Magnetic Force on a Moving Charge

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SUMMARY

The discussion focuses on calculating the magnetic force acting on a charged dust particle moving in a magnetic field. The magnetic field strength is 0.5 T, directed east, while the dust particle has a charge of -8 x 10-18 C and moves downward at a velocity of 0.6 cm/s (0.006 m/s). The magnetic force is determined using the equation F = qv × b, where the velocity and magnetic field are perpendicular, resulting in a force magnitude that requires careful vector analysis to ascertain direction.

PREREQUISITES
  • Understanding of magnetic fields and forces
  • Familiarity with vector cross product calculations
  • Knowledge of SI unit conversions (cm/s to m/s)
  • Basic principles of electromagnetism
NEXT STEPS
  • Study the Lorentz force law in detail
  • Learn about vector cross products and their applications in physics
  • Explore the implications of charge polarity on force direction
  • Review examples of magnetic force calculations involving different charge types
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Students in physics, particularly those studying electromagnetism, educators teaching magnetic forces, and anyone interested in the behavior of charged particles in magnetic fields.

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Homework Statement


A magnet produces a 0.5 T field between its poles, directed to the east. A dust particle with charge q = -8 times 10-18 C is moving straight down at 0.6 cm/s in this field. What is the magnitude and direction of the magnetic force on the dust particle?


Homework Equations



F= qv X b

The Attempt at a Solution



Tried converting .6 cm/s into m/s and solving since the sin (90) would still be 1, but it keeps telling me I am wrong.
 
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Hi, you need to be more explicit as to what you have tried which you said did not correspond to the correct answer. What is your answer? It's a vector isn't it?
 

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