# An electron in a magnetic field

1. Mar 16, 2009

### dphysics

1. The problem statement, all variables and given/known data
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. Relevant 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.

2. Mar 16, 2009

### Queue

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($$\theta$$) and you know B (given) and q, constant. So you get $$sin^{-1}(\frac{ma}{qB}) = \theta$$

Make sense?

3. Mar 16, 2009

### dphysics

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: Mar 16, 2009
4. Mar 16, 2009

### Queue

I get a very similar number for ma/(qb). Is it possible you have a or B written down incorrectly?

5. Mar 16, 2009

### dphysics

I just checked again, and those are the correct values for a and B.

6. Mar 16, 2009

### Queue

OH! F = qvBsin([tex]\theta[\tex]). You missed the v which should account for the 107[\SUP] factor.

7. Mar 16, 2009

### dphysics

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.. =\

8. Mar 16, 2009

### Queue

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).

9. Mar 16, 2009

### dphysics

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..

10. Mar 16, 2009

### Queue

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!