Answer to the Question 07/02

CURRENT IN A WIRE

The question was:

In freshman physics we are told that a current in an "infinite" cylindrical wire flows with constant density J independent of this distance from th e center of the wire. But we also learn how to use Ampere's Law to calculate the magnetic field in the wire. Doesn't the magnetic field act to make the current density non-uniform?


(3/2003) The problem has been solved (15/7/02) by Jared Kaplan from Stanford University, California (e-mail jaredk@stanford.edu) (below we present his (slightly edited) solution), (18/11/02) by Eric T. Lane from Physics Department, University of Tennessee at Chattanooga (e-mail elane@cecasun.utc.edu), and by (4/2/03)Srikant Marakani a graduate student the University of Chicago (e-mail srikant@midway.uchicago.edu). On 24/8/02 Regis Lachaume from UJF, Grenoble (France) ((e-mail lachaume@obs.ujf-grenoble.fr) produced of very simple estimate demonstrating that density variation must be a very small number (see PDF file).

The solution: [This is an edited solution of J. Kaplan. It is presented in MKSA (SI) unit system.] Simplifying assumptions:
-Current is carried exclusively by moving, positive charges, all moving with the same speed
-Negative charges are stationary and uniformly distributed in the wire
-the wire is cylindrical, and current/charge density is a function of the radius only

Let d=density of negative charge, r=radius, e=permitivity, u=permeability, p(r)=density of moving positive charge as a function of r, and v=velocity of positive charge. Thus the current density = p(r)*v. Then we have:
E from negative charge E-=d*r/(2e)
E from positive charge, E+= Integral(p(x)*x*dx,0,r)/(e*r)
B = uv*Integral(p(x)*x*dx,0,r)/r

Using the integral relations for E and B with the symmetry of the setup. In equilibrium, the force on the positive charge should be zero. We allow a force on the negative charge for the moment, because we are assuming it's rigidly held in place.
F on + charge = 0 =(E-+E+)*p(r) + v*p(r)*B
with the convention that a force outwards has positive sign, we get:
I(p(x)*x*dx,0,r) = d*r2/(2e*(1/(e) - u*v2))

An appropriate p(r) is evident at this point. Formally, we can differentiate with respect to r and then solve, yielding:
p(r) = d/(1 - e*u*v2)
Although the precision of the model is questionable, we are led to the result that p(r) is independent of r, thus the current should be uniformly distributed. It is interesting that we sort of see the necessity of special relativity since e*u = 1/c2 means that p(r) = d/(1-(v/c)2), meaning that the total positive charge does not equal the total negative charge exactly. Note that if the positive charge was stationary while the negative charge moved the model would give the same prediction.


Srikant Marakani wrote:

The solution is that the current density is in fact in equilibrium when it is uniform despite the magnetic field in the wire which is proportion to the radius from Ampere's law. Due to the transformation of the current density four vector (or, if you prefer it, the Lorentz contraction) the charge density of the carriers is increased giving rise to an electric field which also increases radially. It is not difficult to see from Maxwell's equations that E=-vxB where v is the drift velocity of the charges and there is no net force on the charge carriers.
Another way to solve it is to look at two fields, both for uniformly charged cylinders of equal but opposite charge density. Lorentz boost one of those fields and then superpose them with the other to get the full field. We start with only electric fields before the boost but end up with both electric and magnetic fields after the boost and superposition which have the required form. [...]
Finally, the best argument (though I am not quite sure it is watertight) I can think of is to use Lorentz invariance. We transform to the reference frame where the background charges are moving with velocity -v so that the original charge carriers are now stationary and the other way around. If there is any change in the radial charge density (except due to the Lorentz contraction) of charge carriers in the original frame, then it should occur for the background charges as well since in the new frame, they are the charge carriers. If the overall mechanical stress is zero, there is no electrical force that will make both the background and carriers move in the same way, so there should be no variation in the current density.


Eric T. Lane also remarked that the presence of confining wire is essential for obtaining the solution (described above). However, "in a plasma, moving charge sets up a magnetic field that generates an attraction among the parts of the current, giving rise to the magnetic pinch effect."
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