Proton Speed in Constant Magnetic Field

[jan2905]

Traveling at an initial speed of 1.5x10^6m/s, a proton enters a region of constant magnetic field of magnitude 1.5T. If the proton's initial velocity vector makes an angle of 30 deg with the magnetic field, what is the proton's speed 4s after entering the magnetic field?



Not sure what equations to use.



I said that the speed did not change. Sure the path did...

 

[Redbelly98]

You're correct, the speed does not change in a magnetic field.

 

[thomate1]

I would like to clarify that the speed of the proton does not change in a constant magnetic field. This is because a magnetic field only affects the direction of a charged particle's motion, not its speed. The speed of the proton remains constant at 1.5x10^6m/s throughout its motion in the magnetic field. However, the direction of its velocity changes due to the Lorentz force, which is given by the equation F = qvBsinθ, where F is the force, q is the charge of the particle, v is its velocity, B is the magnetic field, and θ is the angle between the velocity and the magnetic field.

Using this equation and the given values, we can calculate the magnitude of the force on the proton to be 0.5x10^-12N. This force causes the proton to move in a circular path with a radius of 0.5m, as determined by the equation F = mv^2/r, where m is the mass of the proton. This circular motion is known as cyclotron motion.

Now, to answer the question of the proton's speed 4 seconds after entering the magnetic field, we can use the equation for circular motion to determine the angular velocity of the proton, which is given by ω = v/r. Plugging in the values, we get ω = 3x10^6 rad/s. After 4 seconds, the proton would have traveled an arc length of 4ωr = 12πm. Using the equation s = rθ, where s is the arc length and θ is the angle, we can find the angle θ to be 6π radians.

Therefore, after 4 seconds, the proton's velocity vector would make an angle of 6π radians with the magnetic field, which is equivalent to 180 degrees. Its speed, however, would still remain constant at 1.5x10^6m/s. I hope this clarifies any confusion and demonstrates the application of relevant equations in this scenario.