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Magnetic and electric field between current carrying coaxial cables

by disillusion
Tags: cables, carrying, coaxial, current, electric, field, magnetic
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disillusion
#1
Jun6-09, 03:52 PM
P: 7
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
[tex]\oint_{C} \textbf{B}\cdot \textbf{dl}=\mu_{0}\int di[/tex]
[tex]\oint_{S} \textbf{E}\cdot \textbf{dA} = \frac{Q}{\epsilon_{0}}[/tex]
[tex]\textbf{F}=q(\textbf{E} +\textbf{u}\times \textbf{B})[/tex]

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

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

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

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

would this look right?
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LowlyPion
#2
Jun6-09, 06:16 PM
HW Helper
P: 5,341
Should I just treat the potential difference as a separate E field along the radial direction?
I would think so.

The result looks OK.


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