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Q: There is a uniform time varying magnetic field in a circular region as shown in the figure. find out the potential difference across 2 point along an elliptical path as shown in figure. enter image description here

As far as I am aware, a time-varying magnetic field produces a non-conservative electric field. Hence, the concept of potential difference is invalid in such cases.

As such, is the given question valid? Is is possible to calculate a potential difference in such a case?

Alpha Delta
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The potential difference between the points $A$ and $B$ is, $$V_{AB} = \int_A^B \vec{E}\cdot \vec{d\mathbf{l}}$$ ,where $\vec E$ is induced electric field due to changing magnetic field.

To calculate this integral construct a symmetrical curve joining $A$ and $B$ forming a closed-loop (symmetrical about $AB$). Because the loop and the magnetic field are symmetrical we can safely conclude that $$ \underset{{lower\ curve}}{\int_A^B }\vec{E}\cdot \vec{d\mathbf{l}} = \underset{upper\ curve}{\int_B^A }\vec{E}\cdot \vec{d\mathbf{l}}=V_{AB} $$ So the integral over the whole loop is $$ \oint\vec{E}\cdot \vec{d\mathbf{l}} = \underset{{lower\ curve}}{\int_A^B }\vec{E}\cdot \vec{d\mathbf{l}} + \underset{upper\ curve}{\int_B^A }\vec{E}\cdot \vec{d\mathbf{l}} $$ $$ \oint\vec{E}\cdot \vec{d\mathbf{l}} = 2V_{AB}=-\frac{\text{d}\Phi_B}{\text{d}t} $$ You should be able to proceed from here.

Jatin
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  • Which point is positive, A or B? If neither, how can there be a potential difference. You seem to be suggesting a non-zero potential difference between A and A. – R.W. Bird Oct 15 '20 at 19:26
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    If there is a potential difference between $A$ and $B$ then it should not matter if $V_A >V_B$ or not. $V_{AB}$ can just be negative if that is not the case. The potential difference will be zero if the rate of change of flux is zero through the closed-loop However if there is a non zero magnetic flux through the closed-loop then $V_{AB}$ would most certainly be non-zero. – Jatin Oct 15 '20 at 19:38
  • @Jatin , according to you, $\underset{{lower\ curve}}{\int_A^B }\vec{E}\cdot \vec{d\mathbf{l}} = \underset{{upper\ curve}}{\int_B^A }\vec{E}\cdot \vec{d\mathbf{l}} \ \Rightarrow \underset{{lower\ curve}}{\int_A^B }\vec{E}\cdot \vec{d\mathbf{l}} - \underset{{upper\ curve}}{\int_B^A }\vec{E}\cdot \vec{d\mathbf{l}} = 0 \Rightarrow \underset{{lower\ curve}}{\int_A^B }\vec{E}\cdot \vec{d\mathbf{l}} + \underset{{upper\ curve}}{\int_A^B }\vec{E}\cdot \vec{d\mathbf{l}} = 0 \ \Rightarrow 2V_{AB} = 0 \ \Rightarrow V_{AB}=0$ Did I do something wrong? – Alpha Delta Oct 16 '20 at 02:53
  • Plus non-conservative electric fields do not have potential difference. – Alpha Delta Oct 16 '20 at 02:57
  • https://physics.stackexchange.com/questions/75349/if-electromagnetic-induction-generates-a-potential-difference-in-a-loop-where-a – Alpha Delta Oct 16 '20 at 03:16
  • @AlphaDelta It is true that you cannot define a potential difference for non-conservative electric fields. That means you cannot define a potential $V(\vec{R})$ for electric field $\vec{E}$ such that $\vec{E}=-\nabla V$. Stil we could define the potential difference difference between two points using the integral $\int_A^B \vec{E}\cdot\vec{dl}$. This difference, however, would be path-dependent and cannot be expressed as $V(\vec{r_B})-V(\vec{r_A})$ (because no such $V$ exists) unlike with the case for conservative fields where the difference is path-independent and can be represented as ..... – Jatin Oct 16 '20 at 03:57
  • ...$V(\vec{r_B})-V(\vec{r_A})$ which is dependent on inital and final positions alone. In the equation $\underset{{lower\ curve}}{\int_A^B }\vec{E}\cdot \vec{dl} + \underset{upper\ curve}{\int_A^B }\vec{E}\cdot \vec{dl}=0$, while $\underset{{lower\ curve}}{\int_A^B }\vec{E}\cdot \vec{dl}$ is same as $V_{AB}$, $\underset{upper\ curve}{\int_A^B }\vec{E}\cdot \vec{dl}$ is not $V_{AB}$ . In the upper curve $\vec{E}\cdot \vec{dl}$ is negeative while in the lower region it is positive. So $\underset{upper\ curve}{\int_A^B }\vec{E}\cdot \vec{dl}$ is actually $-V_{AB}$. – Jatin Oct 16 '20 at 04:08
  • Let us continue this discussion in chat. – Jatin Oct 16 '20 at 04:18