Calculating the magnitude of the force on an electron

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SUMMARY

The discussion focuses on calculating the magnitude of the force acting on an electron moving through a uniform magnetic field of 1.2 x 104 T at a speed of 2.0 x 104 m/s. The force is calculated using the equation F = QVB sin(θ), where Q is the charge of the electron. For the scenarios presented, the force is 0.0012 N when parallel, 16.47 N at 45 degrees, and 0 N again when parallel. The right-hand rule confirms that no force acts when the velocity vector is aligned with the magnetic field vector.

PREREQUISITES
  • Understanding of electromagnetic force principles
  • Familiarity with the equation F = QVB sin(θ)
  • Knowledge of the right-hand rule in physics
  • Basic concepts of magnetic fields and their effects on charged particles
NEXT STEPS
  • Study the Lorentz force law in greater detail
  • Explore the implications of magnetic fields on charged particles
  • Learn about the right-hand rule and its applications in electromagnetism
  • Investigate the behavior of electrons in varying magnetic field strengths
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of charged particles in magnetic fields.

jaa72
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Homework Statement


an electron travels at a speed of 2.0X10^4 m/s threw a uniform magnetic field whose magnitude is 1.2x10^4 T. what is the magitude of of the force on the electron if its velocity vector and magnetic vector are
a)parralel
b)45degrees
c)parralel


Homework Equations


Fmax=QVB
F=QVSintheta



The Attempt at a Solution


for a i got .0012
b i got 16.47N
c i got 0
i think i am using the wrong equations
 
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well using the right hand rule, which says that the current,force and magnetic field are at 90 degrees to each other, you'd see that if the velocity direction is parallel to the magnetic field you have no force.

Use th equation F=BQvsin\theta to get the force for part b
 

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