Magnetic Fields and electron

  • Thread starter Dart82
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  • #1
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Homework Statement


An electron is moving through a magnetic field whose magnitude is 9.10x10^-4 T. The electron experiences only a magnetic force and has an acceleration of magnitude 3.60x10^14 m/s^2. At a certain instant, it has a speed of 6.30x10^6 m/s. Determine the angle (less than 90°) between the electron's velocity and the magnetic field.



Homework Equations


Magnetic Force
F = qvBsin(theta)

F=ma



The Attempt at a Solution



F=ma --> mass of electron x acceleration = 3.27x10^-16N

F = qvBsin(theta)
3.27x10^-16N / [(charge of electron) x (6.30x10^6 m/s) x (9.10x10^-4 T)] = sin (theta)

solving like this would make sin theta = 3.56 x 10^11, which is not reasonable at all....What am i doing wrong?
 

Answers and Replies

  • #2
ranger
Gold Member
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I didnt punch in the numbers, but you need to solve for theta not sine theta (assuming the last step you did was the finial step). Example in a right triangle, arcsine(opp/hyp) = theta.
 
  • #3
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i understand what you are saying, however when i use the numbers i have calculated in the arcsine (opp/hyp) = theta equation, i am getting numbers that dont correspond to reasonable angles. This leads me to believe i am screwing up something in the F=ma or F=qvBsin(theta) department. in other words arcsine(F/qvB) should equal theta...right?
 
Last edited:
  • #4
Dick
Science Advisor
Homework Helper
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If it's any encouragement, I plugged your numbers in and got a perfectly reasonable value of sin(theta). Your theory is fine, your plugging is a problem.
 
  • #5
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ok, i'll check it out again
 
  • #6
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amazing...i finally got it! what in the world was i doing to get such wacky #'s. Thanks guys.
i just realized i was putting in the wrong exponent for the charge of the electron. DOH!
 
Last edited:

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