Direction of electron in magnetic field? (1 Viewer)

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

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.


Homework Helper
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!

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving