Voltmeter reading accross 1/4th of ring where current is due to changing magnetic field There is a ring with constant changing perpendicular magnetic field producing emf = 12 V across the ring. (whole ring is an uniform resistance)
If we put an ideal voltmeter connecting 1/4th of the ring then voltange showing will be?
A) 3 V (if we see the 1/4th part between voltmeter)
B) 9 V (if we see the 3/4th part between voltmeter)
C) 0 V (if we consider emf caused by changing field is in the form of infinite batteries and if we traverse the circut then current in infinitesimal resistance will cancel voltage due to infinitesimal battery)
I am confused please help

 A: 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.
A: 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. 
A: The trick is to see the partial “circuit” 1/4arc-Vmeter as a separate circuit. What causes the emf (ie the charges to wanna flow) is actually the CROSSING of the magnetic field lines through the arc.
In the partial circuit the B-field lines cross the arc. And thus LEAVE (cq enter) the partial circuit, and ENTER (cq leave) the arc, the total circuit.
So if you focus on the partial circuit and apply Faradays law, the 1/4arc is where the actual flux change takes place, not in the other parts of the wire that connect the V-meter.
And the 1/4arc is “responsible” for 1/4 of the total flux change in the entire ring.
Maybe this way of looking points you in the right direction?
Kind regards, martin
