1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electron revolving in a circular path

  1. Feb 6, 2014 #1

    utkarshakash

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    An electron is at P at t=0. It is circulating in anti-clockwise direction with a constant angular speed ω along the shown(see attachment) circular path. Magnetic field at Q (CQ=2R, where R is radius of circle) will be recorded as zero at times .............?


    3. The attempt at a solution
    The situation is equivalent to a circular ring in which current is flowing in clockwise direction. I can't really see why will the magnetic field be zero at some time. It should be constant as the current does not change.
     

    Attached Files:

  2. jcsd
  3. Feb 6, 2014 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It's not equivalent to a continuous ring of circulating charge. It's a single point charge moving in a circle. That's different.
     
  4. Feb 7, 2014 #3

    utkarshakash

    User Avatar
    Gold Member

    Ok. I understand that. But how can a moving charge exert magnetic force at any point?
     
  5. Feb 7, 2014 #4

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    A circulating charge of course also generates both an electric and a magnetic field, including radiation ("synchrotron radiation", "bremsstrahlung"). Thus other charged particles are subject to the Lorentz Force according to this electromagnetic field.
     
  6. Feb 7, 2014 #5

    utkarshakash

    User Avatar
    Gold Member

    Can you please give some equations to start with?
     
  7. Feb 7, 2014 #6

    berkeman

    User Avatar

    Staff: Mentor

  8. Feb 7, 2014 #7

    utkarshakash

    User Avatar
    Gold Member

    From the equation you gave above it is clear that B=0 when [itex]\vec{v} \times \vec{r} =0[/itex]

    Let C be origin and CQ be X-axis.
    Then
    velocity of electron at any time t is given as [itex]\omega R (sin \omega t \vec{i} + cos \omega t \vec{j} ) \\ \vec{r} = (2R -R cos \omega t ) \vec{i} - R sin \omega t \vec{j} [/itex]

    Setting cross product equal to 0 gives a relation [itex] 2 cos \omega t = cos 2 \omega t [/itex].

    But from the given 4 options none of them satisfies the above relation.
     
  9. Feb 8, 2014 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I think you have a sign wrong.
    What are the four options?
     
  10. Feb 8, 2014 #9

    utkarshakash

    User Avatar
    Gold Member

    Are you trying to say that [itex]\vec{v} = \omega R ( -sin \omega t \vec{i} + cos \omega t \vec{j} )[/itex]

    The four options are
    [itex] \pi / 3 \omega \\ 5 \pi / 3 \omega \\ 7 \pi / 3 \omega \\ 8 \pi / 3 \omega \\ [/itex]
     
  11. Feb 8, 2014 #10

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes. What does that give you?
     
  12. Jul 4, 2014 #11
    Why is it not equivalent to a continuous ring of circulating charge ? I am not able to differentiate between the two .What is wrong with treating a rotating point charge q as equivalent to current given by i = qω/2π ?

    Many Thanks
     
    Last edited: Jul 4, 2014
  13. Jul 5, 2014 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    A continuous ring of circulating charge wouldn't produce a time-varying field. It always looks the same. A lump of charge moving in a circle will.
     
  14. Jul 6, 2014 #13

    Sorry . I didn't understand much .

    Could you explain in a little simpler terms why a ring of charge is a current but not a point charge ?

    Thanks
     
  15. Jul 6, 2014 #14

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If you have a single charge moving around a circle when you look at the circle at different times then you will see the charge in a different place at each time. This means the fields will be different at different times. If it's a continuous ring of charge smeared around the circle then it looks the same at all times. So the fields don't depend on time. There's just a circulating current.
     
  16. Jul 7, 2014 #15
    Thanks .

    I have understood the difference between a ring of charge and point charge in terms of magnetic field .

    But how do I understand it in terms of the definition of current i.e - rate of charge flow through a cross section ? Don't you think if we go by this definition ,then a point charge is crossing a cross section after a fixed time t = 2π/ω ,hence a current ?

    Thanks for your patience :)
     
  17. Jul 7, 2014 #16

    Delta²

    User Avatar
    Gold Member

    I think (not sure though) that it is about the instanteneous rate of charge flow versus the average rate of charge flow. If q is the charge of the point particle, the average rate of charge flow for Δt=2π/ω will be the same for all the points in circle and equal to qω/2π , however the instanteneous rate of charge flow for a point of the circle at angle phi will be qω/2π only for discrete times t_n=(2nπ+phi)/ω, and will be zero for any other time t.
     
  18. Jul 7, 2014 #17

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, it's a current. It's not a continuous current. You can approximate it as a continuous current if you want the average values of the fields over a whole rotation and not instantaneous values.
     
  19. Jul 7, 2014 #18

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Vibhor,

    You have indicated elsewhere that you are a high school student. Here is a video that reviews how to find the magnetic field of a moving point charge.



    It is at the high school level (AP Physics). See if you can apply the concepts discussed there to the problem you are trying to work. The formula discussed here is the same as given in Berkeman's link. (Post #6)

    [If the charge is moving very fast, then the field becomes much more complicated. But, from the given choices of answers, I don't think you are expected to worry about that. Assume that the equation given in the video is applicable.]
     
    Last edited by a moderator: Sep 25, 2014
  20. Jul 7, 2014 #19

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Although it is not clear from the wording of the problem, the line CQ lies in the plane of the circle.
     
  21. Jul 7, 2014 #20
    Thank you TSny .
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted