Re: Induced E Fields in Solenoids

[palmer@sfu.ca (Leigh Palmer)]

John Mallinckrodt offers a good answer to the question:

> Suppose we look along the axis of the solenoid with B toward us and
> increasing.  Then the emf around *any* closed path within the solenoid
> will be clockwise. Symmetry requires that the E field itself be
> clockwise, directed tangent to circles coaxial with the solenoid, and of
> constant magnitude on these circles.  However, the symmetry is disrupted
> by the introduction of an off-axis conducting loop so it should be no
> surprise that the previous result no longer holds.  It is, nevertheless,
> highly instructive to ask (as you have) *why*.  The answer requires an
> understanding of transient effects in circuits a la Chabay and Sherwood.
>
> Within the conducting loop the current density that exists at every point
> *must* indicate the direction of the local electric field since it is only
> *local* electric fields that can drive current.  Faraday's law still
> demands that the emf and, therefore, the current around the ring be in the
> clockwise direction.

It might be pointed out that this is not quite the answer to the question
that was asked. In general there will be another component of the induced
E field which is transverse to the wire at each point, but which is
balanced exactly by surface charges on the wire (as John points out
below). The local E field within the wire must, of course, be in the
direction of the current density as you say, but in the limit of a very
thin wire the E field must be exactly as it is for the empty solenoid
except for the region within the wire itself. Only if the changing
current in the wire itself is significant will the induced E field differ
from what is calculated for the empty solenoid. The E field everywhere
will then be a superposition of the original empty solenoid E field, the
E field due to the changing current in the loop, and the E field due to
surface charges on the wire.

On reexamining what I have written I see that it is no more transparent
than John's explanation. If Chabay and Sherwood have some clever way of
treating this conceptually I would like to see it. I have no trouble
understanding it myself (I've been teaching E&M for 31 years), but it is
evident that I would have trouble teaching it!

> Now imagine what happens when the B field first starts changing.
> Initially, different parts of the loop will "see" different electric
> fields--those derived from the symmetric analysis.  Thus, in an off axis
> loop, different currents will be driven in different portions of the loop.
> Of course, this leads to progressive "pile ups" and "depletions" of charge
> (i.e., non zero partial time derivatives of charge density) at various
> locations around the loop.  Because of the strength of the electrostatic
> interaction, however, this can't go on for long. Indeed, the process stops
> precisely when charge has been perfectly arranged so that the electric
> field is--everywhere within the loop--in exactly the direction it needs to
> be to ensure current continuity.  The time required for this process to be
> completed depends directly on the resistivity and distributed capacitance
> of the loop in an obvious--if hard to quantify--manner.
>
> I find that complicated situations such as this are very often best
> understood by considering such transient, charge redistribution effects.

I don't understand why you treat this as a transient effect. It should
hold in steady state if the applied B field increases linearly in time.
That is easier for me to understand than the transient problem.

To this I will add that the Hall effect must also be taken into account
somehow; I don't know how! This off-axis loop problem is shmutzig; could
the original poster please tell us why the problem is of interest?

Leigh