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



In other words,
the power should scale like V dot, not like V.

Interesting. If we wrote Q=CV, then V/R gives Q/RC whereas dQ/dt gives CdV/dt which works out approximately to be Qw where w=sqrt(g/L). And indeed RC is different than 1/w by about 20 orders of magnitude. So what you're saying is the charge rearranges much faster than the swing time of the bob.

Numerically the charge is the same as you would get for
a capacitor, but the field-line pattern is dramatically different.

I see the numerical part, as in my calculation above. But how is the field-line pattern different?

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.

I agree with your calculation. It doesn't seem that dirty - one just has to integrate sine squared (subject to the small-angle approximation, as you point out).

So indeed, the damping is small.

Very very small.

I'm surprised. It's completely negligible. Obviously this wasn't a surprise to you. I was willing to believe 1 uJ. But 1e-39 J: that's super tiny. Can you shed any further intuitive way to see that it's basically zero for all practical purposes? Maybe part of the answer would be: are there other systems for which the current goes like V dot?