What Magnetic Field Strength Keeps an Electron on a Straight Path?

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SUMMARY

The discussion focuses on calculating the magnetic field strength required to keep an electron on a straight path between two parallel plates with a potential difference of 110 V. The electron, accelerated from rest through a potential difference of 1.7 kV, achieves a velocity of 24,449 m/s. The key equations used include F=qvB for magnetic force and F=qE for electric force, leading to the relationship vB=E, where E is derived from E=V/d. The participant struggles with the final calculations, indicating a potential error in their math.

PREREQUISITES
  • Understanding of electric fields and potential difference
  • Familiarity with the Lorentz force equation (F=qvB)
  • Knowledge of kinetic energy equations (qU=1/2 mv²)
  • Basic algebra for solving equations
NEXT STEPS
  • Learn how to derive velocity from potential difference in charged particles
  • Study the relationship between electric and magnetic fields in charged particle motion
  • Explore the concept of uniform magnetic fields and their effects on charged particles
  • Practice solving problems involving forces on charged particles in electric and magnetic fields
USEFUL FOR

Students in physics, particularly those studying electromagnetism, educators teaching electric and magnetic fields, and anyone interested in the dynamics of charged particles in fields.

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Homework Statement



In the figure, an electron accelerated from rest through potential difference 1.7 kV enters the gap between two parallel plates having separation 20.0 mm and potential difference 110 V. The lower plate is at the lower potential. Neglect fringing and assume that the electron's velocity vector is perpendicular to the electric field vector between the plates. What uniform magnetic field allows the electron to travel in a straight line in the gap?


Homework Equations


F=qvB
q=charge
v=velocity
B=magnetic field




The Attempt at a Solution



I honestly don't know where to even begin, I know I've got to be missing a formula because distance has got to be used somewhere in there, I'm just interested in getting an idea how to start this problem.
 
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If the electron travels in a straight line it doesn't accelerate upwards/downwards, which essentially means that the sum of all forces acting in the direction of the y-axis is zero. The magnetic force has to be equal, but opposite to that of the electric force.
 
Ok, so that means that F=-F, and i found another formula qvB=(mv^2)/r, but I'm sure that I need to find the velocity first and then use that and substitute and since i have two equations i would be able to solve for F which would allow me to find B, but I'm not sure how to find the velocity by only knowing the potential differences.
 
Forget qvB=(mv^2)/r. This equation describes an electron (or any other particle with a charge q) moving in a circular path, this is not true in your assignment.

If a particle with the charge q is accelerated with the voltage U it will gain kinetic energy according to

qU=\frac{1}{2}mv^2

From this equation you can solve the speed of the electron.
 
Ok so, I found the velocity to be 24,449 m/s and then by combining the equations qvB=F and F=qE I got the equation vB=E so 24449 m/s x B = E and E=V(volts) / d

so I plugged in numbers and 1.27e-10 but wrong... Is my math wrong?
 

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