MadSci Network: Engineering

Re: spinning a copper ring quite different to current in ring

Date: Mon Jul 17 09:23:07 2006
Posted By: Zack Gainsforth, Staff Scientist
Area of science: Engineering
ID: 1151430666.Eg

Hi David,

Practically speaking, B1 = B2.  This is because when the electrons collide
with the lattice, they pick up momentum.  The rate at which momentum is
transfered is a function of the resistance (being a measure of how
frequently the collisions occur).  However, the transfer of momentum is
nearly instantaneous compared to any mechanical acceleration you are likely
to impose on the ring, and so you can consider that the electrons are
moving at the same speed as the lattice.

Therefore, the magnetic field is due to the excess charge moving with the
ring as it spins.  There is really no difference between this and a
stationary ring with a current, at least as far as magnetics is concerned.

Granted, when you first accelerate the ring, the lattice will move first,
and subsequently the electrons will move, and so you can consider that
there will be a *very* brief period when the system is not in equilibrium,
and the excess charge would be *stationary* until the electrons catch up
with the lattice.  Let's investigate that for a second.

The energy of the electrons, would be

E = (N m v^2)/2

which is just (1/2)mv^2 times the number of electrons in the ring.  m is
the mass of an electron, and v is the velocity -- v is assumed to be the
same for all electrons.  It is an approximation, sure, but it won't harm us
here.  Likewise, the current moving through any point in the ring is 

I = A p q v

where A is the cross sectional area of the ring, p is the electron density
(I can't write rho easily here), q is the charge per electron, and v is the
speed they are travelling.

We can use the second equation to get:

v = I / (A p q)

and plug that into our expression for energy:

E = (N m / 2) (I / (A p q))^2

Then notice that the energy of the electrons is changing due to the
collisions with the lattice which is a function of resistance like you
noted.  We can use the relationship for the power dissipated by a current
traveling through a resistance, and the fact that the power dissipated is
the derivative of the energy with respect to time.

P = I^2 R = dE/dt

So that gives (notice that I is a function of time and so we have to use
the chain rule):

I^2 R = (N m / (2 (A p q)^2)) 2 I dI/dt

So you get:

dI/dt = ( R (A p q)^2 / (N m) ) I = k I

This is just an exponential decay with a time constant k:

I = e^(kt) =>  dI/dt = k e^(kt) = k I

We can simplify k a little bit, and we had better do so because it looks at
first glance as though it is dependent on the geometry of the ring.  Notice
that N is the total number of conducting electrons in the ring, so

N = A L p

Where A is still the cross section, L is the length (so A L is the volume
in the limit that the cross section is much less than the length -- if not
you'll have to integrate), and p is the electron density again.

k = R A^2 p^2 q^2 / ( A L p m)


k = (A / L) ( R p q^2 / m)

But the resistance of the whole ring (R) is a function of the shape of the
ring too.  The standard formula for connecting the resitance (R) of a rod
to the resistivity (p) with cross section A and length L is:

p = A R / L

We're not using a rod, but again, so long as L is much greater than A, this
is a good approximation.  (Otherwise, integrate.)

k = p^2 q^2 / m

And this is just a function of the resistivity.  This is somewhat intuitive
because we would expect that adding more copper to the ring would add more
electrons, and so the behavior aught not change.

In our case we're using copper, so you can look in the CRC Handbook, (Lide,
David R, 77th Edition)

The resistivity of copper at 300 K is about 1.7 * 10^-8 ohm-m  (Table 12-40)

The electric charge is 1.6 * 10^-19 C

Electron mass is 9.1 * 10^-31 kg

So k is really small, like 10^-24 seconds, if I didn't make a goof.  That
is to say, the time scale over which the electrons pick up momentum is MUCH
shorter than the timescale over which a pressure wave travels through the
copper.  (About 4*10^3 m/s per

So that means that you can't possibly accelerate the crystal fast enough to
get a differential between the electrons and the lattice which is big
enough to measure.

Hence, practically speaking, B1 = B2.

Hope this helps!


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