Calculating Power Radiated & Energy in Magnetic Loop

[EEWannabe]

Homework Statement

A Loop of height L and initial width X is placed in a magnetic field and the plane of the loop is perpendicular to the field, the loop is connected to a light bulb of resistance R. By applying a force F to the left hand side of the loop it can be compressed at a constant rate dx/dt = -a.

Calculate the power radiated by the lightbulb.
Calculate the total energy emitted by the lightbulb, does it depend on a? Comment on the answer with reference to the force F.

http://img859.imageshack.us/img859/2079/faraday.jpg

Homework Equations

E = - d(flux)/dt

The Attempt at a Solution

So since the flux = BA (sin90) = BA = BLx, d(flux)/dt = BLdx/dt = BLa

So E = -BLa. This means the power through the bulb = E^2/R = (B^2 L^2 a^2) / R.

And the overall energy is Power * Time = ((B^2 L^2 a^2) / R) * (X / a) = (B^2 L^2 a X) / R

And comparing this to the energy used by the force.

P = Fv = BIL a, from above the I is given by E/R = BLa/R so P = B^2 L^2 a^2 / R. So it's the same result.

However! I can't come to terms with this result, I don't see how doing it quickly will use more evergy than doing it slowly. I don't see why to the overall energy the speed, a, makes a difference?

Maybe this is trivial, or maybe I've made a mistake with the algebra, but anyone clearing this up would be most appreciated!

 

[zhermes]

Excellent work EEWannabe, looks great.

However! I can't come to terms with this result, I don't see how doing it quickly will use more evergy than doing it slowly. I don't see why to the overall energy the speed, a, makes a difference?
Maybe this is trivial, or maybe I've made a mistake with the algebra, but anyone clearing this up would be most appreciated!

You've shown that the EMF is proportional to the velocity. I.e. the faster the flux changes, the more voltage is produced. On the other hand the faster it moves the less time there is to dissipate energy. BUT---the power goes as the velocity squared, while the duration decreases linearly with a; thus, overall, the energy radiated still increases with velocity.

Does that help at all?

 

[EEWannabe]

Ah I've gotcha, I guess I was getting confused as I wasn't confident in the maths and I carried on thinking about the argument you outlined.

Thanks a lot for the help!

 

[zhermes]

Happy to help; your derivation was really sharp, couldn't have been done better.