What is the Maximum Magnetic Force on an Electron in a Television Set?

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SUMMARY

The maximum magnetic force on an electron in a television set, accelerated through a potential difference of 23 kV and passing through a 0.29 T magnetic field, can be calculated using the formula F = BQVsin(θ). With θ set to 90 degrees for maximum force, the charge of the electron (q = 1.6 x 10-19 C) and its velocity (v = 592673.9 m/s) derived from the kinetic energy equation, the maximum force is determined to be approximately 2.75 x 10-14 N. The calculation confirms that the method used is correct, although the potential difference must be explicitly included in the analysis.

PREREQUISITES
  • Understanding of electromagnetic force equations
  • Knowledge of electron charge (1.6 x 10-19 C)
  • Familiarity with kinetic energy and potential difference concepts
  • Basic principles of magnetic fields (0.29 T)
NEXT STEPS
  • Study the relationship between potential difference and kinetic energy in charged particles
  • Learn about the Lorentz force and its applications in electromagnetic fields
  • Explore advanced calculations involving magnetic fields and charged particle trajectories
  • Investigate the effects of varying magnetic field strengths on electron motion
USEFUL FOR

Students in physics, electrical engineers, and anyone interested in the principles of electromagnetism and particle dynamics in electronic devices.

kdrobey
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Homework Statement


In a television set, electrons are accelerated from rest through a potential difference of 23 kV. The electrons then pass through a 0.29 T magnetic field that deflects them to the appropriate spot on the screen. Find the magnitude of the maximum magnetic force that an electron can experience.


Homework Equations


B=F/QVSin(theta)


The Attempt at a Solution


Since I'm solving for F, i got it by itself:F=BQVSin(theta). I had theta equal to 90, because a charge experiences Fmax when perpendicular to the field. I had q=1.6x10^-19, which is the charge of an electron. I know that the potential difference is 23 kV, but i cannot figure out how to solve for V
 
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to solve for v, i tried v=(Square root)2q/m. so 2(1.6e-19)/(9.11e-31)=592673.9=v. then F=0.29*(1.6e-19)(592673.9)=2.75e-14
 
You appear to have missed a variable, that of the potential difference over which the electrons were accelerated. Apart from that your method appears quite all right.
 

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