# Voltmeter reading accross 1/4th of ring where current is due to changing magnetic field

Fools rush in where angels fear to tread...

The difficulty is that it is a non-conservative electric field that drives charge round the ring, so we cannot apply the concept of potential difference.

Nonetheless, we can calculate the current to be 12 V /R in which R is the ring resistance, and hence the voltage drops over 3/4 and 1/4 of the ring circumference to be $$\frac{12\ \text V}{R} \times \tfrac34 R = 9 \text V$$ and $$\frac{12\ \text V}{R} \times \tfrac14 R = 3\text V$$ respectively, assuming uniform resistance.

If the voltmeter is positioned as shown, it will read 3 V as there is no changing flux linking the loop that consists of bottom right quarter arc of the ring, voltmeter and voltmeter leads, and therefore no emf in this loop; all we have is the voltage drop across the quarter arc.

You object: suppose we consider the loop consisting of the other 3/4 of the ring, voltmeter and voltmeter leads? Then we have a 9 V voltage drop across the 3/4 arc, but we also have the full emf of 12 V, because the whole of the changing flux is linked with this loop. Therefore the net voltage is 3 V, as we found before.

The correct answer is, C. There is no measurable voltage between any two points in this conducting ring. For each very short segment, the emf is dissipated to current flowing through resistance. Knowing the emf and R you can calculate the current and power, but find no voltage drop. If you cut the ring at some point, then the current will stop and a 12 volt drop will appear across the gap at the cut.