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Electron speed due to magnetic field.

  1. Feb 9, 2010 #1
    1. The problem statement, all variables and given/known data

    electrons emitted by a certain device are observed to travel in a circular path with a radius of 2.0cm when plaves in a uniform magnetic field whose strength is 10MN/C. What is the speed of the elctrons emitted by this device?

    2. Relevant equations
    I know that 10MN/C is "b-bar" as it is called in the book. But i'm not sure what equation to use.

    3. The attempt at a solution
    I attempted to use the equation Fm=q(velocity/c x B).... but im not sure what to use for the magnetic force.
  2. jcsd
  3. Feb 9, 2010 #2
    im not really sure about it , but i think that since these electrons are moving in a circular path in a uniform magnetic field , then using newton's 2nd Law :

    qvB = m v^2/R
  4. Feb 9, 2010 #3
    Yes, the magnetic force provides the necessary centripetal force for the circular motion of the electrons.

    Fm = q (v X B) = m*v2 / r
    Don't really know much about b-bar. The units should be either Tesla or Gauss.
  5. Feb 9, 2010 #4
    I believe that if the given value 10MN/C ( 10 mega newton per coulomb) , then this given value is the electric field (E) ..
  6. Feb 9, 2010 #5
    the B-bar is measured in tesla. and if the electric field vector is the one that is given, how do I find the magnetic field in order to use the equation Fm=mv^2/B if i am trying to solve for v??
  7. Feb 9, 2010 #6
    In your question, it said that "uniform magnetic field whose strength is 10MN/C" but for me that doesnt make sense? how it is a magnetic field with this unit, I know that it should be tesla, weber per meter square or volt.second per meter square but not newton per coulomb ? can you please check the question once more ?
  8. Feb 9, 2010 #7
    The magnetic field for B-Bar is definately 10MN/C
    the book says that vector B is in units of tesla which is: (N/C)(m/s^-1)
  9. Feb 9, 2010 #8
    I dont really get you!! what B-Bar supposed to mean is it the same as B?? and what is M(in 10 MN/c) ? , im sorry if I cant be helpful but if you could clarify these questions then I may can help ..
  10. Feb 9, 2010 #9
  11. Feb 9, 2010 #10
    I have checked the link you provided, but I cant see where it mentions that both E and B can be expressed in N/C unit? and I dont think that is possible for two different quantities to have the same units even if different system ..

    If it is really mentioned somewhere then I apologize , and please quote from where it is mentioned .. & thanks
  12. Feb 9, 2010 #11
    That is the whole point of the CGS system. :) [tex]\vec B[/tex] is defined differently there, so that [tex]\vec B[/tex] and [tex]\vec E[/tex] have the same units. The Lorentz force in CGS units would be:

    [tex]\vec F=q(\vec E+\frac{\vec v}{c} \times \vec B)[/tex]

    See the link for the exact transformations in going from one unit of measurement to the other.
  13. Feb 9, 2010 #12
    hmmm I have never encounter this before .. thanks for bringing this up :) .. and I hope that sarahaha288 would be able now to solve the problem , it is just converting back B to tesla and then substituting in qvB = m v^2/R to get v ..
  14. Feb 10, 2010 #13
    thanks so much! that helped alot.
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