# Would the voltmeters give different readings in a circuit with induced current?

$$\def\vE{{\vec{E}}}$$ $$\def\vD{{\vec{D}}}$$ $$\def\vB{{\vec{B}}}$$ $$\def\vJ{{\vec{J}}}$$ $$\def\vr{{\vec{r}}}$$ $$\def\vA{{\vec{A}}}$$ $$\def\vH{{\vec{H}}}$$ $$\def\ddt{\frac{d}{dt}}$$ $$\def\rot{\operatorname{rot}}$$ $$\def\div{\operatorname{div}}$$ $$\def\grad{\operatorname{grad}}$$ $$\def\rmC{{\mathrm{C}}}$$ $$\def\rmM{{\mathrm{M}}}$$ $$\def\ph{{\varphi}}$$ $$\def\eps{{\varepsilon}}$$

Hi,

The figure shows a circular loop and a changing magnetic field applied perpendicular to it, directed inside the page. Assume the resistance in the loop is constant per unit length. Assume measure of arc AB is 90°. Assume the voltmeters to be ideal and the wires joining the voltmeters to have zero resistance. Would the two voltmeters show different readings? What will they read?

I remember seeing a soution that says V1 will read k/4 and and V2 will read 3k/4. I don't recollect the solution completely but I think it used, $$\oint_{\partial A} \vE \cdot d\vr + \ddt\int_A \vB\cdot d\vA = 0$$

on the 2 loops joining the points A and B with the voltmeters.

I asked my friends for help and they said both the voltmeters would read zero. They replaced parts of the circuit with batteries and removed the magnetic field. Now, I am not sure if the resultant circuit is equivalent to the given situation as I remember seeing a solution which showed otherwise. I am aware this might probably be a basic problem but I am not able to find it on Google as I don't know what exactly to search for.

So, I have come here for help.