How Is Magnetic Force Calculated on an Electron in a Uniform Field?

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SUMMARY

The magnetic force on an electron in a uniform magnetic field is calculated using the formula F = qvBsin(theta). In this discussion, an electron is accelerated by a voltage of 48200 V and enters a magnetic field of 0.183 T. The velocity of the electron is derived from equating potential energy (U = Vq) to kinetic energy (1/2 mv^2), leading to the calculation of force as F = (1.6e-19 C)(0.183 T)(SQRT(2(1.6e-19 C)(48200 V)/9.11e-31 kg))sin(90), resulting in a force of approximately 3.82e-12 N. The calculations are confirmed to be correct under the assumption that the electron starts from rest.

PREREQUISITES
  • Understanding of electromagnetic theory, specifically the Lorentz force law.
  • Knowledge of basic physics concepts such as potential energy and kinetic energy.
  • Familiarity with the properties of electrons, including charge and mass.
  • Proficiency in algebra and square root calculations.
NEXT STEPS
  • Study the Lorentz force law in detail to understand the effects of electric and magnetic fields on charged particles.
  • Learn about the principles of energy conservation in electric fields and their applications in particle acceleration.
  • Explore the concept of uniform magnetic fields and their applications in devices like cyclotrons.
  • Investigate the behavior of charged particles in magnetic fields, including circular motion and radius of curvature.
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This discussion is beneficial for physics students, educators, and professionals involved in electromagnetism, particularly those studying the behavior of charged particles in magnetic fields.

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Homework Statement


An electron in a vacuum is first accelerated by a voltage of 48200 V and then enters a region in which there is a uniform magnetic field of 0.183 T at right angles to the direction of the electron's motion. What is the force on the electron due to the magnetic field?


Homework Equations



F = qvBsintheta

The Attempt at a Solution



U = Vq (potential energy)
1/2 mv^2 (kinetic energy)
I set potential energy equal to kinetic energy to solve for velocity.

Vq = 1/2 mv^2 (then solve for v)
v = SQRT(2qv/m)

F = qB(SQRT(2qv/m))sintheta
Plugging in the numbers, I get
F = (1.6e-19) (0.183T) (SQRT(2(1.6e-19)*48200V/9.11e-31))* sin (90)
F = 3.82 e -12

Is this the right way to do this problem? Thanks in advance for looking it over for me.
 
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You seem to be correct as long as the electron starts off at a velocity of 0, which I'm assuming it does since the problem does not state otherwise. I also get the same answer that you do. Nice job!
 

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