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## Main Question or Discussion Point

Okay, I just took a test where there was a loop with a bar and a resistor in a magnetic field going into the screen as follows

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lxxxxxxxxxxxxxxxlxxx

lxxxxxxxxxxxxxxxlxxx

lxxxxxxxxxxxxxxxlxxx

l____(bar)_____lxxx

lxxxxxxxxxxxxxxxlxxx

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where the bar starts at rest, and begins to accelerate vertically down due to gravity. We were asked to find the final velocity (which I understand), and then AFTER IT HAS REACHED TERMINAL VELOCITY find the power dissipated through the resistor as a function of time and the work done by gravity as a function of time.

At first I set P=IR^2, and solved using the terminal velocity I found. But then I thought that the current isn't changing, because the velocity is constant, so I assumed it would be zero. I am very shaky on this one, and it's probably wrong, right?

Then, for the work done by gravity, I set W(t)=mgh=mgvt. But then I remembered the work-energy theorem, and there is no change in kinetic energy due to its constant velocity, and there is no net force on it anyhow. What are the answers?

____/\/\/\_____

lxxxxxxxxxxxxxxxlxxx

lxxxxxxxxxxxxxxxlxxx

lxxxxxxxxxxxxxxxlxxx

lxxxxxxxxxxxxxxxlxxx

l____(bar)_____lxxx

lxxxxxxxxxxxxxxxlxxx

lxxxxxxxxxxxxxxxlxxx

lxxxxxxxxxxxxxxxlxxx

lxxxxxxxxxxxxxxxlxxx

where the bar starts at rest, and begins to accelerate vertically down due to gravity. We were asked to find the final velocity (which I understand), and then AFTER IT HAS REACHED TERMINAL VELOCITY find the power dissipated through the resistor as a function of time and the work done by gravity as a function of time.

At first I set P=IR^2, and solved using the terminal velocity I found. But then I thought that the current isn't changing, because the velocity is constant, so I assumed it would be zero. I am very shaky on this one, and it's probably wrong, right?

Then, for the work done by gravity, I set W(t)=mgh=mgvt. But then I remembered the work-energy theorem, and there is no change in kinetic energy due to its constant velocity, and there is no net force on it anyhow. What are the answers?