the closed line integral of electrostatic Field Φ E. dI = 0.
Hence electrostatic field cannot cause current flow in the circuit. The source of energy in a circuit may be chemical, mechanical or thermal, or it may result from a changing magnetic field. Let us consider the case of a battery. The chemical process within the battery is responsible for the generation of nonconservative field and changes chemical energy into electrical energy.

Let us consider that the terminal a is maintained by the source at a higher potential than terminal b. There is therefore an electrostatic field Ec at all points between and around the terminals. The source is itself a conductor and if only this electrostatic field exists, the positive charges will move from a toward b and the negative charges from b toward a. The excess charges on the terminals would decrease and hence the pd between them would decrease and eventually become zero. As it does not happen, we conclude that there must also exist at every point within the source and which is equal and opposite to the electrostatic force Fe = qEe. We can define the non-electrostatic field En as Fn = qEn. When the source is an open circuit, the vector sum of these fields at every point must be zero, i.e.,
            E = E_n + E_e = 0     or    E_e = -E_n                         … (1)
. :            ∫^b_a E_e . dI = - ∫^b_a En . dI = ∫^a_b = En . dI.
But            ∫^b_a Ee . dI = Vab    
. :            ∫ba E_n . dI = vab                                 … (2)


The line integral of the non-electrostatic field from b to a is called the electromotive force (EMF) represents the work done by the non-electrostatic (chemical) forces during the transfer of a u nit charge from the negative to the positive terminal.

According to Eq. (2), En is “non-conservative”, because the line integral of En around a closed path is not zero.
The emf of the source equals the open circuit potential difference.
We should not confuse between terms potential difference and emf. The p.d., is the line integral of an electrostatic field. The emf is the integral of a non-electrostatic field. The electrostatic field, hence the potential emf of the source is a constant independent of the current and this represents the property of the source.
For a closed circuit, i.e., when two terminals of a source are connected by a conductor, the resultant field E within the source is E = Ee + En.
. :    ∫ab E.dI = ∫^a_b E_e . dI + ∫^a_b En . dI
Or    Ir = - V_ab + E.                                        … (3)
Here, r is the internal resistance of the source. Here the direction of integration is the direction of current within the source.
Let R be the resistance of the conductor connecting the terminals a and b. Then Eq. (2) can be written as
    Ir = - IR + ε    or     I = ε / (R + r)                            … (4)

The potential difference across the external resistance R in this case is smaller than the emf ε of the battery by an amount ir.

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