I have been reading about the electric field induced on a metal ring due to an increasing magnetic field directed along the axis of the ring and the electric field lines are found to form concentric circular loops. This implies that the electric field formed is not conservative. Why is it so? Are electric field due to free charges and those due to changing magnetic field two entirety different things? Please provide an intuitive and less mathematical explanation as I am just a high school student with knowledge of basic calculus.


1 Answer 1


If a vector field $\mathbf E$ is the gradient of a conservative potential $\phi$ then its curl must be zero i.e.

$$ \nabla \times \mathbf E = 0 $$

because the curl of the gradient of any continuously twice-differentiable scalar field is always the zero vector. However Maxwell's equations tell us:

$$ \nabla \times \mathbf E = -\frac{\partial \mathbf B}{\partial t} $$

and in the presence of time dependent magnetic fields the right hand side is not zero so there is no conservative scalar potential from which the electric field is derived.

Looking at it another way, we can write the electric field as a function of the usual scalar potential and the magnetic vector potential:

$${\bf E} = -\nabla \phi -\frac{\partial {\bf A}}{\partial t}$$

So once again we see that if the presence of time dependent magnetic fields the electric field isn't simply due to a conservative potential.

  • 2
    $\begingroup$ Unlikely to be helpful given the last sentence of the question. $\endgroup$
    – ProfRob
    Commented Aug 15, 2017 at 16:44
  • 2
    $\begingroup$ @RobJeffries: Gautham and I discussed this in the chat room, which I think was productive. This answer is more of a for the record post. If you want to venture a more intuitive approach please do. $\endgroup$ Commented Aug 15, 2017 at 16:45
  • $\begingroup$ swell hat and glasses, john. merry xmas to you. $\endgroup$ Commented Dec 25, 2017 at 0:52

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