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Re: [Phys-L] amusing electrostatics exercise

Let the current in the original wire be I. Replace the original wire with two skinny (negligible diameter) wires, wire C and wire G. Wire C lies along what was the centerline of the original fat wire with the hollow tube in it. Wire G lies along what was the centerline of the hollow tube. Wire C has a current of I*A/(A-a) in the same direction as that of the original current and wire G has a current I*a/(A-a) in the opposite direction, where A is what the cross-sectional area of the original wire without the hole would be and a is the cross-sectional area of the hole. The ends of wire C are connected to one power supply. The ends of wire G are connected to another power supply. Both power supplies are in about the same location as the original power supply that was causing the current in the wire with the tube through it. In that region of space that would be outside the surface of the original wire if it were still there, the magnetic field is, to a very good approximation, at locations far from the power supplies, identical to the magnetic field due to the original current carrying wire.

-----Original Message-----
From: Phys-l [mailto:phys-l-bounces@phys-l.org] On Behalf Of John Denker
Sent: Wednesday, February 27, 2013 2:34 PM
To: Phys-L@Phys-L.org
Subject: Re: [Phys-L] amusing electrostatics exercise

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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