Direction of electron in magnetic field? (1 Viewer)

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

An electron moves with speed 4.00 105 m/s in a uniform magnetic field of 3.2 T, pointing south. At one instant, the electron experiences an upward magnetic force of 1.00 10-14 N. In what possible directions might the electron be moving at that instant? Give your answers as angles clockwise from south (from 0° to 360°), in increasing degrees.

(I have to find two answers in degrees, both clockwise from south).

2. Relevant equations

F = qvBsin(theta)

3. The attempt at a solution

F = qvBsin(theta)

sin(theta) = F / qvB

= (1x10-14) / [(1.6x10-19)(4x105)(3.2)]

= .0488

theta = sin-1(.0488)

theta = 2.8

So my first angle was approximately 2.8. I subtracted that from 270 degrees ("south") to get my final angle of 267.2 degrees. Then I subtracted 2.8 from 180 to get my second angle, which was approximately 177.2. I subtracted THAT number from 270 degrees to get my final second angle of approximately 92.8 degrees. I thought that 267.2 and 92.8 were my answers, but they are coming up as incorrect. I also tried rounding to a whole number, which didn't work either.
 

Delphi51

Homework Helper
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10
I wouldn't "subtract from 270".
The 2.8 is already the angle from south.
According to the left hand rule, it has a component to the left, so it must be 2.8 degrees counterclockwise from south. That would 357.2 degrees clockwise from south.
 
It was 182.8 and 357.2. Thanks!
 

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