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Kloiper
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Homework Statement
Our teacher isn't very descriptive:
A ring of radius "a" and resistance "R" is placed at the center of a long solenoid with "n" turns (assume the solenoid is longer and wider than the ring) with its axis lined up with that of the solenoid. Find the amount of work done to rotate the ring so its axis is perpendicular to the solenoid's.
Homework Equations
For a solenoid, [itex]\vec{B}= μ_0nI[/itex]
[itex]\phi = \int{\vec{B}\bullet da}[/itex]
[itex]V=-\frac{d\phi}{dt}[/itex]
The Attempt at a Solution
Unfortunately, he didn't say anything about a current, voltage, E or B field. (or H or D fields for that matter). So my first reaction was to make up a current "I" flowing through the solenoid.
This would cause a magnetic field [itex]\vec{B}=μ_0nI[/itex], and thus a flux of [itex]\phi=μ_0nI\pi a^2[/itex].
From there, I used [itex]\frac{dW}{dt} =-VI=\frac{d\phi}{dt}I[/itex], meaning [itex]dW=d\phi I[/itex], but I'm not sure which I that would be describing.
If it's I in the solenoid, which I'm pretty sure it's not, the answer works out fine because it is a constant and multiplies in just fine.
If it's I induced in the ring, we get [itex]dW=d\phi \frac{V}{R}=-\frac{d\phi^2}{dt}\frac{1}{R}[/itex]. Since we know the change in flux will be from [itex]\phi=μ_0nI\pi a^2[/itex] to [itex]\phi=0[/itex], we know [itex]d\phi[/itex]. But regardless of that, work never depends on time. Why would my term for work have some time variable in it?
I also thought that magnetic fields did no work, which means the work would be 0, which confuses me as to why our professor would ask us about something that doesn't happen.
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