My explanation turned out to have a gaping hole in it.
First, background:
-- A wire carrying current has a "self inductance", which is fixed
per unit length.
-- The self inductance of a wire is larger for a ferromagnetic wire
than a nonmagnetic wire, so the effects described here are larger
for ferromagnetic bearings than for stainless steel (nonmagnetic)
bearings.
That sounds good.
Now the bad part:
Imagine a ring, connected to a current source, with a sliding contact
(i.e., a ball bearing) running around the inside of the ring, going
counter clockwise (the "positive" direction, in common usage).
As I observed, as the contact moves, the current on the ring "behind"
it must decrease, while the current on the ring "in front of it" must
increase. Current "resists" decreasing or increasing, so we'd expect
the current change to "lag behind" as the contact moves. That would
tend to make the current "follow" the contact.
However, there's a second effect, which I neglected: As the contact
moves past a particular point on the ring, the current in that (tiny)
portion of the ring REVERSES. Of course, current in a wire resists
changing and that includes short wires, too. This effect will tend to
force the current "behind" the wire to decrease more rapidly than it
might otherwise do. That's not so good -- it's acting against the
motor.
An analysis of the two effects seems to indicate that they exactly
cancel. We'll now show the analyses of those effects which led to
that conclusion.
* * * *
We'll start with the back EMF at a sliding contact.
Imagine a wire, fixed at one end, with a sliding contact. Assume there
is a constant current, "I", through the wire, running from the fixed
terminal to the sliding contact. It looks something like this:
Fixed
Terminal
O-------------------------------
^
|
Sliding Contact
Moving at velocity "v"
As the contact slides along the wire at velocity "v", there will be a
voltage induced between the fixed terminal and the contact. Call that
voltage V. Then, if the self-inductance of the wire per unit length is
"\rho", we'll have:
V = - \rho * v * I
as I show, far below, in footnote [1].
Now if we bend the wire into a ring, the current, I, divides into a
portion going down one side and a portion going down the other side.
The *total change* in current at a particular point as the sliding
contact (i.e., ball bearing) passes that point on the wire will still be
I, however, so the voltage induced by the movement of the contact around
the ring will still be V = \rho * v * I.
If we have "k" contacts, instead of just 1, then the current going
through any particular contact will be I/k rather than I. The voltage
induced by each one, being a purely local affair, will therefore be
\rho*v*I/k. But since there are k of them, the total induced voltage
will be
V(total) = - k * \rho * v * (I/k) = - \rho * v * I
and in fact it doesn't make any difference how many balls there are in
the bearing. The "induced EMF" will be -\rho*v*I.
Now, that voltage points BACKWARDS around the ring, versus the
direction the balls are rotating, and in fact that's the WRONG
DIRECTION if we want to induce a useful B field.
Furthermore, the voltage change when we go all the way around the ring
must sum to zero. So, what's providing the "balancing" voltage? If
it were just resistance of the wire, then there'd be a current going
wrong-way round the ring caused by this voltage, and the motor would
stop dead.
But there's a second induced voltage here. Let's assume we have one
contact, rotating around the ring in the positive direction. Let the
current going from the fixed terminal to the sliding contact in the
positive direction be I1, and the current going from the sliding
contact to the fixed terminal be I2. Current I1 will be decreasing as
the contact moves, and current I2 will be increasing, and that, with
the inductance of the ring material, will induce a voltage on the two
segments, which will clearly be in the positive direction.
As I'll show below, in footnote [2], that voltage is equal to
W = + \rho * v * I
and it exactly balances the "contact voltage" and we are back at
square 1.
