In article , ShLampen wrote:
In article , (Stewart
Pinkerton) writes:
So what? Electromotive force is still generated, which causes the
wires to vibrate.
I would bet that most of this motion (if any) is caused by the PHYSICAL
vibration of the speaker itself. Are you saying you can feel it at the
amplifier output?
Actually, the force can be calculated.
This is a high-school level physics calculation. I'll avoid the
derivation and simply show that the force between two parallel
conductors carrying a current is calculated as:
F/L = u0/(4 pi) 2I^2/r
where F/L is the force per unit length, u0 is the permability of
free space, close enough for this purpose, I is the current
(and the I^ term assumes that the currents are of equal
magnitude) and r is the distance separating them.
Indeed, the definition of an ampere, the fundamental unit of
current, is in terms of the force between two parallel wires
carrying equal currents:
"An ampere is that constant current which, when flowing in
each of two infinitely long parallel wires that are 2 meter
apart in a vacuum produces a force on each wire of
2*10^-7 newton per meter of length"
Semat, Fundamentals of Physics, Holt, Rinehart &
Winston, 4th ed.
Now, 1 amp is sufficient current to produce 8 watt into a
nominal 8 ohm loudspeaker load, so that gives us a good basis on
which to make some calculations. The separation of the
conductors in a typical 10 gauge zip-cord style speaker wire is
on the order of 5 mm, or 0.005 meters. Since the force on each
goes as the inverse separation, we would expect the force top be
200 times that if the conductors were separated by 1 meter, so
that the total force is:
200 * 2 * 10^-7 newtons/meter
or 4 * 10^-5 newtons/meter per conductor, or a total net force
of 8 * 10^-5 newtons/meter. That's 0.00008 N/m
the force exerted by a 1 gram weight due to the acceleration of
gravity at the earths surface is, (since F = m*a)
F = 0.001 * 9.8 m/sec^2
or 9.8 * 10^-3 newtons. That's 0.0098 N, call is 0.01 N.
Let's put this into perspective. A standard U. S. one cent coin
(a penny) has a mass of 3 grams and weighs in, therefore, at 0.3
newtons. To generate the equivalent force of that 1 ampere
current along 1 meter of wire, we'd have to take that penny,
chop it into about 350 individual pieces, draw each piece
(weighing about 8 micrograms each) into a wire 1 meter long, and
carefully lay it on top of the insulation so that the wire is
compressed under that heavy burden.
Now, using the softest PVC insulation imaginable, let's estimate
its mechanical compliance is on the order of about .1 mm/N/cm of
length, or, thus, about 0.001 mm/N/m of length (just did a VERY
rough measurement). We can now calculate the total displacement
of the conductors in such a wire. simply:
x = F * C
where x is the displacement in meters, F is the applied force in
newtons and c is the mechanical compliance in meters/newton. In
our case, since F = 8 * 10^-5 n/m and C = 10^-6 m/n/m, then
x = 8*10^-5 n * 10^-6 m/n/m
x = 8 * 10^-11 meters
that's 80 TRILLIONTHS of a meter. The radius of a hydrogen atom
is on the order of 10^10 meters, so we're talking about
movements comparable to atomic dimensions for fairly sizeable
currents.
--
| Dick Pierce |
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