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Magnetic force, finding angle between velocity and magnetic field

  1. Jun 23, 2008 #1
    1. The problem statement, all variables and given/known data

    a 3 microcoulomb charge having a speed of 10m/s is fired into a region of constant magnetic field where B is given by:

    B = 0.3i - 0.4j

    a) the magnitude force on the charge is found to be 2*10^-6 newtons. what is the angle between v and B?

    b) if the direction of v is changed so that the charge moves along the +z direction at a speed of 10m/s, what would be the magnitude and direction of the magnetic force in this case?

    i am supposed to get 7.66 degrees for part a, and 15*10^-6 newtons at 36.9 degrees for part b

    2. Relevant equations

    radius r = mv/qB where m is mass, v is velocity, q is charge, B is magnetic field

    electromagnetic force, F = qE + qv X B where X indicates cross product, E is electric field

    3. The attempt at a solution

    part a:

    using F = qE + qv X B ----> find theta
    F = qE + qv X B
    2*10^-6 = (3*10^-6E) + ((3*10^-6)(10)(0.3i - 0.4j)sin(theta))
    sin(theta) = 2*10^-6 - 3*10^-6E/((3*10^-6)(10)(0.3i - 0.4j))
    theta = inverse sin((2*10^-6 - 3*10^-6E)/(9*10^-6i - 1.2*10^-5j)) holding electric field E constant

    is this the correct approach to find theta?

    i havent started part b yet, but i am assuming theta changes to (theta from part a + 90 degrees) and i just substitute that new theta into the electromagnetic force equation to solve for force.

    is my approach for part a, correct? how about part b? or am i using the wrong equation?

    i am supposed to get 7.66 degrees for part a, and 15*10^-6 newtons at 36.9 degrees for part b

    any help appreciated
    Last edited: Jun 23, 2008
  2. jcsd
  3. Jun 23, 2008 #2
    hint: the magitude of v is already given, you have to calculate the magnitude of B then find theta by:
    F = q(v X B) = q|v||B|sin(theta)

    hope this helps

  4. Jun 23, 2008 #3


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    Homework Helper

    While your general statement of the force on an electric charge is correct, there is no electric field implied in this problem. You can just set E = 0.
  5. Jun 23, 2008 #4
    oh you a right, that equation was right in front of me, must've forgotten.

    F = qvB sin(theta) --> find theta
    2*10^-6 = (3*10^-6)(10)(0.3i -0.4j) sin(theta)
    sin(theta) = (2*10^-6)/(9*10^-6i - 1.2*10^-5j)
    theta = inverse sine [(2*10^-6)/(9*10^-6i - 1.2*10^-5j)]
    theta = inverse sine[(2*10^-6)/(-3*10^-6)]
    theta = inverse sine(-0.667)
    theta = -41.83 degrees

    i must've done something wrong with the denominator in terms or i and j. i am supposed to get 7.66 degrees. what do i need to fix?

    as for part b, how will the fact the charge moves along the +z axis, affect the components of the given magnetic field B?
  6. Jun 24, 2008 #5


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    Homework Helper

    The magnetic field B = 0.3i - 0.4j. Its magnitude is [(0.3)^2 + (0.4)^2]^1/2 = 0.5 T
    Using this find the angle.
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