***************************************
Footnote [1]
***************************************
Here's my rather sloppy derivation of the induced contact voltage V,
which comes from macroscopic changes in current in microscopic pieces
of a wire as the contact slides past:
Let l = length of wire, from the terminal to the sliding contact point
(that's "ell" not "one")
Let L = *total* inductance of wire = \rho * l
Let V = voltage induced by the sliding of the contact
Let i = current through a tiny piece of the wire past which
the contact moves
Then we have:
dL = \rho dl { tiny inductance of the "added" tiny piece
of wire when the contact moves dl units }
di = change in current in "dl" as contact passes = I
{ Note that the "infinitesimal" change di, which is
the change in current in the "newly added" piece of
wire "dl", is actually the fixed macroscopic
value "I". }
dl = v * dt { Change in length is velocity of the contact,
times the time which elapses }
And then:
V = - dL * di/dt // inductance * rate of change in current
= - \rho * dl * I/dt // di -> I, dL->\rho * dl
= - \rho * v * dt * I/dt // velocity = dl/dt
= - \rho * v * I // and we cancel the dt's
and so the voltage induced by the sliding of the contact is
V = - \rho * v * I
I played kind of fast and loose with the infinitesimals there, though,
so I'm not sure that's all 100% kosher. Formula looks reasonable,
however, and it appears to match what we get from conservation of
energy considerations, given that the power here is going into the B
field around the wire.
******************************************
Footnote [2]
******************************************
Here's my derivation of the portion of the EMF which comes from
microscopic changes to the current in macroscopic parts of the ring.
Make the radius of the ring 1, so we can ignore it.
Let:
omega = resistance of the ring per unit length
rho = inductance per unit length
I = total current from terminal to sliding contact
I1 = current going from the terminal to the contact in the
positive direction
I2 = current going from the sliding contact to the terminal,
*also* in the positive direction. Note that if the
ring voltage is positive, then I2 < 0 < I1.
theta = angle at which the contact is currently located
Note that if the contact is moving in the positive
direction, then theta is increasing (dtheta/dt > 0)
R1 = resistance of the ring from the terminal around to
the sliding contact, going positively
R2 = resistance of the ring from the sliding contact to
the terminal, going positively again
L1 = inductance of the segment of the ring between the
terminal and the sliding contact, going positively
L2 = inductance of the segment of the ring from the
contact to the terminal, going positively
Then we have a bunch of stuff:
1) I = I1 - I2
R1 = theta * omega
R2 = (2 pi - theta) * omega
L1 = theta * rho
L2 = (2 pi - theta) * rho
I1*R1 = -I2*R2
2) I1 = (1 - 2pi/theta)*I2
3) I2 = theta/(theta - 2pi) * I1
Plugging (3) in to (1) we get
I1 = (1 - theta/2pi)*I
and plugging (2) into (1) we get
I2 = - theta/2pi * I
and if we subtract I2 from I1 we get back I which is nice, means we
may not be totally lost.
Taking derivatives,
dI1/dt = - 1/2pi * I * dtheta/dt
dI2/dt = - 1/2pi * I * dtheta/dt
Note that they're equal, and both decreasing, but I1 starts large and
goes to zero, while I2 starts at zero and goes down, as the contact
goes around the ring.
Now the induced voltages are as follows, where I'm using W instead of
V because I used V up above already:
W1 = - L1 * dI1/dt
= rho * theta * 1/2pi * I * dtheta/dt
W2 = - L2 * dI2/dt
= rho * (2pi - theta) * 1/2pi * I * dtheta/dt
and the total induced voltage by this mechanism is:
W = W1 + W2
= rho/2pi * I * dtheta/dt * [2pi - theta + theta]
= rho * I * dtheta/dt
= rho * v * I
where v=dtheta/dt, with our assumption that the radius of the ring is
one. (Of course the ring radius doesn't affect the conclusions; it
depends only on our choice of units.)
If we have multiple sliding contacts, I'm not certain but I think the
currents from them "superpose" linearly and the induced voltage is the
sum of the induced voltages. Since with k contacts the current to each
is divided by k, the result is that, once again, the number of balls
in the bearing doesn't matter; the induced voltage is the same.
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