Magnetic field solar wind question

[johnson.3131]

The solar wind is a thin, hot gas given off by the sun. Charged particles in this gas enter the magnetic field of the Earth and can experience a magnetic force. Suppose a charged particle traveling with a speed of 8.20 106 m/s encounters the Earth's magnetic field at an altitude where the field has a magnitude of 1.10 10-7 T. Assuming that the particle's velocity is perpendicular to the magnetic field, find the radius of the circular path on which the particle would move if it were each of the following.
(a) an electron


(b) a proton


Relevant Equations:

F=I*L*B

F=qvB

charge electron and proton= +/- 1.6E-19

ATTEMPT: i plugged in the charges into q then then velocity and the magnetic field and got the answer, now I'm confused on how to find the radius.
please help. thanks!

 

[LowlyPion]

The Force vector is going to be qV X B

The cross product gives you a force perpendicular to it's motion.

If the B field is pointing out of your paper and the Velocity is up, then I get V X B as being a force vector right directed to the motion.
(It's not important that you see this to solve the problem, just that it may be useful to visualize.)

Given that you have found the |F| in the B field, then the curvature should be given by centripetal acceleration shouldn't it?

|F| = m|V²|/R

 

[nlightner]

I would like to clarify that the equations you have listed are not relevant for finding the radius of the circular path of a charged particle in a magnetic field. The correct equation to use is the Lorentz force equation, which is F = qvB, where F is the force experienced by the charged particle, q is the charge of the particle, v is the velocity of the particle, and B is the magnetic field strength.

To find the radius of the circular path, we can use the centripetal force equation, which is F = mv^2/r, where m is the mass of the particle and r is the radius of the circular path. By equating the Lorentz force and the centripetal force, we can solve for the radius:

F = qvB = mv^2/r

r = mv/qB

Now, let's plug in the given values for the electron and the proton:

(a) For an electron:

r = (9.11 x 10^-31 kg)(8.20 x 10^6 m/s) / (1.6 x 10^-19 C)(1.10 x 10^-7 T)

r = 0.0477 meters

Therefore, the radius of the circular path for an electron would be approximately 0.0477 meters.

(b) For a proton:

r = (1.67 x 10^-27 kg)(8.20 x 10^6 m/s) / (1.6 x 10^-19 C)(1.10 x 10^-7 T)

r = 2.19 meters

Therefore, the radius of the circular path for a proton would be approximately 2.19 meters.

It is important to note that these calculations assume that the charged particle's velocity is perpendicular to the magnetic field. If the angle between the velocity and the magnetic field is not 90 degrees, the radius of the circular path would be different and can be calculated using the same equations. I hope this helps clarify the calculation process.