Electrons accelerated between two plates

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SUMMARY

Electrons are accelerated between two plates to an energy of 7.1 keV, entering a magnetic field of 0.004 T directed along the y-axis. The relevant equations include the Lorentz force equation F=qVBsin(t), the radius of curvature r=(mv/qB), and kinetic energy KE=1/2mv². The calculated velocity of the electrons is 5 x 10⁷ m/s, leading to a force of 3.2 x 10⁻¹⁴ N acting on the electrons. The coordinates of point K, reached after a quarter-circle trajectory, need to be determined based on these calculations.

PREREQUISITES
  • Understanding of kinetic energy and its calculation (KE=1/2mv²)
  • Familiarity with the Lorentz force equation (F=qVBsin(t))
  • Knowledge of circular motion in magnetic fields (r=(mv/qB))
  • Basic principles of electromagnetism and electric fields
NEXT STEPS
  • Calculate the coordinates of point K using the derived velocity and force values.
  • Explore the implications of magnetic field strength on electron trajectories.
  • Investigate the effects of varying the energy levels of accelerated electrons.
  • Learn about the engineering ethics code in relation to collaborative homework practices.
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Physics students, electrical engineers, and anyone studying electromagnetism and particle dynamics in electric and magnetic fields.

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Homework Statement



Electrons are accelerated between points A and B to the energy of 7.1keV, see attachment. The coordinate system was selected to have the origin at the startup point A. The electrons enter an area between two parallel plates where the magnetic field is established by running a current through the coils. It has magnitude of 0.004T and pointed in the direction of the y axis. After making a quarter of a circle, they reach a certain point K in between the plates. Report the coordinates x,y and z (in centimetres) of the point K.

I'm not sure what to do beyond what I've done already...

Homework Equations


F=qVBsin(t)
F=q(vXB)
r=(mv/qB)
KE=1/2mv^2


The Attempt at a Solution



KE=1/2mv^2
v=sqrt((2(1.135x*10^-15)/9.1*10^-31)
v=5*10^7 m/s

F=Bqv
F=.004(1.602*10^-19)5*10^7
F=3.2*10^-14 N
 

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u must go to u of s hahah and your an idiot your going against the engineering ethics code
-Blake B
 
really dough brain? how is this different than you and your friends getting together to work on assignments?
 

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