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A particle accelerating in both magnetic and electric fields

  1. Mar 26, 2010 #1
    1. The problem statement, all variables and given/known data

    A proton (q = 1.6×10−19 C, m = 1.67×10−27 kg) moves at 300 ms−1 along a line midway between the x and z axes. There is a magnetic field of 1T along the x axis and an electric field of 2000 Vm−1 along the y axis. What is the force on the proton and what is its acceleration?


    2. Relevant equations

    Scalar product: a·b=abcosθ=axbx +ayby +azbz.

    Vector product: a×b = i(aybzazby)+j(azbxaxbz)+k(axbyaybx), |a×b| = absinθ.

    Forces on a charged particle in electric and magnetic fields: F = qE + qv × B.

    3. The attempt at a solution

    First off, my main assumption is that there shouldn't be any relativity involved in this question since v=300m s-1 is not particularly fast. However when I arrived at the answer I have calculated below, that is quite a lot of acceleration! I didn't end up needing to use the scalar and vector product formulae that were provided on the question sheet, so I fear that I am missing something!

    I have drawn a diagram to show how I have interpreted the question, which is attached. The proton is traveling with initial velocity v=300m s-1 along a line where x=z and y=0 (which I assume is what my lecturer means when he says "along a line midway between the x and z axes).

    So the total force acting on the particle F=FE+FB where FE is the force due to the electric field of 2000 Vm−1 and FB is the force on the proton due to the magnetic field of 1 T.

    So F=qE+qvB

    But what is v? Because only the component of motion perpendicular to the magnetic field (the velocity along the z-axis) affects the force, v=300sin(Pi/4)=[tex]150\sqrt{2}[/tex] m s-1.

    Using the right hand slap rule for the magnetic field, I can see that the force exerted on the proton by the magnetic field is in the same direction as that exerted by the electric field on a positive particle, that is, straight along the y-axis. Therefore I should be able to just add these two forces.

    Therefore F = 1.6×10−19 C × 2000 Vm-1 + 1.6×10−19 C × [tex]150\sqrt{2}[/tex] m s-1 × 1 T = 3.54×10−16 N along the y-axis.

    Then using a=F/m = (3.54×10−16 N) / (1.67×10−27 kg) = 2.12×1011 m s-2 in the y-direction.

    Does all of this working seem to make sense to you? I can't shake the feeling that I have missed something - why would my lecturer give me the scalar and vector product formulae if I did not have to use them?
     

    Attached Files:

  2. jcsd
  3. Mar 27, 2010 #2

    ehild

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    Homework Helper
    Gold Member

    Your method is all right and you do use the scalar and vector products :).

    When you get the velocity components, you take the scalar product of the velocity vector v with the unit vectors. So vz= v˙k .

    The magnetic force is the vector product qvxB= q(vxi + vzk) x Bi=qvzj.

    ehild
     
  4. Mar 27, 2010 #3
    Ah I see it now! Those operations just seemed to me like "common sense", I guess I didn't analyse my method enough. Typical. Thank you for your clarification :)
     
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