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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.
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.