Radiation fields from a rotating ring with current

[fluidistic]

Homework Statement

A ring with radius R has a constant current and is rotating around a diameter with constant angular velocity omega.
1)Calculate the radiation fields far from the ring.
2)What's the direction of the polarization for an observer along the axis of rotation?
3)What are the directions corresponding to maximum and minimum radiated power?
4)Calculate the force required to maintain a constant angular velocity.

Homework Equations

That's one big problem, I don't find any helpful equation whatsoever despite efforts.

The Attempt at a Solution

I am completely stuck on part 1, I've been skimming through both Jackson's and Zangwill's books as well as searching the web (found https://www.physicsforums.com/threads/electric-dipole-radiation-from-a-spinning-current-loop.800676/).
I am not really understanding the proceedure for part 1). The true expressions for the E and H fields are too complex to calculate so we make an approximation for far fields and we also make another approximation? I.e. we will consider only the electric dipole radiation assuming it is different from 0. If it is worth 0, then we will consider the magnetic dipole radiation assuming it is different from 0. If it is worth 0, then we will consider the radiation from the electric quadrupolar term? Is this all correct?
So I start with the electric dipole term, $\vec p = \int \vec r ' \rho (\vec r ' ) d^3 r'$. In my problem there's no explicit rho (only J) but I'm not sure it's enough to conclude that $\vec p=\vec 0$. I don't find any definition of the electric dipole term in terms of the current.

Now for the magnetic dipole term, it's worth $\frac{1}{2}\int (\vec r \times \vec J)d^3r$. My problem is that I couldn't find out how to write the current in neither spherical nor Cartesian coordinates. I'm currently at page 413 of Jackson's 3rd edition and I'm at a loss. Even if I had calculated $\vec m$ I wouldn't know how to proceed.
Thanks for any help, really.

 

[mfb]

Looks good so far. There is no non-zero charge density anywhere, so no electric contributions.

Finding $\vec m$ should be the easiest way to proceed. It is possible to evaluate the 3D integral with a suitable current distribution, but that could get messy.

 

[fluidistic]

Looks good so far. There is no non-zero charge density anywhere, so no electric contributions.

Finding $\vec m$ should be the easiest way to proceed. It is possible to evaluate the 3D integral with a suitable current distribution, but that could get messy.

Thank you so much for the reply!
Ok, if the ring wasn't moving the magnetic moment would be worth $\vec m_0=I_0\pi R^2\hat x=m_0\hat x$ but if it's rotating around the z-axis then $\vec m(t)=m_0[\cos (\omega t)\hat x + \sin (\omega t) \hat y]$. Since it is different from 0, I shouldn't bother by calculating the quadrupole term (which I guess would vanish), because this magnetic dipole term is the one that dominates in the radiation zone.
From it, it's not clear to me how to compute the fields. From Zangwill's book, he says it's more convenient to work with $\alpha (\vec r , t)=\frac{\partial \vec A_{rad}(\vec r ,t)}{\partial t}$ than with the potential vector $\vec A$ in order to get the fields. He writes a few pages later that $\alpha (\vec r , t)=\frac{d^2}{dt^2}\vec p _{ret} + \frac{1}{c}\frac{d^2}{dt^2}\vec m _{ret} \times \hat r + \frac{1}{c}\frac{d^3}{dt^3}\vec Q _{ret} \cdot \hat r+...$.
In my case $\vec \alpha (\vec r, t)\approx \frac{-\omega ^2 m_0}{c}[\cos [\omega (t-r/c) + \sin [\omega (t-r/c)]]]\times \hat r$.
If I use $\hat r=\cos\phi\sin\theta \hat x+\sin\theta\sin\phi\hat y+\cos\theta\hat z$, then I get \vec \alpha (\vec r, t)=-\frac{\omega^2 m_0}{c} \{ \cos [\omega (t-r/c) (\sin\phi \sin\theta \hat z - \cos \theta \hat y)] +\sin [\omega (t-r/c)(\cos\theta \hat x - \cos \phi \sin \theta \hat z)] \}.
Now I can compute the fields with $c\vec B_{\text{rad}}=-\hat r \times \frac{\mu_0}{4\pi r} \vec \alpha (\vec r ,t)$ (I am unsure of this formula, I couldn't find it in the book but that the only way to get the fields I see, by using the alpha vector) and $\vec E _{\text{rad}}=-\hat r \times c \vec B _{\text{rad}}$. Does this look correct?

 

[mfb]

I don't know those formulas, but they look reasonable.

 

[fluidistic]

I see, thanks.
I'm getting a really huge expression for B... it's intimidating and I see no simplification although it's quite possible that I'm missing some simplification.
I get c \vec B_{\text{rad}}=\frac{\omega^2 m_0}{c}\frac{\mu_0}{4\pi r} \{ \left [ \sin ^2 \theta \sin ^2 \phi \cos [\omega (t-r/c)] - \cos \phi \sin ^2 \theta \sin \phi \sin[\omega (t-r/c)] + \cos ^2 \theta \cos [\omega(t-r/c)] \right ] \hat x + \left [ (-)\cos \phi \sin \phi \sin ^2 \theta \cos [\omega(t-r/c)] + \cos ^2 \phi \sin ^2 \theta \sin [\omega (t-r/c)] + \cos ^2 \theta \sin [\omega (t-r/c)] \right ] \hat y + \left [ (-)\cos \phi \sin \theta \cos \theta \cos [\omega (t-r/c)] - \sin \theta \sin \phi \cos\theta \sin [\omega (t-r/c)] \right ] \hat z \}, honestly I fail to see the point of the exercise if that's the result they expected...