Charged particle at 16° experiences force F; find angle for 1.5F

[osustudent2010]

Hi, I am having trouble with my last physics homework question (below):

When a charged particle moves at an angle of 16° with respect to a magnetic field, it experiences a magnetic force of magnitude F. At what angle (less than 90°) with respect to this field will this particle, moving at the same speed, experience a magnetic force of magnitude 1.5F?

I'm not sure how to solve it. I've tried using F(mag)= qvB (paralell)

But this doesn't work! Any help would be appricated (ASAP)!

 

[whozum]

You know that F = qv (cross) B, or qvBsin(theta).

qvBsin(16) is F1

1.5F1 = F2
1.5F1 = 1.5qvBsin(16)

F2 = qvBsin(theta)


So basically you want to solve 1.5qvBsin(16) = qvBsin(theta)

The problem states that q,v, and B are constant.

 

[thepatientmental]

Hi there,

Thank you for reaching out for help with your physics homework question. It seems like you are on the right track with using the equation F(mag) = qvB (parallel). However, there are a few steps you need to take to solve this problem.

First, we need to find the relationship between the angle of the particle and the magnitude of the magnetic force. This can be done by using the trigonometric function sine. Since we know the angle is 16°, we can use the equation sin(16°) = F/1.5F to find the angle that will result in a magnetic force of 1.5F. This gives us an angle of approximately 24.5°.

Next, we can use this angle in the F(mag) = qvB (parallel) equation to solve for B. Since the speed and charge of the particle are constant, we can set up a ratio of the magnetic forces at the two angles:

F(mag)1.5F = qvB(sin(24.5°)) / F = qvB(sin(16°))

Solving for B, we get B = 1.5sin(16°)/sin(24.5°) = 0.906 T.

Therefore, at an angle of approximately 24.5° with respect to the magnetic field, the particle will experience a magnetic force of 1.5F. I hope this helps and if you have any further questions, please don't hesitate to ask. Good luck with your homework!