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