Electron moving through a uniform magnetic field

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An electron moving at 9.00 x 10^5 m/s in a 2.0 T magnetic field experiences an upward force of 2.00 x 10^-14 N, leading to two possible movement angles from south. The first angle calculated is 356.02°, derived from 3.98° counterclockwise from south. The second angle is correctly identified as 183.98°, which arises from the understanding that sin(180° + θ) results in a negative value, indicating a downward force, while sin(180° - θ) retains the positive value. This highlights the importance of the sine function in determining the direction of the force relative to the electron's motion. Understanding these angles is crucial for analyzing the behavior of charged particles in magnetic fields.
bioradical
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The Problem:
An electron moves with speed 9.00 105 m/s in a uniform magnetic field of 2.0 T, pointing south. At one instant, the electron experiences an upward magnetic force of 2.00 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.


The only equation needed should be F= lql * VBsin(theta)


I used the above equation and found the angle to be 3.98° counterclockwise of South. So my first answer was 360-3.98=356.02° (which was correct). My problem came with finding the angle at the second position. I initially thought that it would be 180-3.98, but, through guess work, I found that the correct answer is 180+3.98= 183.98. I'm not sure why this is right.

Can someone help me understand the concept behind the second possible position of the electron?

Thank you!
 
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Welcome to PF!

Hi bioradical! Welcome to PF! :smile:
bioradical said:
So my first answer was 360-3.98=356.02° (which was correct). My problem came with finding the angle at the second position. I initially thought that it would be 180-3.98, but, through guess work, I found that the correct answer is 180+3.98= 183.98. I'm not sure why this is right.

Because what matters is sinθ.

sin(180º + θ), which was your first try, is minus sinθ (so the force would have to be down) …

but sin(180º - θ) is sinθ :wink:
 
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