Magnetic field curvature question

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SUMMARY

The minimum radius of curvature for an alpha particle (He) moving at a velocity of 2.0 x 10^-6 m/s in a magnetic field of 2.9 x 10^-5 T can be calculated using the formula for magnetic force, F = qvBsin(theta). The calculated magnetic force is 9.2916 x 10^-18 N. To find the radius of curvature, one must relate this force to the centripetal force acting on the particle, which is essential for determining the particle's circular trajectory.

PREREQUISITES
  • Understanding of magnetic force equations, specifically F = qvBsin(theta)
  • Knowledge of centripetal force concepts in circular motion
  • Familiarity with the properties of alpha particles
  • Basic algebra for manipulating equations and solving for unknowns
NEXT STEPS
  • Study the relationship between magnetic force and centripetal force in circular motion
  • Learn how to derive the radius of curvature formula for charged particles in magnetic fields
  • Explore the implications of different magnetic field strengths on particle trajectories
  • Investigate the behavior of other charged particles in magnetic fields, such as electrons and protons
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Students studying electromagnetism, physics educators, and anyone interested in the dynamics of charged particles in magnetic fields.

jalen
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Homework Statement



What is the minimum radius of curvature for an alpha particle, He, moving at 2.0x10^-6m/s in a magnetic field of 2.9x10^-5T?

Homework Equations



F=qvBsin(theta)...then F=BILsin(theta)

The Attempt at a Solution



F=qvBsin(theta)
=(1.602x10^-19)(2.0x10^6)(2.9x10^-5T)sin90
=9.2916x10^-18

F=BILsin(theta)...
 
Physics news on Phys.org
You have a value for the Magnetic Force on the particle, but no you need to somehow relate that value to the radius of the particles trajectory in order to find your answer.

HINT:

The magnetic force is making the particle move in a circle. So, what type of force does that make it?
 

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