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Motion in a Magnetic Field

  1. Feb 28, 2013 #1
    1. The problem statement, all variables and given/known data
    A charged particle of mass m = 7.2X10-8 kg, moving with constant velocity in the y-direction enters a region containing a constant magnetic field B = 3.3T aligned with the positive z-axis as shown. The particle enters the region at (x,y) = (0.62 m, 0) and leaves the region at (x,y) = 0, 0.62 m a time t = 565 μs after it entered the region.

    2) What is Fx at time t1 = 188.3 μs after it entered the region containing the magnetic field.
    3) What is Fy at a time t1 = 188.3 μs after it entered the region containing the magnetic field.
    4) What is q, the charge of the particle? Be sure to include the correct sign.

    3. The attempt at a solution
    answer from preceding question: 1.72E+03 m/s
    i do understand that to find the components, i need to find out what the angle is at t=188.3 microseconds. but how do i get the distances?

    Attached Files:

  2. jcsd
  3. Feb 28, 2013 #2
    What shape will the particle travel in the uniform magnetic field?
    Last edited: Feb 28, 2013
  4. Feb 28, 2013 #3
    I can trace that the particle will move in a circular counterclockwise direction, but when I apply the right hand rule (finger pointing in velocity direction, curling toward magnetic field -- out of the screen) the force direction seems to point outward, not the expected inward direction
  5. Feb 28, 2013 #4
    Attached is an image showing you how to do the right hand rule. Let your thumb be the magnetic field, your index be the velocity vector, and the direction of your middle finger to be the force vector.

    What does that tell you about the sign of the charge moving in the field?

    Attached Files:

    Last edited: Feb 28, 2013
  6. Feb 28, 2013 #5
    that makes complete sense. i don't know how i got that mixed up. yes, so the force vector will be perpendicular to the velocity vector, pointing towards negative x (or inwards towards the origin)
  7. Feb 28, 2013 #6
    Yes, and you need to know the coordinates of the charge at t=188.3us, correct?

    Here is a hint, it takes the charge 565us to cover 1/4 of the circle. How much of this segment is covered in 188.3us?
  8. Feb 28, 2013 #7
    in fact it is this first step where i'm unclear. i can't apply a square coordinate system, since the movement is circular. but winging it, if i set (90 degrees/ x)=(565/188.3), i get x = 30 degrees. is that close?
  9. Feb 28, 2013 #8
    Here, I attached an image. Can you apply similar triangles and go from here? Remember, the resultant of the forces components points towards the center of the circle; hence, along the hypotenuse of the triangle.

    Attached Files:

  10. Feb 28, 2013 #9
    i gather both components will have negative direction then? sorry, our prof isn't very helpful and he didn't even give us a hint on thinking like this, with triangles. so i got Fnet= (m*v^2/r), and multiplied by cos(30) and sin(30) for the x and y components, respectively, but my answers are marked wrong.
  11. Feb 28, 2013 #10
    Yes, both vector components will be negative. What answers did you get? Are you positive that your value for velocity is correct?
  12. Feb 28, 2013 #11
    1.72E+03 m/s was the answer marked correct for a preceding question. using F= mvsquared/r, i got a force magnitude 1.33E-17. the x component was 1.15E-17, and y component=6.67E-18. i manually added the negative directions.
  13. Feb 28, 2013 #12
    I would check your math on those. I got something completely different. I can't really see where you went wrong if you don't show me your work.
  14. Feb 28, 2013 #13
    lols sorry, im using excel to do my math. got the numbers sorted out. but i had a general question: shouldn't the negatives be a part of the answer by default, or would i add the negative direction using intuition? i'm asking because the same sign issue carried over to the final question asking about the charge of the particle. of course while doing homework i can be guided to the correct notation, but during a test i'd probably not have that chance.
  15. Feb 28, 2013 #14
    You know the charge is traveling in a circle. In order to remain travelling in a circle, the resultant force must be acting on the particle, pointing towards the center of the circle. In your notes you should have rules for determining the directions for both positive and negative charges. Can you tell whether or not the charge for this given problem is positive or negative?
  16. Feb 28, 2013 #15
    it was found to be negative.
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