Magnetic field solar wind question

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SUMMARY

The discussion focuses on calculating the radius of the circular path of charged particles, specifically an electron and a proton, when they encounter the Earth's magnetic field. The magnetic field strength is given as 1.10 x 10-7 T, and the particles travel at a speed of 8.20 x 106 m/s. The relevant equations include F=qvB and the centripetal force equation |F| = m|V2|/R. Participants clarify that the force acting on the particles is perpendicular to their motion, influencing their circular trajectory.

PREREQUISITES
  • Understanding of electromagnetic force, specifically Lorentz force
  • Knowledge of centripetal acceleration and its relationship to circular motion
  • Familiarity with the properties of charged particles, including charge values of electrons and protons
  • Basic proficiency in vector mathematics, particularly cross products
NEXT STEPS
  • Calculate the radius of circular motion for charged particles in a magnetic field using the formula R = mv/qB
  • Explore the effects of varying magnetic field strengths on particle trajectories
  • Investigate the implications of particle charge and mass on their motion in magnetic fields
  • Learn about applications of magnetic fields in particle accelerators and astrophysics
USEFUL FOR

Students and professionals in physics, particularly those studying electromagnetism, astrophysics, or particle physics, will benefit from this discussion.

johnson.3131
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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!
 
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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
 

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