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Charges in Magnetic Fields from Surfing Physics book

  1. May 4, 2017 #1

    alv

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    • Member advised to use the homework template for posts in the homework sections of PF.
    I am a teacher and I can't work this one out. I am trying to answer (a). I have only included the further parts of the question (b, c and d) so you can see why some parts of the question are included.

    A single positively charged particle, mass 4.6 x 1027 kg, enters a 4.0 T magnetic field into the page at 2.5 x 106 m/s. The field covers an area of 0.1 m x 0.1 m. It enters at 90 deg, 0.05 m up from the bottom left corner.
    (a) Calculate the force on the particle while it is in the field.
    (b) Calculate the radius of the path the particle takes in the field.
    (c) Calculate the speed with which the particle exits the field.
    (d) Where will the particle exit the field?

    There is no indication as to what the positive charged particle is. It's too massive for a proton. So I don't know what the charge on it is. I can't see how would be meant to assume that it has the charge of a proton therefore I don't know what the charge is.

    I have tried to use F = mv2/r but I don't know the radius.
    I could use F = qvB sinθ but as I said, I don't know the charge
    I have calculated E from v = E/B so E = vB = 2.5 x 10^6 x 4 = 1 x 10^7 but then I don't know what to do with it.
    There is F = BIl sinθ but I don't know what the current is (or length for that matter).

    The answer is 1.8 x 10-12 but I don't know how to get it and need to so I can show my students.
    Your help would be greatly appreciated.
     
  2. jcsd
  3. May 4, 2017 #2

    andrevdh

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    Homework Helper

    if I assume one elementary charge the answer comes to 1.6 x 10-12 N?
     
  4. May 4, 2017 #3

    cnh1995

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    The information is insufficient. They should mention the charge on the particle.

    Also, the mass of the particle given makes it heavier than the earth. I believe the power of 10 should be -27 instead of +27.
     
  5. May 4, 2017 #4
    If you interpret it as a single positive (elementary) charge on the particle, the answer should be ##1.6\times 10^{-12}## N, as andrevdh wrote. If you don't know the charge nor the trajectory of the particle, it's not possible to calculate the force. And you need the force to calculate the trajectory. But, then, the answer provided is a bit off.

    Where did you get E=vB from?
     
  6. May 4, 2017 #5

    gneill

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    Staff: Mentor

    "A single positively charged particle" no doubt means a particle with a single positive elementary charge; There's no reason why they would have to specify that a particle is a single particle.

    The mass of the particle, assuming that the power of ten is -27 and not 27 as pointed out by @cnh1995, is 2.75 proton masses, which doesn't strike me as corresponding to any one known particle. So presumably it's a fabrication by the problem's author.

    It's possible that over time the details of the problem's given values have been changed to "refresh" the problem for reuse. We occasionally see examples like this where an answer-key value is incorrect either due to a typo or because it wasn't updated when the problem was revamped.

    I suggest that you make reasonable assumptions and from there do the calculations in detail to show your students. The radius of curvature for a charged particle moving in a magnetic field is a well known calculation and comes up often.
     
  7. May 4, 2017 #6

    alv

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    Thanks so much for these replies everyone. Much appreciated :smile:
     
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