# Induced Electric Field

1. Oct 26, 2008

### ElectromagStudent

Hello all,

In Faraday's law we have that the variation of the magnetic flux in time
will generate an induced electric field that loops on itself.

My question concerns this induced electric field.

If we imagine a bar magnet moving through a circular surface and consider
the electric field induced on the circle at the boundary of this surface.
(Note: the circle may be imaginary or an actual circle made of wire)

1. What is the electric potential of this induced field (measured on the
rest frame of the surface) for one complete circular loop?

2. Since it is a loop, if we placed an imaginary test charge on it, what
would be the energy that this charge would acquire (or work done to move
this charge around the loop)? You can also imagine a superconducting wire
with no resistance going around the induced electric field.

Basically I don't understand the ideal of an electric field looping on
itself. In my mind when we move along an electric field line we must have a
delta in the potential. So is this delta in potential uniquely defined? What
if I do one lap or two or infinite laps? what would the delta potential be?

2. Oct 26, 2008

### Bossavit

>the variation of the magnetic flux in time
>will generate an induced electric field (...)
>My question concerns this induced electric field.
>What is the electric potential of this induced field (...)

That's the point: There is NO electric potential for this induced
entail curl E = 0, instead of Faraday's curl E = - d_t B, where the
r.h.s. is nonzero in the situation you describe.

3. Oct 27, 2008

### ElectromagStudent

So when this electric field appears due to the d_t B if I place a test
charge there and circulate the field I will do no work?

"Bossavit" <bossavit@lgep.supelec.removethis.fr> wrote in message
news:ge176p$ku0$1@fb07-hees.theo.physik.uni-giessen.de...
>
> >the variation of the magnetic flux in time
> >will generate an induced electric field (...)
> >My question concerns this induced electric field.
> >What is the electric potential of this induced field (...)

>
> That's the point: There is NO electric potential for this induced
> field. If we had one, i.e., if we had E = - grad Phi, that would
> entail curl E = 0, instead of Faraday's curl E = - d_t B, where the
> r.h.s. is nonzero in the situation you describe.

4. Oct 28, 2008

### Roland Franzius

ElectromagStudent schrieb:
> Hello all,
>
> In Faraday's law we have that the variation of the magnetic flux in time
> will generate an induced electric field that loops on itself.
>
> My question concerns this induced electric field.
>
> If we imagine a bar magnet moving through a circular surface and consider
> the electric field induced on the circle at the boundary of this surface.
> (Note: the circle may be imaginary or an actual circle made of wire)
>
> 1. What is the electric potential of this induced field (measured on the
> rest frame of the surface) for one complete circular loop?
>
> 2. Since it is a loop, if we placed an imaginary test charge on it, what
> would be the energy that this charge would acquire (or work done to move
> this charge around the loop)? You can also imagine a superconducting wire
> with no resistance going around the induced electric field.
>
> Basically I don't understand the ideal of an electric field looping on
> itself. In my mind when we move along an electric field line we must have a
> delta in the potential. So is this delta in potential uniquely defined? What
> if I do one lap or two or infinite laps? what would the delta potential be?

The Maxwell equations are a linear set, so for the fundamentals one is
able to understand everything in a linear setting.

Assume a constant magnetic field rising with linear in time t with
constant rate b

B = t (0,0,b).

This fixes the induced part of the electric field by the law of
induction (t = c*time)

dB/dt + curl E = 0

Near to the origin of space and time you may choose eg a circular
electric field around the origin

E = 1/2 (-y, x, 0)

Of course, electric fields around other origins or superpositions with
constant fields yield the same curl

E = 1/2 (-(y-y0) s , (x-x0)(1-s)), s,x0,y0 some real parameter.

So the actual electric field in such a situation will be specified by
boundary or symmetry conditions of the real situation. This may be a non
constant magnetic field to fix a midpoint. Or you have boundary
conditons of conducting surfaces - eg the superconducting ring with a
slit to measure the electric line integral along the wire aka the
induced voltage over the slit.

Of course, there exists a 4-potential for this situation.

A = (A0,A1,A2,A3) = 1/2 t (0 , -(y-y0) s, (x-x0)(1-s), 0)

with B and E given by

B = curl spacepart A = curl(A1,A2,A3)

E = spacepart d/dt A

Scalar potentials for B and E do not exist, since line integrals of
(-y,x,0) depend on the contour of integration chosen.

In a closed normal conducting wire the time changing magnetic field is
transferring energy to the electron gas by induction heating up the wire.

In a superconducting ring the rising magnetic field - assumed starting
at zero strength - is inducing a surface current distribution.

The resulting magnetic field is a topologically determined solution of
the time-linear Maxwell boundary problem with the effect, that the
magnetic flux through the superconducting torus up to its outermost
contuour is exactly zero. The inner part of the torus is free of any
fields.

In the superconducting surface Maxwells equation are satisfied by an
exponential decaying distribution of the surface current and the
shielded outer magnetic field (Meissner effect).

