- #1
987oscar
- 7
- 0
Homework Statement
These is a problem from my textbook. You have an infinite solenoid with n turns per unit longitude, radius "a" and a stationary current I. In the axis there exits a uniform line charge with lineal density λ. Compute total electromagnetic momentum (lineal and angular).
Homework Equations
Poynting vector: $$\mathbf{S}=\frac{1}{\mu_0}\mathbf{E}\times\mathbf{B}$$.
Momentum density: $$\mathbf{g}=\mu_0\epsilon_0\mathbf{S}$$
The Attempt at a Solution
Well, I computed electric and magnetic fields as usual.
Electric field :
$$\mathbf{E}=\frac{\lambda}{2\pi\epsilon_0 r}\mathbf{\hat{r}}$$
Magnetic field (only exits inside solenoid):
$$\mathbf{B}=\mu_0 nI\mathbf{\hat{z}}$$
From here I find Poynting vector:
$$\mathbf{S}=\frac{-\lambda n I}{2\pi\epsilon_0 r}\mathbf{\hat{\phi}}$$
and momentum density:
$$\mathbf{g}=-\frac{\lambda n I}{2\pi\epsilon_0 c^2}\frac{1}{r}\mathbf{\hat{\phi}}$$
From here I have angular momentum ##\mathbf{l}=\mathbf{r}\times\mathbf{g}##:
$$\mathbf{l}=-\frac{\lambda nI}{2\pi\epsilon_0 c^2}\mathbf{\hat{z}}.$$
Integrating inside the solenoid for length 1 i find total angular momentum per unit longitude:
$$\mathbf{L}=\frac{-\lambda n Ia^2}{2\epsilon_0 c^2}\mathbf{\hat{z}}$$
When I try to compute total momentum per unit longitude I find:
$$\mathbf{G}=-\frac{\lambda n I}{2\pi\epsilon_0 c^2}\int_0^1dz\int_0^{2\pi}d\phi\int_0^a\frac{1}{r}rdr\mathbf{\hat{\phi}}=\frac{-\lambda nIa}{\epsilon_0 c^2}\mathbf{\hat{\phi}}$$.
Now the solution provided is ##\mathbf{G}=0## because momentum ##\mathbf{g}## forms closed lines ¿? Where is the mistake in my calculation?. I have been thinking about it but I can't find. Any idea? (##\mathbf{L}## calculation seems to be correct)
Thank you in advance!