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An electron in a magnetic field

  • Thread starter dphysics
  • Start date
1. Homework Statement
The acceleration of an electron in a magnetic field of 87 mT at a certain point is 1.268×1017 m/s2. Calculate the angle between the velocity and magnetic field.


2. Homework Equations
F = q dot v cross B
F = qBsin(theta)
F/m = a

3. The Attempt at a Solution

I calculated the force on the electron using F/m = a rearranged to F = ma, then plugged F into F = qBsin(theta). I came up with sin-1(8243000) = theta, which is obviously incorrect.

Am I making a calculation error, or is there an error with my formulas?

In advance, any help is much appreciated and thank you.
 
34
0
There is a mild error in your formula, at least a bit. You know F = ma, you know the mass of the electron (it's a constant) and you know a since it's given. You also know F = qB sin([tex]\theta[/tex]) and you know B (given) and q, constant. So you get [tex]sin^{-1}(\frac{ma}{qB}) = \theta[/tex]

Make sense?
 
Hmm, that is what I have above, at least I think.

Because when you substitute F = ma into F = qBsin(theta), you end up with ma = qbSin(theta), which can be arranged to sin-1(ma/qb) = Theta.

The values I used:
m = 9.1E-31 kg
q = 1.609E-19 C
B = 87E-3 T
a = 1.268E17

I still end up calculating 8243000, and when I try to take the inverse sine of that I get a domain error.
 
Last edited:
34
0
I get a very similar number for ma/(qb). Is it possible you have a or B written down incorrectly?
 
I just checked again, and those are the correct values for a and B.
 
34
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OH! F = qvBsin([tex]\theta[\tex]). You missed the v which should account for the 107[\SUP] factor.
 
Ah alright, that does make sense then. How would I go about calculating the velocity though?

qvb = F
F = ma
qvb = ma
v = 8243000

qvB/ma = sin(theta)

qvB/ma = 1, giving you theta = 90 degrees.

This unfortunately is incorrect.. =\
 
34
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qvb = F? That's missing a sin(theta) isn't it, that would then explain how you get sin(theta) = 1.

I think there's an equation we're not looking at that we should be because otherwise we've two unknowns and only one question. Or perhaps velocity should be a given (which is how I see the question posed elsewhere online).
 
It mentions in an earlier part of the problem that the velocity is 9.40×106, however it doesn't specify if that remains constant throughout the problem.

Assuming for a moment that it does, lets try to plug that into the equation:

F = qvBsin(theta)
F = 1.609E-19 * 9.40×106 * 87E-3 (I'm assuming 87 mT = 87 E-3 T)

And, assuming that I can use F = ma (which I'm not sure about)

I can substitute ma = 9.1E-31 * 1.268E17, divide that by qvB

End up with sin-1(.8769)
= 61.27 deg

Which is the correct answer! Thanks a lot for your help, sorry I missed the velocity, probably would have helped from the beginning..
 
34
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No worries; for all the online things I've worked with like that all the givens are constant unless stated otherwise.

Welcome, and good work!
 

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