Period of an Electron Around a Magnetic Field

Click For Summary
SUMMARY

The discussion focuses on calculating the period of an electron's motion in a uniform magnetic field of 0.152 teslas. The initial attempt incorrectly used the speed of light (3.00E8 m/s) for the electron, leading to an erroneous period of 1.86E-6 seconds. The correct approach involves substituting the expression for velocity in terms of the period, resulting in the formula T = 2(pi) * M / (Q * B). This yields a correct period of 2.35E-10 seconds for the electron's circular motion.

PREREQUISITES
  • Understanding of electromagnetism principles, specifically Lorentz force.
  • Familiarity with circular motion equations and centripetal acceleration.
  • Knowledge of fundamental constants such as electron charge (1.6E-19 C) and mass (9.11E-31 kg).
  • Basic algebraic manipulation skills for solving equations.
NEXT STEPS
  • Study the derivation of the Lorentz force equation in electromagnetism.
  • Learn about the relationship between magnetic fields and charged particle motion.
  • Explore the concept of cyclotron frequency and its applications in physics.
  • Investigate the effects of varying magnetic field strengths on charged particles.
USEFUL FOR

Physics students, educators, and anyone interested in the dynamics of charged particles in magnetic fields will benefit from this discussion.

bleedblue1234
Messages
108
Reaction score
0

Homework Statement



An electron enters a uniform magnetic field of 0.152 teslas such that the electron follows a circular path. What is the period of the electron's motion? Electrons have a mass of about 9.11 x 10-31 kilograms each. Hint: Since the radius of the electron's path is not given, it must cancel out of the equations.


Homework Equations



Fm = Q * V * B

F = M * A

A = V^2/r

V= 2(pi)r/T


The Attempt at a Solution



Q * V * B = M * A
Q * V * B = M * V^2/r
Q * 2(pi)r/T * B = M* V^2/r
Q * 2(pi)/T * B = M * V^2
(1.6E-19) * 2(pi)/T = (9.11E-31) * (3.00E8)^2
Solve for T
T = 1.86E-6 s
But it is wrong...
 
Physics news on Phys.org
How can you use v=3.00E8? Electrons can't possibly travel at the speed of light, and even if they could, the question doesn't say that they do.

Try substituting 2(pi)r/T in the v on the right side as well. Better yet, cancel out one v in Q * V * B = M * V^2/r, then substitute 2pi*r/T into the remaining v.
 
Cool got it... here is what I did if somebody else would like to see...

Q * V * B = m * v^2/r

Q * B = M * V / R

Q* B = M * (2(pi)r/T)/r

Q * B = M * (2pi)/T

T = 2(pi) * M / Q * B

I got 2.35E-10 seconds... worked great... thanks
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
791
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
11
Views
2K
Replies
14
Views
2K
Replies
6
Views
1K