The emf of the battery source can be defined as the potential difference between two electrodes when no current is flowing. Now how I can apply this definition of the emf which is produces by change in the magnetic flux to the coil. Which two points I have to take for defining the emf produce in closed loop. Suppose 2 volt emf is produces in the closed loop when I change the magnetic flux linked with it. Now in which two points in the loop the potential difference 2volt is created.
1 Answer
The emf as defined between two electrodes is the difference in electric potential $V$ between the two electrodes. If $A$ and $B$ denote the locations of the two electrodes, then emf can be defined as
$$ \mathcal E = V(B) - V(A) = -\int_A^B \vec E \cdot \mathrm{d}\vec l \tag{1} $$
For situations involving changing magnetic fields however, $\vec E$ is no longer an irrotational field
$$ \nabla \times \vec E = - \frac{\partial \vec B}{\partial t} \neq 0 $$
which means it can't be given by a potential
$$ \vec E \neq - \nabla V $$
Which further means one can't unambiguously define the line integral of $\vec E$ between two points (the potential difference between two points). Nothing stops you from talking about the line integral along a specific curve, however. In the case of a conducting loop in a changing magnetic field, you only care about the field along the loop, since that is what causes the flow of current. So you define the emf produced in the loop as
$$ \mathcal{E} = \oint \vec E \cdot \mathrm{d} \vec l \tag{2} $$
where the integral is taken along the closed loop.
The two definitions of emf then, are distinct but related, which becomes transparent through the use of line integrals. In the electrostatic case, $(1)$ is always well defined and $(2)$ is always zero. With changing magnetic fields, $(1)$ ceases to be well defined, and $(2)$ may be nonzero. In particular, you cannot define any "emf" (line integral of $\vec E$) via two points, since the integral is path-dependent.
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$\begingroup$ Is it means that definition of emf that it is potential difference between the electrode can not be applied in the closed loop $\endgroup$– KeshavCommented Dec 23, 2017 at 12:34
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$\begingroup$ Yes. Electric potential, and potential difference, is only defined if the fields don't change with time. $\endgroup$– StygCommented Dec 23, 2017 at 13:31