Motion of a point charge in a magnetic field

[apt8]

Homework Statement

A proton with a speed 1.00x10^6 m/s enters a region with a uniform magnetic field of 0.800 T, and points into the page. The proton enters the region at an angle of 60º Find the exit angle.

Homework Equations

F=qvBsin(theta), F=qvxB

The Attempt at a Solution

I know that the charge of a proton is 1.6x10^-19C so I solved for the magnetic force.
F=(1.6x10^-19C)(1.00x10^6m/s)(.800T)sin(60) =1.109x10^-13N
I don't understand how I can get the exit angle now.

 

[anathema]

Hello! Thank you for posting your question on the forum. It seems like you have made some good progress with your attempt at finding the exit angle. To find the exit angle, we can use the equation F=qvBsin(theta) and set it equal to the magnetic force you calculated. Then, we can solve for the exit angle, theta.

F=qvBsin(theta)
1.109x10^-13N = (1.6x10^-19C)(1.00x10^6m/s)(.800T)sin(theta)

Dividing both sides by (1.6x10^-19C)(1.00x10^6m/s)(.800T) gives us:

sin(theta) = 1.109x10^-13N / ((1.6x10^-19C)(1.00x10^6m/s)(.800T))

Using a calculator, we can find that sin(theta) = 0.866. To find the exit angle, we can take the inverse sine (or arcsine) of both sides:

theta = sin^-1(0.866) = 60.02º

Therefore, the exit angle is approximately 60.02º. I hope this helps clarify the steps needed to find the exit angle. Let me know if you have any further questions. Good luck with your studies!