In addition to what John Denker wrote, this puzzle can also be looked at from a somewhat different angle - that of boundary conditions necessary to obtain a unique solution. As John points out, the problem is issentially that of solving differrential equations, which for a given situation is differential form of Faraday's law. In its initial form, the region of changing magnetic field is finite, which is equivalent to the requirement that the field falls off fast enough in all directions (or at least in a plane perpendicular to the field) as we go to infinity. And this is the boundary condition necessary for corresponding differential equations involving curl B. Only under these conditions we can obtain a unique solution, which is especially straightforward for an arrangement with axial symmetry. The moment you make the changing field spatially uniform in infinite Eucledean space, you retain its axial symmetry, but loose the unique position of the symmetry axis. Accordingly,
we loose the uniqueness of solut
There is an electrostatic analogue of this problem without this aggravating temporal feature, and nonetheless similarly lacking uniquieness. Consider an electric charge uniformly smeared over all space so that the volume charge density is finite and constant everywhere. Then the same argument as in Roger Haar's initial problem shows that we cannot assign any definite electric field to any point in space. Again, this would be an attempt to get a unique solution to differential equation without proper boundary conditions. In rhis case the equation would be a differential form of Gauss's law, which implies that a charge distribution occupis a finite region enclosed by a surface no matter how twisted but of finite area. In terms of the field this would mean that the field falls off fast enough in all directions as we go to infinity, - only under these conditions we would get a unique solution to equations div E = ro/epsilon, and curl E = 0, or the corresponding Poisson equati
on for potential. The solution i
Again, the absense of such solution in this case does not contradict physics, because the condition under which we try to find the solution is not physical.
Moses Fayngold,
NJIT
-----Original Message-----
From: Forum for Physics Educators on behalf of John Denker
Sent: Fri 4/8/2005 7:18 PM
To: PHYS-L@LISTS.NAU.EDU
Cc:
Subject: Re: Odd EM problem
On 04/08/05 18:50, Roger Haar wrote:
>
> We came across an odd EM problem involving the
> electric field induced by time varying magnetic
> field. If it makes you happy, let it be large or
> infinite in exten. Let it point vertically down,
> be uniform, and be increasing in magnitude.
That's called a betatron field.
> For
> any given loop in the horizontal plane it is easy
> to calculate the induced EMF around the loop and
> from that you think you could figure out the
> electric field at any point on the loop and thus
> one could calculate the electic field at any spot
> for a given time.
>
> But here is the problem, first consider a loop
> in the horizontal plane with radius of 1 meter and
> located 1 meter to the right of the origin ( the
> center is at (1,0,0) and use this loop to find
> the electric fiel at the origin. Next calcuate
> the electric feld at the orign by consider a
> similar loop but with its center at (-1,0,0). In
> one case the electric field seems to point in the
> positive y direction while in the other it points
> in the negative y direction.
The answer is simple yet profound: you cannot
calculate the E field from the given information.
There is a gauge symmetry. Moe can choose one
gauge, and Joe can choose another, and there's
no way to label either choice right or wrong.
A related fact is that given the circulation around
a loop, you cannot infer the field at any point on
the loop without additional assumptions. Sometimes
you can arrange it so that the loop sits at a place
of symmetry, so that the field is everywhere
tangential, and in this special case you can infer
the field ... but not in general.
The way to think about all such problems is to
appeal directly to the Maxwell equation. I like
the Clifford algebra representation
del F = 4 pi J [1] http://www.av8n.com/physics/maxwell-ga.htm#eq-max-exp
but you can get the same result with an equally-
negligible amount of work in the older div/grad/curl
representation.
The statement of the problem tells us that the RHS
of equation [1] is zero, and on the LHS the term
involving the time derivative of Bx is nonzero. By
looking for corresponding terms, we see that there
are two (and only two) other terms that can be of
interest:
the x derivative of Ey
minus
the y derivative of Ex
and all you know is that *some* weighted combination
of those two adds up to cancel the magnetic term.
That is *all* you can know from the given statement
of the problem.
For the gauge and origin I have chosen, at the 12:00
position on the diagram the field points one way; at
the 3:00 position it points another way, etc. etc.
Note that the B field is uniform and translationally
invariant. You can check that the circulation around
any loop you care to draw on the diagram is translationally
invariant.