# Electric field angular momentum

## Homework Statement

A straight line with charge density ##\lambda ## is in the middle of a large isolation cylinder, that can rotate around it's axis (line). Moment of inertia for that cylinder per unit of length is ##l## and electric charge density applied on the cylinder is ##\frac{\lambda }{2\pi a}## where ##a## is the radius of the cylinder. At first we have a magnetic field ##B##, which is than slowly turned off. Calculate the angular velocity ##\omega ## after the magnetic filed is turned off. [Feynman paradox]

## The Attempt at a Solution

Angluar momentum before $$\Gamma =\varepsilon _0 \int EB dV=\frac{\lambda l B a^2}{2}$$ And after...? Yup, i have i guesstion here. I know that angular momentum should have two terms, one ## J\omega## and the other one that comes from the magnetic filed induced due to the rotation of the charged cylinder. I do have a couple of question concerning that second term.

To get that second term, I have to calculate the magnetic filed ##B_2## due to the rotation of the cylinder with angular velocity ##\omega ##.
I have two options but I can't decide which one is right or which one is not. Here they are:
1)
Simply $$B=\frac{\mu _0 NI}{l}=\frac{\mu _0}{l}\frac{\lambda l \omega}{2\pi}$$ where I used ##N=1## and ##I=\frac{e\omega }{2\pi}##. I don't really understand this option - but it is the one we wrote at university. My solution differs a bit:
2) Let's ignore that ##B## and ##j## are vectors for a moment $$\nabla \times B=\mu _0 j$$ $$\nabla \times B=\mu _0 (\sigma 2\pi l a)(r\omega)$$ if ##l## is the length of the cylinder. Integrating the last equation and using stokes law on the LHS leaves me with $$\int B ds =\mu _0 \sigma 2\pi la\omega \int r 2\pi r dr$$ which sadly leads to a different result than option 1). because $$B=\frac{\mu _0\lambda l \omega a^2}{3}$$

I seriously doubt we got it all wrong at university, so my question here is: Why is my solution wrong? And could somebody explain the option 1.) - especially where this ##I=\frac{e\omega }{2\pi}## comes from.

The charge on the cylinder is only on the surface and not throughout the entire cylinder. This changes $$\nabla \times B=\mu _0 (\sigma 2\pi l a)(r\omega)$$ to $$\nabla \times B=\mu _0 (\sigma 2\pi l a)(a\omega)$$ and leads to the same result as in option 1.).