Accelerated electron through a circular path

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SUMMARY

An electron accelerated through a potential difference of 5.0 kV achieves a velocity of 3 x 107 m/s. When this electron enters a transverse magnetic field, it moves in a circular path with a radius of 0.20 m, resulting in a magnetic field magnitude of 8.52 x 10-4 T. The time taken for the electron to complete one rotation can be calculated using the formula t = 2πr/v, where r is the radius and v is the velocity.

PREREQUISITES
  • Understanding of electron acceleration and potential difference
  • Knowledge of circular motion in magnetic fields
  • Familiarity with the formula for rotational kinetic energy
  • Basic proficiency in physics equations involving velocity and time
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  • Study the principles of electron acceleration in electric fields
  • Learn about the Lorentz force and its effect on charged particles in magnetic fields
  • Explore the derivation of the formula for time period in circular motion
  • Investigate the applications of magnetic fields in particle accelerators
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Physics students, educators, and anyone interested in the dynamics of charged particles in electric and magnetic fields.

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Homework Statement


An Electron is accelerated through a potential difference of 5.0KV
(i) Find it's velocity.
(ii) It is then passed through a region of transverse magnetic field where it moves in a circular path of radius 0.20m. Find the magnititude of the field.
(iii) Find the time taken to for the electron to complete one rotation.


Homework Equations


For (iii)
Rotational KE = ½ Iw2
max velocity = 3x10^(7)
Magnitude of the field = 8.52x10^(-4)
potential difference = 5.0kV
radius = 0.20m
t = ?


The Attempt at a Solution


(i) I got 3x10^(7)
(ii) I got 8.52x10^(-4)
(iii) I'm really not sure how to approach this question any hints in the right direction would be great.
 
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the magnetic field does not change the speed of the electron.
so its velocity would be what u calculated in (i).

then t = 2*pi*r/v.
as simple as that.
 

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