# Calculating Force using the Maxwell Stress Tensor

1. Apr 21, 2015

### RawrSpoon

1. The problem statement, all variables and given/known data
Calculate the force of magnetic attraction between the northern and southern hemispheres of a uniformly charged spinning spherical shell, with radius R, angular velocity ω, and surface charge density σ. Use the Maxwell Stress Tensor

2. Relevant equations
$$F=\oint \limits_S \! \vec{T} \cdot da - \epsilon_0 \mu_0 \frac {d}{dt} \int \limits_V S d \tau$$
Because we're dealing with steady currents, the second term goes to zero.

I know that with a sphere, we have
$$da=r^2 sin \theta d \theta d \phi$$

I also know that
$$B= \begin{array}{11} \frac {2 \mu_0 \sigma R \omega}{3}\hat{z} & inside \\ \frac {2 \mu_0 m}{3 r^3}(2 cos \theta \hat{r} - sin \theta \hat{\theta}) & outside \end{array}$$
where
$$m=\frac{4}{3} \pi \sigma \omega R^4$$

3. The attempt at a solution
I think I find Tzz since Txz and Tyz would be 0 because Bx and By are zero?

I'm honestly very confused by tensors in general, I have a pretty good idea about what they do and how they work, but I don't really know how to work them myself. Please point me in the right direction, I'm very lost as to what to do and I don't want to seem like I'm being lazy but I've broken my head all day trying to figure out what to do. The solutions manual doesn't give me an answer that makes me feel like I understand what's going on, and I can't really find anywhere that breaks it down enough for me to get that aha moment where I get it.

Thank you very much for all help :)

Last edited: Apr 21, 2015
2. Apr 27, 2015

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Apr 27, 2015

### Goddar

Hi. The most useful formula i know for the Maxwell stress tensor is (in Gaussian units) the matrix form:
T = (1/4π)[EE + BB – ½I(E2+B2)],
Where I is the identity matrix and , for example:
EE = (Exi +Eyy +Ezk )2
= (Ex)2ii + (Ey)2jj + (Ez)2kk + ExEy(ij + ji) + ExEz(ik + ki) + EyEz(jk + kj)
Now in your case you have r = R and the symmetry will make almost all integrals vanish at the end so once you've converted all unit vectors in terms of i, j and k watch for integrals with terms cosφ or sinφ from 0 to 2π because you can discard them right away.
This type of calculation is typically tedious but straightforward: symmetry indeed cancels some terms but better going through the whole machinery once in a lifetime...