Am I doing this torque on dipole problem right?

  • Thread starter Thread starter schattenjaeger
  • Start date Start date
  • Tags Tags
    Dipole Torque
schattenjaeger
Messages
176
Reaction score
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
 
Physics news on Phys.org
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:
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?)
 
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.
 
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top