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Charged particle in Electric and Magnetic field

  1. May 8, 2016 #1
    1. The problem statement, all variables and given/known data

    A particle of specific charge q/m is projected from the origin of coordinates with initial velocity ##v_1\hat{i}+v_2\hat{j}## in space having uniform electric field and magnetic field as ##-E\hat{j}## and ##-B\hat{j}## respectively. The particle will definitely return to the origin once if

    A) ##\frac{v_1B}{\pi E}## is an integer
    B) ##\frac{v_2B}{\pi E}## is an integer
    C) ##\frac{\sqrt{{v_1}^2+{v_2}^2}B}{\pi E}## is an integer
    D) ##\frac{q}{m}\frac{\sqrt{{v_1}^2+{v_2}^2}B}{\pi E}## is an integer


    2. Relevant equations

    Lorentz force = ##q\vec{v}## x ##\vec{B} + q\vec{E}##

    3. The attempt at a solution

    The force on the charged particle will be ##\vec{F} = qBv_y\hat{i} - qE\hat{j}- qBv_x\hat{k}## .

    There will be a force on the particle in negative z- direction which means particle will never return to origin . I am not sure how to move forward .

    Please help me .

    Thanks
     
    Last edited: May 9, 2016
  2. jcsd
  3. May 9, 2016 #2

    BvU

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    If so, it acquires a velocity component in that direction and your Lorentz force changes !
     
  4. May 9, 2016 #3
    Ok . So what should be my next step ?
     
  5. May 9, 2016 #4

    BvU

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    What kind of trajectories can you think of in a field like that ? Which would let the partcle go through the origin more than once ?
     
  6. May 9, 2016 #5
    Sorry , I have no idea .
     
  7. May 10, 2016 #6

    ehild

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    Check the vector product. Y component of velocity does not give component of force with y component of magnetic field.
    Without the electric field, the particle would move in the ( x,z) plane, perpendicular to the magnetic field, along a circle.
    Recall that ##\vec F= m \frac {d \vec v }{dt}##. Write it in components.
     
    Last edited: May 10, 2016
  8. May 10, 2016 #7
    Hello ehild ,

    Thanks for replying.

    Sorry . That was a typo . ##\vec{F} = qBv_z\hat{i} - qE\hat{j}- qBv_x\hat{k}##

    Why ? There is an initial component of velocity along Y-axis .

    ##m a_x = qBv_z##
    ##m a_y = -qE##
    ##m a_z = -qBv_x##
     
  9. May 10, 2016 #8

    ehild

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    Ah, yes.


    Write the acceleration components as time derivatives of the velocity components, and solve the system of differential equation.
     
  10. May 10, 2016 #9
    You mean I should solve the following three equations .

    $$m \frac{dv_x}{dt} = qBv_z$$
    $$m \frac{dv_y}{dt} = -qE$$
    $$m \frac{dv_z}{dt} = -qBv_x$$

    They look bit difficult to solve ?
     
  11. May 10, 2016 #10

    ehild

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    No. You can solve the y component at once. The x, z components correspond to circular motion. (Why?) Take the derivative of the first equation and substitute ## \frac{dv_z}{dt} ## into the third one.
     
  12. May 10, 2016 #11
    Ok . So what should I do once I get vx,vy,vz ? How do I impose the condition that particle passes the origin ? Should I need to find general coordinates of the particle ?
     
  13. May 10, 2016 #12

    ehild

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    Integrate the velocity components to get the coordinates with the initial condition x=y=z = 0. Find the condition that they are the same at a later time.
    No need to use other coordinates.
     
  14. May 10, 2016 #13
    What is same at a later time ?
     
  15. May 10, 2016 #14

    ehild

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    x and y and z all are zero again.
     
  16. May 10, 2016 #15
    Ok . I tried finding vx . ##v_x = ±v_1sin(ωt+φ)## where ω = qB/m and φ = nπ/2 ( n is odd) . Is this alright ?
     
  17. May 10, 2016 #16

    ehild

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    You can write it as vx=v1 cos (wt). And what is vz?
     
  18. May 10, 2016 #17
    Please explain .
     
    Last edited: May 10, 2016
  19. May 10, 2016 #18

    ehild

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    You left your solution with an arbitrary odd n. Choose n =1, then vx=v1cos(wt) will fulfill the initial condition.
     
  20. May 10, 2016 #19
    Ok .

    So, ##x=\frac{v_1}{ω}sinωt## ??
     
  21. May 10, 2016 #20

    ehild

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    Yes. Go ahead.
     
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