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Homework Help: Magnetic force

  1. Oct 22, 2012 #1
    Once again I can't seem to do this simple problem, not sure why.

    hate mechanics :(

    1. The problem statement, all variables and given/known data

    Charged particle enters a region containing constant magnetic field and leaves it after 708 micro seconds, what is Fx, the x component of the force after being 236 micro seconds long in the constant magnetic field
    Also it travels a quarter circle(not sure if the right word, basically a half of a semicircle)

    we're given:

    B=1.2 T
    R=0.95 m
    m=5.7*10^-8 kg

    2. Relevant equations


    3. The attempt at a solution

    I calculated the constant velocity when the charge is in the magnetic field, the only problem i have is calculating Vx and Vy after 236╬╝sec, particularly i cant find the angle, one way i tried solving it is realizing that 236/708 is 1/3, so distance that particle has traveled is

    (2piR/4)/3 which i found to be 0.49742...then idk, i thought maybe the angle would be 60 deg or 30 deg since it has travelled 1/3 of the way so i tried finding Vx=V*cos(30) or Vx=V*cos(60) but it didn't work out, i think i know that on halfway the angle is 45(Vx=Vy), but that's it, basically the equation i got is

    Fx=m*(V*cos(theta))^2/R and I can't find theta :(
  2. jcsd
  3. Oct 22, 2012 #2

    Simon Bridge

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    When the particle enters the B field, it has circular motion.
    So you can use the kinematics of circular motion to find the angle that rotates through by the time it leaves the field.
  4. Oct 23, 2012 #3
    The force stays perpendicular to the motion so the speed stays constant. A quarter circle is 90 degrees so a third of it is 30 degrees.
  5. Oct 25, 2012 #4

    Simon Bridge

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    Though you are encouraged to think of angles in radians.

    The size of an angle is the distance around the circumference of a unit circle that is inside the angle.

    The total distance around the whole circle is ##2\pi##.
    A half circle is therefore ##\pi## and a quarter circle is ##\pi/2##.
    The units here are "radius units" or "radians" for short.

    Use these and a lot of physics equations will make more sense - like the arc-length inside angle ##\theta## at distance ##r## is ##s=r\theta##. If you used degrees you have to say that ##s=\pi r/180## .

    A lot of trig starts to make sense as well ... like the tangent of the angle is the distance along the tangent to the unit circle that is inside the angle, and the sine is the length of the cord that is inside the angle.
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