Potential in induced fields and faraday law

Question

The term "potential (emf) " has no meaning in non conservative fields. Then why is it included in faraday 's law of induction which deals with induced electric field which is a non conservative field.

Actually, the term potential has quite confused me. :

1. In the explanation of

$\oint\vec{E}\cdot\mathrm{d}\vec{S} = -\frac{\mathrm{d}\phi}{\mathrm{d}t}$

My book (resnick halliday physics part 2) has written that :

Because this flux is changing with time, an induced emf will appear around the loop.

What does the italicised phrase mean? In a conservative field we think of potential difference between two points. How is potential developed around the loop considered?

Answer 1

In a conservative field we think of potential difference between two points.

That's correct and it is reasonable to do so since, in the case of a conservative electric field, the line integral of the electric field between two points is path independent, i.e., the work associated with moving a unit test charge between the two points is the same no matter the path taken.

In particular, if the path is a closed loop (such that the beginning and end points are the same point), the work is zero.

This all implies that we can define a scalar potential that the electric field is the negative gradient of.

How is potential developed around the loop considered?

If the electric field is non-conservative, the line integral of the electric field between two points is, in general, path dependent and so the work associated with moving a test charge between the two points depends on the path taken.

Importantly, this implies that it is possible for the work associated with moving a unit test charge (once) around a closed loop to be non-zero; this work is the emf around the loop.

For this reason, when the electric field is non-conservative, we cannot define a scalar potential that the electric field is the negative gradient of.

Answer 2

Because this potential, $U$, is the potential difference, $\Delta V$, and not the absolute potential because the absolute potential cannot be defined.

For an electrostatic/conservative field $$\oint\vec{E}\cdot\mathrm{d}\vec{S}=0$$ since, while performing a circular integral, it is always performed over a circuital path and hence it like integrating a function from point $a$ to point $a$, hence the result would be zero, irrespective of which curve in the universe you use to complete the circle, ie.. in the conservative field $$\Delta V_{ab}=-\int_{a}^{b}\vec{E}\cdot\mathrm{d}\vec{S}\quad\text{since}\quad \vec{E}=-\nabla V$$

But, in a non conservative field $$\vec{E}\neq-\nabla V,\quad \mu\epsilon\frac{\mathrm{d}\vec{E}}{\mathrm{d}t}=\nabla\times\vec{B}-\mu\vec{J}\quad\text{and}\quad\nabla\times\vec{E}=-\frac{\mathrm{d}\vec{B}}{\mathrm{d}t}$$

It means that in a nonconservative field, the electric lines of force form closed loops just like the magnetic field(just like magnetic lines of force are caused from dipoles and changing electric fields and form closed loops, non conservative electric fields are caused from changing magnetic fields and current carrying wires and form closed loops). Conservative lines of force are derived of electric mono-charges and do not form closed loops. Thus the induced emf appears around the closed loop