2
$\begingroup$

Problem Diagram

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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.