Electron moving through a uniform magnetic field

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SUMMARY

An electron moving at a speed of 9.00 x 105 m/s in a uniform magnetic field of 2.0 T experiences an upward magnetic force of 2.00 x 10-14 N. Using the equation F = |q| * v * B * sin(θ), the angle of the electron's movement was calculated to be 3.98° counterclockwise from south, resulting in a second position angle of 183.98° instead of the initially assumed 177.02°. The correct interpretation hinges on the properties of the sine function, specifically that sin(180º + θ) results in a negative value, while sin(180º - θ) retains the positive value.

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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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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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