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.

 

[rwisz]

I would like to commend you for your attempt at solving this problem and for asking for clarification on any discrepancies you may have encountered.

First of all, your calculations for the power radiated by the lightbulb and the total energy emitted are correct. However, I believe the confusion lies in the interpretation of the results and the relationship between the force F and the speed a.

The power radiated by the lightbulb is dependent on the square of the speed a, meaning that the faster the loop is compressed, the more power is radiated by the lightbulb. This is because the rate of change of flux (d(flux)/dt) is directly proportional to the speed a. This also explains why the overall energy emitted is also dependent on the speed a.

Now, let's consider the force F. The power used by the force is also dependent on the speed a, as you have correctly calculated. However, this does not mean that doing it quickly uses more energy than doing it slowly. The total energy used by the force is the same regardless of the speed a, as you have shown in your calculations. This is because the force F is applied over a shorter period of time when the speed a is higher, but the overall energy used by the force remains the same.

In summary, the speed a affects the power radiated by the lightbulb and the power used by the force, but it does not affect the total energy emitted or used. The overall energy is dependent on the force F and the initial dimensions of the loop (height and width), but not on the speed a.