Am I doing this torque on dipole problem right?

  • #1
schattenjaeger
178
0

Homework Statement



You've got a circular loop with a steady current I and radius 'a' a distance r from a square current carrying loop with sides of 'b' and current I, r >>> a or b(and they're arranged in such a way as you can think of the circular loop's dipole as pointing up, and the square loop point right)

What's the torque ON the square loop caused by the circular loop, and the square loop's final orientation assuming it's free to move

Homework Equations


A couple of ways to get B, I figure the curl of A, and I have A for a dipole, which will I use with the circular loop

Then the torque on a dipole in a B field is B x m, where m is the dipole, in this case the square loop's dipole

Err, I guess that was my attempt pretty much, I just need to make sure that's the right way to do it
 

Answers and Replies

  • #2
schattenjaeger
178
0
Just a thought here, buuuut isn't the B field of a circular current just a straight line going through its center(orientated with the right hand rule)? So...would there even be a torque on the square loop?

Edit: No wait that's totally wrong
 
Last edited:
  • #3
schattenjaeger
178
0
Think I got it, I'll just put in my own work for shiggles

Well no, I don't know Latex worth a damn, suffice it to say I used the formula for B for a magnetic dipole sensed by another dipole, with the angle between them being pi/2, and ultimately I got

N=-I^2ua^2b^2/(4r^3) in the x-hat direction, which I believe means it'd rotate until it was antiparallel with the circular loop's moment, then theta =0, and no more torque(though I guess technically it'd swing past but blah blah ultimately they're anti-aligned?)
 
  • #4
Meir Achuz
Science Advisor
Homework Helper
Gold Member
3,605
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The magnetic dipole moment of each loop is just mu=IXarea/c (in gaussian units). Then, the torque is calculated just as if they were electric dipoles.
torque=muXB, where B is torque due to the other dipole.
 

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