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Magnetic Fields. Finding the angle of a moving charge.

  1. Mar 14, 2009 #1
    1. The problem statement, all variables and given/known data

    An electron is moving through a magnetic field whose magnitude is 8.70 X 10-4 T. The electron experiances only a magnetic force and has an acceleration of magnitude 3.50 X 1014 m/s2. At a certain instant, it has a speed of 6.80 X 106 m/s. Determine the angle [tex]\theta[/tex] (less than 90o) between the electron's velocity and the magnetic field.

    2. Relevant equations

    F=ma

    B = F / qo v (sin[tex]\theta[/tex])


    3. The attempt at a solution

    We know the mass of the electron= 1.673 X 10-27
    So we plug that in the equation F=ma=(1.673 X 10-27)(3.5 X 1014 m/s2=5.86 X 10-13

    Then we use the magnetic field equation and solve for sin[tex]\theta[/tex]

    sin[tex]\theta[/tex] = F / q v B = (5.86 x 10-13)/((8.7 X 10-4)(6.8 X 106)(1.6 X 10-19)) = 618.608

    That is wrong because the books answer is 19.7o. I know I am supposed to take inverse of the answer to find the angle. But I did my calculation correct and I am still getting the same answer...is there anything that I am doing wrong?


    Thanks
     
    Last edited: Mar 14, 2009
  2. jcsd
  3. Mar 14, 2009 #2

    gabbagabbahey

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    Really?!:wink:
     
  4. Mar 14, 2009 #3
    Sorry, The mass of an electron is 9.1 X 10-31

    After all the calculation, I get the right answer! Thanks for letting me know! I did a silly mistake!
     
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