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Magnetic Force on moving ion?

  1. Feb 9, 2005 #1
    Ok, an ion experiences a magnetic force of 6.2X10^-16 N when moving in the positive x direction, but no force when moving in the positive y direction. What is the magnitude of magnetic force exerted on the ion when it moves in the x-y plane along the line x=y? Assume the ions speed is the same in all cases.

    I know that for this problem F=qvB sin theta. Because the ion travels on the line x=y, then the theta angle is 45. Other than that I have no clue. :confused:

    Thanks again for any help. :smile:
     
  2. jcsd
  3. Feb 9, 2005 #2

    StatusX

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    Homework Helper

    break it up into an x component and a y component velocity and find the force for each.
     
  4. Feb 10, 2005 #3
    It says the speed is the same in all cases.
     
  5. Feb 10, 2005 #4
    so what?
    break it up into an x component and a y component velocity and find the force for each.
    for sure v_x and v_y is not equal to v.... but can you find v_x and v_y in term of v using the P******* theorm? (really have no idea how to spell that word)
     
  6. Feb 10, 2005 #5
    StatusX means use vector notion instead of F=qvB sin theta.

    [tex]
    F= q \vec{v} \times \vec{B} =( v_y B_z - v_z B_y )\hat{x} + (v_z B_x - v_x B_z )\hat{y} + (v_x B_y - v_y B_x)\hat{z}
    [/tex]

    Some terms will be zero though if [tex]\vec{B}[/tex] is perendicular to [tex]\vec{v}[/tex]
     
  7. Feb 10, 2005 #6

    Doc Al

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    Staff: Mentor

    This should tell you two things:
    (1) The direction of the magnetic field (B)
    (2) The value of qvB (apply [itex]F = qvB sin \theta[/itex], where [itex]\theta[/itex] is the angle between v and B)

    Once you've figured out (1) and (2), all you need to find your answer is to again apply [itex]F = qvB sin \theta[/itex] with this new angle. What angle does the ion velocity make with the B field when it travels the line x=y?
     
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