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?


1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – R.W. Bird
    Commented Oct 15, 2020 at 19:26
  • 1
    $\begingroup$ 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. $\endgroup$
    – Jatin
    Commented Oct 15, 2020 at 19:38
  • $\begingroup$ @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? $\endgroup$ Commented Oct 16, 2020 at 2:53
  • $\begingroup$ Plus non-conservative electric fields do not have potential difference. $\endgroup$ Commented Oct 16, 2020 at 2:57
  • $\begingroup$ physics.stackexchange.com/questions/75349/… $\endgroup$ Commented Oct 16, 2020 at 3:16

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