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Velocity of a beam

  1. Oct 5, 2005 #1
    Hi,

    My Question:

    What is the velocity of a beam of electrons taht go undeflected and magnetic fields of magnitude 8.8 x 10^3 V/m and 3.5 x 10^.9 T, respectively? what is the radius of the electron orbit if the electric field is turned off?

    My Work:

    V= E/B
    V=(8.8 x 10^3 V/m)/(3.5 x 10^.9 T)
    V=2.51 x 10^-6 m/s

    and for the radius

    r=mv/qB
    r=(9.11 X 10^-31 kg)(2.51 x 10^-6 m/s)/(1.6 x 10^-19 coul)(3.5 x 10^-3 T)
    r= .004m

    Are these correct :confused:

    Thank You :smile:
     
  2. jcsd
  3. Oct 6, 2005 #2

    Andrew Mason

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    If B = 3.5 x 10^9 T (not 10^.9), and there is an electric field of 8.8e3 V/m, both perpendicular to the direction of v, and to each other, your answer is correct, although that seems like a very slow electron beam (it is very strong magnetic field). I would check the given values again. (You are using 3.5 x 10^-3 T in the next part, which works out to v = 2.51 x 10^6 m/sec). You should show your reasoning as well: e.g.
    For a straight line path, force = 0 so [itex]q\vec{v} \times \vec{B} = q\vec {E} => v =E/B[/itex]

    Where do you get 3.5 x 10^-3 T?. Again, show your reasoning and you won't be confused:
    Centripetal force is supplied by the Lorentz force:[itex]F_c = mv^2/r = qvB[/itex] so [itex]r = mv/qB[/itex].
    AM
     
  4. Oct 6, 2005 #3

    mukundpa

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    going right, (calculation part not seen.)
     
  5. Oct 8, 2005 #4
    Hi,

    I'm sorry for replying so late but I recalculated the answer for the question as asked by Andrew Mason

    and came up with 2.51 x 10^12 m/s, is this answer resonable.

    Also for the second portion of the question:

    I can still use this equation

    r=mv/qB

    and come up with the right answer?

    Thank You:smile:
     
  6. Oct 9, 2005 #5

    Andrew Mason

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    Unfortunately, this is faster than the speed of light i.e. it is not reasonable. You should provide us with the magnetic field value. I suspect it is 3.5 x 10^(-3) T. based on your answer in the second part. If that is the case, the speed should be 2.51 x10^6 m/sec which is a reasonable value.

    Of course. But you have to use the correct value for v. If you substitute v = E/B,

    [tex]r = mE/qB^2[/tex]

    AM
     
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