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Magnetic and electric field between current carrying coaxial cables

  1. Jun 6, 2009 #1
    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?
     
    Last edited: Jun 6, 2009
  2. jcsd
  3. Jun 6, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    I would think so.

    The result looks OK.
     
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