# Is it possible to find potential difference between two point in case of induced EMF created by a time-varying magnetic field?

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.

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?

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.
• 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. Commented Oct 15, 2020 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? Commented Oct 16, 2020 at 2:53