Calculate the time it takes the frequency to decrease

[WooHoo41]

Homework Statement

A thin copper ring rotates about an axis perpendicular to a uniform magnetic field H_{0}=200G. Its initial frequency of rotation is \omega_{0}. Calculate the time it takes the frequency to decrease to 1/e of its original value under the assumption that the energy goes into Joule Heat. Copper has resistivity of 1.7 x 10^-8 ohm-meters, and a density of 8.9 g/cm.

Homework Equations

none given

The Attempt at a Solution

The issue I'm running to at the time is that there is no mention of the radius of the loop. I think that would be important. Then I thought since we were given the resistivity, I could apply the Joule heating law and set it equal to the change in omega.

I just fail to see how this problem can be completed without the cross sectional area because from there I can go ahead and find the magnetic moment and approach everything from that direction but I feel I need the radius as a start... Sorry if this is a shoddy attempt, my brain is fried, haha.

 

[StatMechGuy]

You may find that the radius cancels out... Go ahead and solve the problem with arbitrary variables and then see if anything nice comes of it.

 

[WooHoo41]

Thanks that's definitely a good start, whether its a wild goose chase or not, I've been busy for the last hour. I'll let you know if I reach a dead end, or the answer. Anything else you can add to the problem would help greatly.

Thanks for the nudge.

 

[WooHoo41]

Oh, it was also brought to my attention that the way that the ring rotates isn't exactly clear. I'll try to explain this the best I can:

Set the H to be going in the positive Z direction. It rotates about the Y axis into the X axis. Say Z is up, Y is left and right, and X comes out of the screen. The top of the ring rotates INTO the monitor while the bottom rotates out of the screen.

It makes it reasonable say that the flux is is changing with omega perhaps,

Phi=HAsin(w) ?

thats kinda what I have been toying with now, Trying to find the current induced within the ring based off of the Electromagnetic Force. If I find the current in the ring, I can find the magnetic moment and the accompanying torque. Hopefully this is at least getting me closer.

 

[maverick280857]

You probably meant \sin(\omega t)

 

[WooHoo41]

You probably meant \sin(\omega t)

ha, very true. wouldn't be the same without good ol' t in there.