1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    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?

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


    User Avatar
    Homework Helper
    Gold Member

  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!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook