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Induced electric field ambiguities

Physics Asked by Physicsa on December 28, 2020

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 :
$$ϵ=∮overrightarrow{E}.doverrightarrow{l}=-frac{dɸ}{dt}$$
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….

One Answer

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 $2pi r E$.

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 $E(r) = rC/2$.

Answered by Matteo on December 28, 2020

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