W9_parallel_resonance.eps

262 Week 8: Faraday’s Law and Induction

This is the only possible sign that can correctly cause energy to be conserved as a charge is pushed
around the loop without gaining or losing net energy in a circuit; the charge has togainenergy from
the induced field andloseenergy into Joule heating of the resistance.

Where, exactly, is the field induced? What is it (in detail) inside of the conductor? This depends
on the resistivity and current density associated with the entire conductive pathway, since we know
that Ohm’s Law is written as:
E~=J~ρ (565)

at all points inside the current carrying conducting pathway. Whereρis zero, there is no field at all.
Whereρis not zero, there must be a field pushing the charges through the resistive conductor there.
The cumulative work done by that field equals the rate that work appears as heat in the resistor.

The best that can be said, then, is that the field appears in theentire loop, not “across the rod”
or “across the resistor” (which isn’t even moving) or “along the rails” (which might actually be a
part of the net resistance, as might be the rod). This means that the induced electric fieldforms a
closed loop. This does not violate Gauss’s Law for Electrostatics – we can add any electric field loops
we like to the electrostatic field loops it describes and they will not contribute to the net electric flux
through any closed surfaceS– but it does make one of our rules for visualizing electric field lines
obsolete. Electrostaticfields begin and end on electric charges, but induced electrodynamicfields
apparently can form closed loops, not beginning or ending on any charge!

This does have a significant impact on how we write theelectric potentialassociated with the
electric field. Recall that we defined aconservative forceas one where:
∮

C

F~·d~ℓ= 0 (566)

for all closed loopsCone can draw in space. The electrostatic field was conservative – if we let
F~=qE~and factored and cancelledq, we got:
∮

C

E~·d~ℓ= 0 (567)

Theinducedelectrodynamicfield that appears in the loop, however, isnot conservative! It has
a nonzero integral around the loop:

∆Vind=

∮

C

E~ind·d~ℓ=BLv 6 = 0 (568)

We recall that the whole point of a conservative field and its associated potential was thatE~=−∇~V
(encapsulating Newton’s Second Law) in cases where the work donegoing around a closed loop didn’t
depend on the path taken. This new result more or less means that the work donedoesdepend on
the path taken, but in a very special way. It also does indeed mean thatE~isno longer going to be
equal to the negative gradient of the electrostatic potential! We are going to get anadditionalpiece
that depends in some way on the magnetic field and the loop itself!

My goodness, things are getting complicated! Perhaps it is time to make just two more observa-
tions and then finish off this particular problem before coming back tothe equation that it seems
to imply. The first observation is that (given constantBandLin the picture above):

|∆Vind|=

∮

C

E~ind·d~ℓ=BLv=d(BLx)

dt

(569)

(becausev=dxdt) and, noting thatA=LXis the area inside of the loop we can write this as

|∆Vind|=

∮

C

E~ind·d~ℓ=d(BA)

dt

(570)