I am currently in highschool and don't know maxwell's equations much. However I do know about Faraday's law and recently we were taught about Electromagnetic Induction. I am having a few problems in wrapping the concept of Induced electric fields around my mind. I am told that : $$ϵ=∮\overrightarrow{E}.d\overrightarrow{l}=-\frac{dɸ}{dt}$$ Now suppose there is a cylindrical region of radius $R$ where magnitude of magnetic field varies with time as $$B=Ct$$but is uniform through space. We place a circular conducting loop of radius $r (<R)$ perpendicular to the axis such that magnetic field lines are perpendicular to the plane(going inwards) of the loop and its center lies on the axis. We can easily solve the above line integral of electric field due to symmetry and get this result : $$ |\overrightarrow{E}|=\frac{rC}{2}$$ Also, by symmetry this electric field is circular i.e it's direction is tangential to position vector $\overrightarrow{r}$ assuming the common center as origin. For the sake of argument lets take it as clockwise. So this gives me the magnitude and direction of induced electric field as a function of position vector in this plane. But now suppose that I displace the loop towards the right by $d(d<R-r)$ , I can again follow the above steps and get induced electric field as a function of position vector in the plane.

So this means that induced electric field depends upon our placement of loop ? Or is there some superposition that I can't see ?

I had thought that it should be independent of even the existence of a loop much less the placement because that seems intuitive.

Do note that the electric field magnitude for outside the cylinder varies inversely with $r(>R)$ and the exact expression is $\frac{R^2C}{2r}$.

All this talk was about the plane of the loop but what happens in other planes ? Are there field lines shaped like infinite solenoids of varying radii and superimposing each other ?

Please help me....

  • $\begingroup$ r is just the radius of the loop, not its position with respect to some arbitrary origin. $\endgroup$
    – my2cts
    Commented Aug 27, 2020 at 19:46
  • $\begingroup$ I am saying that r is the position of a point in plane of the loop with center of loop as origin. The vector tail is at the center and the head is on that point. $\endgroup$
    – Physicsa
    Commented Aug 28, 2020 at 2:31
  • 1
    $\begingroup$ Would you accept this attempt to bring out the paradox? Assume that 𝑅 is large enough compared with 𝑟 that we can move the ring about freely without its leaving the region of uniform field. Suppose that the ends of a diameter of the ring are at X and Y. Translate the ring in a direction parallel to XY so that the diameter-end that was at X is now at Y. If the induced $\vec E$ is in the same sense (clockwise, say) round the ring before and after the translation, the direction of the field at Y has reversed. $\endgroup$ Commented Aug 28, 2020 at 12:48
  • $\begingroup$ @PhilipWood do you mind if I add this in the question ? $\endgroup$
    – Physicsa
    Commented Aug 28, 2020 at 14:39
  • $\begingroup$ No objection at all. $\endgroup$ Commented Aug 28, 2020 at 18:29

1 Answer 1


But now suppose that I displace the loop towards the right by d(d<R−r) , I can again follow the above steps and get induced electric field as a function of position vector in the plane.

I think this is the problem of your argument. Symmetry comes first, so if you displace the loop from the cylinder axis, then you can't expect the electric field to be constant in all points of the loop. Consequently you can no longer simplify the integral to $2\pi r E$.

Notice that in principle you can chose such a loop and still Faraday-Lenz law holds, but you encounter a computational problem concerning the integral, while if you choose a loop that reflects the symmetry of the problem everything gets much easier.

Finally, no matter the choice of the loop, the electric field lines for $r<R$ are circles centered about the cylinder axis and the modulus is $E(r) = rC/2$.

  • $\begingroup$ So, if i didn't know the distribution of magnetic field , I cannot uniquely determine Induced electric field ? And you're implying that the integral has same value for any position of loop inside the magnetic region, I think this would be an amazing property of the field. Is there some concrete mathematical proof ? $\endgroup$
    – Physicsa
    Commented Aug 28, 2020 at 2:26
  • $\begingroup$ If the magnetic field is not uniform then still there is a unique electric field configuration. The problem is that you can't find it with paper and pen, you can only solve it numerically. Yes, I am implying that in a constant magnetic field, the circulation of the electric field is constant no matter the position of the loop (if you don't change the radius). However this physical law is an experimental fact, I don't think you can proove it mathematically $\endgroup$
    – Matteo
    Commented Aug 28, 2020 at 7:46

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