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Magnetic and electric field between current carrying coaxial cables 
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#1
Jun609, 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? 


#2
Jun609, 06:16 PM

HW Helper
P: 5,341

The result looks OK. 


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