Magnetic and electric field between current carrying coaxial cables

[disillusion]

Homework Statement

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?

Homework 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})

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?

 

[LowlyPion]

Should I just treat the potential difference as a separate E field along the radial direction?

I would think so.

The result looks OK.