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Electron in a uniform magnetic field

  1. Feb 7, 2009 #1
    5. Consider an electron moving in the +x direction with speed vx, in a uniform magnetic
    field B = B ˆz.
    (a) What type of trajectory does the electron take? What happens qualitatively to
    its velocity in the x direction as its velocity in the y direction increases? How
    does its kinetic energy change?
    (b) Now imagine that the electron is in a conducting material that obeys Ohm’s law,
    and is subject to an electric field that ensures that its x velocity is always exactly
    vx. What type of trajectory does it take? What happens to its velocity in the x
    direction as its velocity in the y direction increases? How does its kinetic energy
    change?
    (c) When the electron’s energy changes, what force does the work?

    Ok, really just wanna confirm if my answers are correct.

    a). Clockwise circular trajectory, as Vy increases Vx decreases to the total veloty remains constant, kinetic energy remains constant.

    b). Clockwise spiral trajectory, Vx remains constant, total velocity increases and therefore kinetic energy increases. Not too sure if the trajecoty is correct though. Since Vx is constant, then it V total may return to Vx then its directly on the x-axis, and then it woulb be like a harmonic oscillator in the y direction.

    c). The electric field.
     
  2. jcsd
  3. Feb 8, 2009 #2

    gabbagabbahey

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    a) and c) look correct to me :approve: but your trajectory for b) seems off :frown:

    Perhaps you can take me through your reasoning for why you chose that trajectory....what direction does the magnetic force point when the electron just starts its motion? How about a little later? Do the x and z-components of its velocity ever change? what happens to the y-component?:wink:
     
  4. Feb 8, 2009 #3
    Yeah, this was the part that was the trickiest to me.

    I'm not really sure what they mean about the x component always being Vx. I guess what I though at first was that Vx was constant, and that Vy changed dependnet on the abngle from the y-axis aroung the xy-plane.

    Now i'm inclined to think its more of a clockwise eliptical trajectory.
     
  5. Feb 8, 2009 #4

    gabbagabbahey

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    That seems reasonable to me!....but that isn't really a spiral path is it?:wink:
     
  6. Feb 8, 2009 #5
    No, I meant like a planets orbit style, not spiral.
     
  7. Feb 8, 2009 #6

    gabbagabbahey

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    It isn't an orbit either....v_x is constant, so it's always getting further away from where it started in the x-direction......

    If E is in the x direcition, B is in the z-direction and v_x is a constant, what does the Lorentz force law tell you about the acceleration in the y and z directions?

    Hint: [tex]\vec{F}=m\vec{a}=-e(\vec{E}+\vec{v}\times\vec{B})=-e\left( E(x)\hat{x}+\begin{vmatrix}\hat{x} & \hat{y} & \hat{z}\\ v_x & v_y & v_z \\ 0 & 0 & B \end{vmatrix}\right)[/tex]
     
  8. Feb 8, 2009 #7
    So if Vx is constant, and as you say it trvels further in the x direction everytime it passes the x-axis, and I think its Vy is changing such that it never gets any further away in the y direction, then its kinda like an ellipse, but that growing in the x direction.
     
  9. Feb 8, 2009 #8

    gabbagabbahey

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    Huh?!:confused:

    Take my advice and actually calculate the y-component of the acceleration.
     
  10. Feb 8, 2009 #9
    ok,

    -e(Ex + (VyB - VxB), but I'm not sure what that says, unless Vx and Vy are both constant. Sorry I don't know how to interpret this.
     
  11. Feb 8, 2009 #10

    gabbagabbahey

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    Huh?! How did you get this?:confused:

    [tex]\vec{F}=m\vec{a}=-e(\vec{E}+\vec{v}\times\vec{B})=-e\left( E(x)\hat{x}+\begin{vmatrix}\hat{x} & \hat{y} & \hat{z}\\ v_x & v_y & v_z \\ 0 & 0 & B \end{vmatrix}\right)=-e\left( (E(x)+v_y B)\hat{x}-v_x B\hat{y}+0\hat{z} \right)[/tex]

    [tex]\implies a_x=\frac{-e}{m}\left(E+v_y B \right), \quad a_y=\frac{e v_x B}{m}, \quad a_z=0[/tex]

    Now your told that [itex]v_x[/itex] is constant, so evidently [itex]E[/itex] is varied in such a way that [tex]a_x=\frac{-e}{m}\left(E+v_y B \right)=0[/tex] always.

    What can you say about [itex]a_y[/itex]....what does that mean for [itex]v_y[/itex]?
     
  12. Feb 8, 2009 #11
    So that means that Vx is constant, so the distance from the origin in the x direction is increasinfg at a constant rate, where as the acceleration in the y direction is constant, so Vy is increasing at a constant rate, and distance from the origin in the y direction is increasing faster than in the x direction. Everytime it pases the positive y-axis, its distance has grown exponentially?

    As for what I got with that equation, I just did not write in the x hat and y hat.
     
  13. Feb 8, 2009 #12

    gabbagabbahey

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    Correct, but that's exactly what the question told you to assume.

    Why would it pass the y-axis more than once? And why would it's distance increase exponentially?

    If it starts at the origin, doesn't it just get further and further away? It is moving at a constant speed in the x-direction and constantly accelerating in the y-direction. If you want to give a name to that sort of trajectory, I would call it a "quasi-parabolic trajectory away from its origin" --- x(t) is a straight line, while y(t) is a parbola; hence the word "quasi-parabolic" describing r(t)

    You should always write the unit vectors when talking about a vector. a=-e(Ex + (VyB - VxB) is meaningless, and a_y=-e(Ex + (VyB - VxB) is incorrect.
     
  14. Feb 8, 2009 #13
    So how would you know that the electron makes at least one revolution around the origin?
     
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