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Charged particles moving through a magnetic field

  1. Apr 15, 2016 #1
    1. The problem statement, all variables and given/known data
    A collection of charged particles move through a magnetic field at an angle to the field lines.
    Calculate the velocity of the particle if it is an electron moving at 30 degrees to the magnetic field of strength 3.4mT, causing it to experience a force of 4.7x106-18N

    2. Relevant equations

    F=BqVSin(Theta)
    3. The attempt at a solution
    4.7x10^-18 / ((3.410^-3)(1.602x10^-19)(Sin(30))) = 8733.44m/s
     
  2. jcsd
  3. Apr 15, 2016 #2

    gneill

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    Staff: Mentor

    Your method looks fine, but your result is incorrect. Check that your calculator is set to use degrees rather than radians for angles. Or, just recall what the value of sin(30°) is and use it: it's a very common angle and it's sine and cosine really should be memorized.
     
  4. Apr 15, 2016 #3
    Okay great thank you
     
  5. Apr 15, 2016 #4

    rude man

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    Your answer is off by a factor of about 2. If you ignored the sin in the denominator you'd be close but still off by about 1%.
     
  6. Apr 21, 2016 #5
    but if i take the sin out the dont i take out the angle altogether? and then doesnt that equation without sin just assum the angle is 90?
    How do i go about correcting my answer so it is spot on?
    Thanks
     
  7. Apr 21, 2016 #6

    gneill

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    |sin(30 rad)/sin(30 deg)| ~= 2. Did you check your calculator deg/rad setting as I suggested?

    Type in sin(30) right now. What do you get?
     
  8. Apr 21, 2016 #7
    Yes i did, and in the end i git 17257.84m/s (by changing to deg)
     
  9. Apr 21, 2016 #8

    rude man

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    No, you need to leave the sin term in. I just observed that without it you'd be close, but that was just a coincidence and to give you a hint as to what the answer might be.

    Your problem, simply, is your math! And BTW you stated in your original post that " ..experience a force of 4.7x106-18N .. ". What does that mean?
    BTW setting your calculator to radians instead of degrees is not the problem either.
     
  10. Apr 21, 2016 #9
    The forced experienced is just what the question stated.

    re-entering everything in and i get 17279.4m/s does this seem more accurate?
     
  11. Apr 21, 2016 #10

    gneill

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    That result looks good to me. Be sure to round to the appropriate number of significant figures before submitting your result.
     
  12. Apr 21, 2016 #11

    rude man

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    Yes, surely does!
     
  13. Apr 21, 2016 #12

    rude man

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    @gneill, seems your suggestion to check for rads rater than sines was correct after all. :oops:
     
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