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Re: [Phys-L] amusing electrostatics exercise

I don't understand the terms of discourse here. The original question had
to do with a long straight wire (with a hole in it). Yes, there has to be a
return path. Who said anything about a short wire?

Here's another, different physical realization:

There's a long straight wire carrying a current; no hole. The current must
be chosen such that the current density is the same as in the proposed wire
that has the hole. There are issues with such devices, but there's a long
tradition of having them anyway. Pick a location in space. Use Ampere's law
to calculate the vector magnetic field at that location.

Destroy that wire. Take a smaller wire, the size of the proposed hole, and
place it where the hole would have been in the first (solid) wire. Run an
appropriate amount of current through it in the opposite direction
(appropriate means same current density as in the first wire). At the same
location as before, use Ampere's law to calculate the vector magnetic field
at that location.

Add vectorially the two results. That will be the field at that location of
the original proposed device. I haven't the slightest idea how charge
conservation comes into this discussion.

Bruce


On Wed, Feb 27, 2013 at 12:33 PM, John Denker <jsd@av8n.com> wrote:

On 02/27/2013 11:57 AM, Bruce Sherwood wrote:
Whether my current distribution is realizable or not is irrelevant to
using
it to calculate what the field would be, since the fields will be the
same
as in the stated situation. You don't have to have physical charges for
the
use of mirror charges in electrostatics to be valid. However, here's a
way
to do it:

Fill the hole with a large number of very thin wires. Half of them have
current running to the left, half to the right, whereas the outer
material
can be solid and has current running to the right. The currents of course
have to be chosen to be consistent with that in the outer metal. With
fine
enough wires, this is the functional equivalent of my distribution.

I assume the rightward fine wires represent the original current,
and the leftward fine wires serve to null out the current in the
hole.

My objection still stands: The leftward fine wires violate conservation
of charge. Electromagnetic calculations based on a non-conserved charge
are meaningless. See below for more details.

Remark in passing: The method of images does not violate conservation
of charge. Au contraire, it provides a convenient way of accounting
for charges that might otherwise be hard to account for.

I don't understand the charge conservation issue.

It's quite an important issue. Charge conservation is implied by
the Maxwell equations. If you have violated conservation, you have
violated the Maxwell equations. If you go down this road, you're not
doing physics.

If there is an issue,
there's a problem with a single very long current-carrying wire, a
problem
we typically sweep under the rug (or we consider a coax cable).

For the very long straight wire, we can /successfully/ sweep away
the problem by stipulating a return-wire very very far away. It is
then self-consistent to say that the field of a single wire falls
of with distance, so the return-wire can be neglected.

HOWEVER with a short wire that tactic does not work. There is some
serious physics here that you ignore at your peril.

In particular, the Biot-Savart law gives the field for an arbitrary
short piece *of an actual circuit* but the validity of the law is
contingent on the rest of the circuit being there. You cannot
apply the law to some wacky subset of the circuit and then go home.

If you don't believe me, try proving the validity of the Biot-Savart
without assuming conservation of charge aka continuity of current.

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