Electric & Magnetic fields, application of Lorenz's Law to -eV electron

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SUMMARY

The discussion centers on calculating the electric field (E) required to ensure that an 830-eV electron moving along the negative x-axis remains undeflected in a velocity selector with a magnetic field (B) of 0.0130 T. Using Lorenz's force equation, the user attempts to equate E to the negative of the cross product of velocity (v) and magnetic field (B). The calculated velocity of the electron was found to be 4.268837e16 m/s, which is significantly higher than the speed of light, indicating a miscalculation. The user is advised to re-evaluate their calculations, particularly the conversion of kinetic energy to velocity.

PREREQUISITES
  • Understanding of Lorenz's force equation
  • Familiarity with electric and magnetic fields
  • Knowledge of kinetic energy and its relation to velocity
  • Basic vector operations, including cross products
NEXT STEPS
  • Review the conversion of kinetic energy from electron volts to velocity using the formula KE = 0.5mv^2
  • Learn about the principles of velocity selectors in electromagnetic fields
  • Study the implications of relativistic speeds on particle motion
  • Explore the correct application of vector cross products in physics problems
USEFUL FOR

Physics students, educators, and anyone studying electromagnetism or particle dynamics, particularly in the context of velocity selectors and Lorenz's Law.

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Homework Statement



A velocity selector consists of electric and magnetic fields described by the expressions vector E = E k-hat and vector B = B j-hat, with B = 0.0130 T. Find the value of E such that a 830 -eV electron moving along the negative x-axis is undeflected?


Homework Equations





The Attempt at a Solution




So I know i want to find E such that an electron with velocity = -v i-hat is undeflected. I also know form Lorenz's force that:

F = q[E + (v x B)] = 0

So i really just need to make E equal to the cross product (v x B) but opposite sign. I did this using using the fact that 830 -eV is the kinetic energy, set this equal to 0.5mv^2 and solved for v to be 4.268837e16 m/s.

Next I computed the cross product to get a value of -5.549e14 k-hat. So E should be equal to this but with opposite sign, not working however for my online submission, any suggestions as to where I am going wrong?
 
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4.268837e16 m/s is ridiculous. The speed of light is about 3e8 m/s. I'd check that part of the calculation first.
 
right right, i should have noticed that
 

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