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Homework Help: Angles and Magnetic Fields

  1. Mar 6, 2010 #1
    1. The problem statement, all variables and given/known data

    A particle with a charge of −2.50×10-8 C is moving with an instantaneous velocity of magnitude 40.0 km/sec in the xy-plane at an angle of 50.0° counterclockwise from the +x axis.

    1) What is the magnitude of the force exerted on this particle by a magnetic field with magnitude 2.00 T in the -x direction?

    2) What is the direction of the force exerted on this particle by a magnetic field with magnitude 2.00 T in the +z direction?
    Please, enter your answer as a counterclockwise angle from the +y direction to the direction of the force.

    3) What is the magnitude of the force exerted on this particle by a magnetic field with magnitude 2.00 T in the +z direction?

    2. Relevant equations

    F=|q|VBsin(θ)


    3. The attempt at a solution

    I thought 1) would be: 0 N because the force is in the -z direction and I have no idea conceptually what I doing wrong here.
    And I thought part 3) was (-2.50x10-8)(4x105)(2)(sin(50)) but I was wrong and I'm not sure why.
    Part 2) I have no idea what to do or how to get there.

    I'm completely new here, but I've looked at past question and that has really helped a lot, so I was hoping more interaction would be even better.

    Thanks guys, and please be as detailed as you can, I'm not sure why I'm not getting this...
     
    Last edited: Mar 6, 2010
  2. jcsd
  3. Mar 6, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi inovermyhead! Welcome to PF! :smile:

    (have a theta: θ and try using the X2 tag just above the Reply box :wink:)

    The force from a magnetic field B is q(v x B).

    In particular, the magnitude in 1) isn't zero (it will never be zero unless θ is zero), and the angle in 3) isn't 50º.
     
  4. Mar 6, 2010 #3
    Re: Welcome to PF!

    Thank you for the quick link, I hope I fixed all the formatting errors.

    About the problem:

    What happens to the force if it isn't "flowing" in that direction?
    I get that it's not zero but if it isn't in that direction, what's going on with it?

    I know the force is F=|q|VBsin(θ), but what are you doing to the angle if the force is -z and they are asking about +z?
     
  5. Mar 7, 2010 #4

    tiny-tim

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    Hi inovermyhead! :smile:

    (just got up :zzz: …)
    Think of the particle as "cutting through" the magnetic field lines …

    if it's moving parallel to them, it doesn't cut any, so the force is zero …

    if it's moving perpendicular to them, the force is maximum …

    generally, the force is proportional to sine of the angle. :wink:
    θ = 180º, so sinθ = 0, and the force is zero, same as if they were both +z.

    (but that's not the case in any of your three questions)
     
  6. Mar 7, 2010 #5
    Errr... okay:

    I know that one is: (2.5x10-8)(4x104)(2)(sin(50))
    **I had messed up my km to m conversion, and feel remarkably stupid***
    and
    I know that three is: (2.5x10-8)(4x104)(2)(sin(90))

    but I'm still unsure what to do with part 2.
    I thought it would be 230° from the y axis and was wrong and now I'm at a complete loss. So am I just getting my angles wrong?
     
  7. Mar 7, 2010 #6

    tiny-tim

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    Let's see …

    the force will be in the direction of q(v x F),

    which is the direction of -v x z,

    so it's perpendicular to both v and z, either up-left or down-right.

    Which? :smile:
     
  8. Mar 7, 2010 #7
    Down right is what I assumed it was, which makes sense to me, at the 230 from the +y-axis, but it wasn't right.
     
  9. Mar 7, 2010 #8

    tiny-tim

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    No, because it's in the direction of -v x z, which is z x v;

    and z x x = y, so "z x " rotates x (and anything else in the plane) anti-clockwise …

    so it'll be up-left. :smile:
     
  10. Mar 7, 2010 #9
    *Bangs Head on Desk*

    Okay. So it's 50...

    Thank you for all of your help, I really appreciate it.
     
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