Magnetic Force of an electron

  • Thread starter xxkylexx
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  • #1
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An electron is accelerated through 2550 V from rest and then enters a region where there is a uniform 1.70 T magnetic field. What are the maximum and minimum magnitudes of the magnetic force acting on this electron?



F = qvB
F = (mv^2)/R
F = qE




I know q = 1.6*10^-19, B = 1.7, and V = 2550. In order to use F = qvB, all I need is the velocity. I guess I'm just not seeing how the 2550 V is relating into this whole scheme of things. Also, I'm not sure why there is is going to be a F.max and a F.min, and not just one F?

Appreciate any help.

Thanks much,
Kyle
 

Answers and Replies

  • #2
7
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Ok, I figured out F.max by just saying:

.5mv^2 = qV

F.max = qvB


However, I'm not sure what F.min is and the difference between F, F.max, and F.min?
 
  • #3
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Bump. Any idea on the F.minimum?

Thanks,
kyle
 
  • #4
1
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Minimum

The minimum is 0, when the particle is moving in the direction of or opposite the direction of the magnetic field
 

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