Confusion regarding emf

Question

As we know that emf of the power supply can be defined as the potential difference between the electrodes when now current is flowing. Suppose the emf of the battery is 2 volt it means the potenrial difference between the two electrodes is 2 volt wjen no current is flowing. How to use this definition in a closed loop in which emf is created by changing magnetic flux? Suppose the emf created in a loop is 4 volt when magnetic flux is changing. What does it mean? In which two points the loop the potential difference is 4 volt?

Answer 1

Suppose the emf created in a loop is 4 volt when magnetic flux is changing. What does it mean?

First, the loop doesn't not have to be physical, i.e., the loop may simply be a closed path in space.

If the emf $\mathcal{E}$ for this closed path is $4\,\mathrm{V}$, then $4\,\mathrm{J}$ of work is associated with (slowly) moving a unit test charge along the path once.

If the closed path is instead formed by a wire of total resistance $R$, the current circulating will be $I = \frac{4\,\mathrm{V}} {R}\,\mathrm{A}$

If a small segment of the wire is then removed, there will be no current circulating and the voltage across the ends of the wire will be $4\mathrm{V}$. This must be the case because there can be no electric field within the wire (otherwise, there would be a current) but there must still be $4\,\mathrm{J}$ of work associated with moving a test charge 'round the loop.

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Can you elaborate on the last paragraph?

The mobile charge within the conductor distributes in such a way that the electric field inside the conductor, due to the charge distribution and the (steadily) changing magnetic flux threading the loop, is zero.

Since there is no electric field within the conductor, moving a test charge around that portion of the loop within the conductor involves no work.

However, the (total) electric field between the ends of the wire where the small segment was removed need not be and is not zero; The emf is $4\,\mathrm{V}$ and so it must be that the electric field there is such that moving the (unit) test charge around the remainder of the loop through the gap involves $4\,\mathrm{J}$ of work.

Answer 2

In the case of the continuous closed conducting loop through which there is changing magnetic flux, the electrons are urged found the loop by an electric field which, according to the Faraday-Maxwell equation, curls round the changing magnetic field. An electric field arising in this way is not conservative, and the concept of potential difference cannot be applied to it. Nevertheless the emf drives a current, given by $I=\frac{\mathscr{E}}{R}$ through the loop.

But, you might ask, what would a voltmeter read if we connected it to two points in the loop (without breaking the loop)? Would it read the voltage (current times resistance, $R_1$) in the major part of the loop or the minor part (resistance $R-R_1$)? This would depend on where the voltmeter is positioned in space, as we'd need to consider the rate of change of flux through the (different) loops containing the voltmeter! [This is nicely explained in "Electromagnetism Book 1: an Introduction to Maxwell's Equations" by John Bolton.]

If there is a break in the conducting loop, negative charge will, within a very short time, pile up on one end, leaving the other end positive. The electric field arising from these essentially static charges is a conservative one, so we can now talk about the p.d between the ends of the loop! The charges stop building up when the resultant of the non-conservative and conservative electric fields inside the loop becomes zero.