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Re: [Phys-l] drop a metal cylinder through a solenoid



On 03/24/2012 07:11 PM, Carl Mungan wrote:

That works out to be B^2 g^0.5 L^1.5 d^3 / p = 1 uJ [1]

So indeed, the damping is small.

I get something quite a bit smaller.

Equation [1] looks fishy to me because it does not depend on ε0.

.... (V^2/R) [2]

Equation [2] also looks fishy because most of the V is opposed by
the induced electric field. For example, consider the ultra-low
frequency limit, where the bob's velocity is virtually constant.
The induced voltage is some nonzero constant, whereas the current
is zero, so the power should be zero ... not V^2/R. In other words,
the power should scale like V dot, not like V.

I actually calculate the charge. I don't call it a capacitor, but
rather a dielectric with an induced charge. The surface charge is
determined by the requirement that it null the field inside the
conductor. Numerically the charge is the same as you would get for
a capacitor, but the field-line pattern is dramatically different.

Given the charge I can calculate the I^2 R heating, which turns out
to be waaaay smaller than V^2 / R. Subject to a couple of dirty
shortcuts, I get 2.6e-38 watts average power dissipation. I checked
my work using the Maxima computer-algebra system. The code is at
http://www.av8n.com/physics/bob-mag-damping.max
so you can play with it if you want.

So indeed, the damping is small.

Very very small.

========

Here's something that may make the result more intuitive: Calculate
the RC time for charge to redistribute itself as necessary from place
to place within the bob. Compare this so the timescale of the mechanical
oscillation. To the extent that this is a small number, the power
dissipation will be small squared.