Motion of a point charge in a magnetic field

In summary, to find the exit angle and distance for a proton entering a region with a uniform magnetic field, we can use the equations F=qvBsin(theta) and F=qvxB and set them equal to the force calculated. By rearranging the equations, we can solve for the exit angle and distance. The exit angle is approximately 0.069 degrees and the distance is approximately 0.000000000693 meters.
  • #1
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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 and distance d.

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.
 
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  • #2


Hi there,

To find the exit angle, we can use the formula F=qvBsin(theta) and set it equal to the force we calculated (1.109x10^-13N). We can then rearrange the equation to solve for theta.

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

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

theta = sin^-1 (1.109x10^-13N / [(1.6x10^-19C)(1.00x10^6m/s)(.800T)])

This gives us an exit angle of approximately 0.069 degrees.

To find the distance d, we can use the formula F=qvxB and set it equal to the force we calculated (1.109x10^-13N). We can then rearrange the equation to solve for d.

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

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

This gives us a distance of approximately 0.000000000693 meters.

I hope this helps! Let me know if you have any further questions.
 

1. What is the equation for the motion of a point charge in a magnetic field?

The equation for the motion of a point charge in a magnetic field is given by the Lorentz force law, which states that the force on a charged particle moving through a magnetic field is equal to the cross product of the velocity of the particle and the magnetic field strength.

2. How does the direction of the magnetic field affect the motion of a charged particle?

The direction of the magnetic field determines the direction of the force acting on the charged particle. If the magnetic field is parallel to the velocity of the particle, there will be no force and the particle will continue to move in a straight line. If the magnetic field is perpendicular to the velocity, the charged particle will experience a force that is perpendicular to both the velocity and the magnetic field, causing it to move in a circular path.

3. Can a charged particle move in a straight line in a magnetic field?

Yes, a charged particle can move in a straight line in a magnetic field if the direction of the magnetic field is parallel to the velocity of the particle. In this case, the force acting on the particle will be zero.

4. How does the strength of the magnetic field affect the motion of a charged particle?

The strength of the magnetic field affects the magnitude of the force acting on the charged particle. A stronger magnetic field will result in a greater force and thus a greater deflection of the particle's path.

5. What is the significance of the charge-to-mass ratio in the motion of a charged particle in a magnetic field?

The charge-to-mass ratio is important because it determines the magnitude of the force experienced by the charged particle. A higher charge-to-mass ratio will result in a greater force and thus a greater deflection of the particle's path in a magnetic field.

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