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Electron [N] moving in a magnetic field [up]

  1. Apr 26, 2014 #1
    1. The problem statement, all variables and given/known data

    A magnetic field of 0.0200 T (up) is created in a region

    a. Find the initial magnetic force on an electron initially moved at 5.00 x 10^6 m/s [N] in the field

    b. What is the radius of the circular path? Make a sketch showing the path of the electron.

    2. Relevant equations

    Fm = q v b sinΘ

    Fc = (m v^2)/r

    3. The attempt at a solution

    a.

    F_M=q v β

    F_M=(1.6*10^(-19) C)(5.00*10^6 m/s)(0.00200 T)

    F_M=1.6*10^(-15) N

    my issue is the direction of the force. how do you apply the right hand rule to a charged particle moving in the same direction as the field? doesn't the particle need to travel perpendicular to the field to interact with it? because it's an electron i know i have to reverse it or use the "left hand rule" but i'm still utterly confused since the particle is moving in the same direction as the field.

    b.

    Fnet = Fm

    in circular motion Fnet = Fc

    Fc = Fm

    (m v^2)/r = q v β

    r = m v/q β

    r = (9.1 * 1^-31 kg)(5.00 x 10^6 m/s)/(1.6 x 10^-19 C)(0.00200 T)

    r = 0.142 m

    r = 14.2 cm
     
  2. jcsd
  3. Apr 26, 2014 #2

    rude man

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    If the B field is in the same direction as the velocity of the electron then there is no force acting on the electron.

    Don't use a 'left-hand' rule. Use the right-hand rule and reverse the direction of the thumb if the scalar is negative, as in this case where F = -q v x B, q = +1.6e-19C.

    I like to set up an xyz coordinate system and use vector notation. That IMO is often a good idea when several vectors are involved.

    Also decide if B = 0.02T or 0.002T, and use B rather than b or β.
     
  4. Apr 29, 2014 #3
    sorry it's definitely 0.0200 T.

    so if the B field is in the same direction as the velocity of the electron...why are they even asking the question and why is there an follow-up part b to the question? I really don't understand this question. like conceptually i get it but i don't think it's possible and it can't be a trick question since there is a part b to it.

    lastly sorry i've just watched a lot of youtube videos and khang academy videos and they used that "left hand rule.
     
  5. Apr 29, 2014 #4

    rude man

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    They did not say that the velocity and B fields are pointed in the same direction. They said the B field points up and v points north. That was your assumption I think.
    Well, as I said, it's not necessarily wrong but I would not want to start getting confused as to which hand to use in a given situation. Do what you're comfortable with.
     
  6. Apr 29, 2014 #5
    ok thank you for your advice. so if the B field points "up" do you think that if we had a x y z co ordinate system, up would refer to the positive z direction and north would be the positive y direction?
     
  7. Apr 29, 2014 #6

    rude man

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    EDIT: (sorry!)
    I would pick x north, y east and z up.

    You can pick any coordinate system so long as it's right-handed. An example of a left-handed system would be x north, y up and z west.

    In unit vector terms, make sure that i x j = k.
     
    Last edited: Apr 29, 2014
  8. Apr 29, 2014 #7
    ok thanks. so specifically in relation to that question, they are talking about a 3-dimensional plane correct? (up) being the 3rd dimention or axis, (north) being the 2nd, and (east) for example being the 1st?
     
  9. Apr 29, 2014 #8

    rude man

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    There is no such thing as a 3-dimensional plane. There is a 3-dimensional coordinate system.

    Any time you're dealing with a moving charge in a B field (F = q v x B) you are automatically dealing with a 3-d system unless v and B are in perfect alignment.
     
  10. Apr 29, 2014 #9
    ok makes sense now thank you!
     
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