# Magnetic and electric field between current carrying coaxial cables

1. Jun 6, 2009

### disillusion

1. The problem statement, all variables and given/known data

Two long concentric, cylindrical conductors of radii a and b (a<b), are maintained with a potential difference V and carry equal but opposite currents I.
An electron, with velocity u parallel to the axis, enters the evacuated region between the conductors and travels undeviated. Find an expression for |u|.

I am not sure I understood the question very well and would like to see what others make of the question.
Since the current is flowing so there would be an E field along the direction of the current, which would not affect the electron. Should I just treat the potential difference as a separate E field along the radial direction?

2. Relevant equations
$$\oint_{C} \textbf{B}\cdot \textbf{dl}=\mu_{0}\int di$$
$$\oint_{S} \textbf{E}\cdot \textbf{dA} = \frac{Q}{\epsilon_{0}}$$
$$\textbf{F}=q(\textbf{E} +\textbf{u}\times \textbf{B})$$

3. The attempt at a solution
The B field is curling around the inner cylinder.
Using Ampere's Law,
$$B=\frac{\mu_{0}I}{2\pi r}$$
for a<r<b
r = distance from centres of cylinders

E field (electrostatics) of concentric cylinders
$$E=\frac{Q}{2\pi\epsilon_{0}r}$$
where Q is charge per unit length, assuming length>>r

Following from above
Capacitance
$$C=\frac{2\pi\epsilon_{0}}{log_{e}(\frac{b}{a})}$$ per unit length
so using Q=CV, get
$$E=\frac{V}{r log_{e}(\frac{b}{a})}$$

now force F=q(E+u^B)
so
$$|\textbf{u}|=\frac{|\textbf{E}|}{|\textbf{B}|}$$
and so
$$|\textbf{u}|=\frac{2\pi V}{\mu_{0}I log_{e}(\frac{b}{a})}$$

would this look right?

Last edited: Jun 6, 2009
2. Jun 6, 2009

### LowlyPion

I would think so.

The result looks OK.