These topological implications of Maxwells equations cannot be
understood locally in terms of fields and potentials. They are deeply
connected to the quantum mechanical description of electric charges and
currents and the aspect of electromagnetic fields as a locally gauge
field for linear and angular momentum and of charged quantum particles.

The kind of solutions of Maxwell equations with superconducting
boundaries depend on the topological genus of surface. There is a
correspondence of the vector space of current distributions on surfaces
without poles and zeros on the one hand and the number of holes in the
surface of the other hand.

Any spherelike object has no holes and so there is no nonsingular
current distribution to shield an external or internal magnetic field:
Spheres, cylinders, cubes dont show super currents.

Any topologically toruslike object has two linear independent current
distibutions as solutions of Maxwells equations.

The first solution - a flow along the rim around the axis without any
carge winding through the hole - is the boundary (and the physical
source) of external magnetic fields with internal zero field.

The second - a flow through the hole without any charge rotating arund
the axis - generates an internal magnetic field with zero field in the
outer space like a closed ore infinite selenoid.

Funny things and a branch of pure mathematics. It shows the way by which
the many degrees of freedom, the Maxwell equations don't fix, can be
used to model the charge and current field distibutions in ideal boundaries.

--

Roland Franzius

5. Oct 28, 2008

### J. J. Lodder

ElectromagStudent <test@hotmail.com> wrote:

> Hello all,
>
> In Faraday's law we have that the variation of the magnetic flux in time
> will generate an induced electric field that loops on itself.
>
> My question concerns this induced electric field.
>
> If we imagine a bar magnet moving through a circular surface and consider
> the electric field induced on the circle at the boundary of this surface.
> (Note: the circle may be imaginary or an actual circle made of wire)
>
> 1. What is the electric potential of this induced field (measured on the
> rest frame of the surface) for one complete circular loop?
>
> 2. Since it is a loop, if we placed an imaginary test charge on it, what
> would be the energy that this charge would acquire (or work done to move
> this charge around the loop)? You can also imagine a superconducting wire
> with no resistance going around the induced electric field.
>
> Basically I don't understand the ideal of an electric field looping on
> itself. In my mind when we move along an electric field line we must have a
> delta in the potential. So is this delta in potential uniquely defined? What
> if I do one lap or two or infinite laps? what would the delta potential be?

What you need to understand is that there is
in general no potential.
There is a simple undergraduate experiment to demonstrate this.
Take a long thin coil, and feed it altenating current.
Loop a wire (or a string of resistors) around it.
Alternating current will be induced.
What is the potential over a given resistor?
When you try to measure it with an AC voltmeter
you will find that the result depends
on the way the leads are arranged.
Hence no unique well defined potential difference.

Jan

6. Oct 28, 2008

### J. J. Lodder

ElectromagStudent <test@hotmail.com> wrote:

> So when this electric field appears due to the d_t B if I place a test
> charge there and circulate the field I will do no work?

The E-field will do work.
To get down to practicalities you might try to understand
how a tokamak device works.

Jan

--
A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet?

> "Bossavit" <bossavit@lgep.supelec.removethis.fr> wrote in message
> news:ge176p$ku0$1@fb07-hees.theo.physik.uni-giessen.de...
> >
> > >the variation of the magnetic flux in time
> > >will generate an induced electric field (...)
> > >My question concerns this induced electric field.
> > >What is the electric potential of this induced field (...)

> >
> > That's the point: There is NO electric potential for this induced
> > field. If we had one, i.e., if we had E = - grad Phi, that would
> > entail curl E = 0, instead of Faraday's curl E = - d_t B, where the
> > r.h.s. is nonzero in the situation you describe.

7. Oct 28, 2008

### Bossavit

>So when this electric field appears due to the d_t B if I place a test
>charge there and circulate the field I will do no work?

No, there is some work involved indeed. But it cannot be expressed in
terms of some electric potential, which in the situation you describe
does not exist.

Consider this highly idealized setup: A ring, made of some magic
material which contains only ONE electron and will not let it escape
outside, but will let it go freely around the loop (no drag). At t =
0, B = 0 and E = 0 all over, and the electron is idle, somewhere.
For t > 0, let B change, by the effect of some outside agency. (Your
hand moving a magnet, for instance.) This creates an electric field E,
which obeys the equation rot E = - dB/dt, among others. The charge is
pushed by E, and accelerates. After the first lap, which takes some
time T > 0 to complete, it has acquired kinetic energy, equal [give or
take a minus sign that I'm too lazy to think about] to the flux of B,
at time T, through the loop.

This energy has been pumped out from the agency, whatever it is, that
caused the built-up of B. [You may be able to devise an experiment in
which you'd *feel* the difference it makes, when you move the magnet
around, whether some conductor is close to you or not.]

The force on the electric charge has two parts, Coulomb's and Lorentz's,
but only the former does work.

You can't determine the force from Faraday's equation alone, extra
tells you about its *curl*. But when the charge performs a closed loop,
you still can compute the *work* performed by this force, which is that
thing called "circulation" of E around the loop, and depends on the
curl of E only, by the Stokes theorem. After Faraday's law, this
circulation is the flux of dB/dt across the loop, which you know if
you know how B evolves in time.