How Do Electric and Magnetic Fields Behave Around a Moving Electron?

  • Thread starter Thread starter gstrosx
  • Start date Start date
  • Tags Tags
    Field
Click For Summary
SUMMARY

The discussion focuses on the behavior of electric and magnetic fields around a moving electron with a speed of 3 x 10^6 m/s at an angle of 60 degrees. The electric field direction is determined using the right-hand rule, resulting in an "out of the screen" orientation. The magnetic field direction is initially misidentified but is clarified to be directed East. The magnitudes of the electric and magnetic fields are calculated using the formulas |E| = Q/(4πε₀r²) and |B| = (μ₀/4π)(qv × r)/r².

PREREQUISITES
  • Understanding of electric fields and Coulomb's law
  • Familiarity with magnetic fields and the Lorentz force
  • Knowledge of vector cross products in physics
  • Proficiency in applying the right-hand and left-hand rules for field direction
NEXT STEPS
  • Study the application of the right-hand rule in electromagnetism
  • Explore the concept of electric field strength calculations
  • Learn about the magnetic field produced by moving charges
  • Investigate the relationship between electric and magnetic fields in electromagnetic theory
USEFUL FOR

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

gstrosx
Messages
4
Reaction score
0

Homework Statement



An electron is moving with a speed of 3*10^6 m/s. What are the directions (NE, N, NW,..., In the screen, out of the screen, zero magnitude) of the electric and magnetic field. What is the magnitude of the electric and magnetic field?
Code:
     moves this direction 3e6
     /
    / angle = 60 degrees
electron-------------------------Observation location, d = 3e-10


Homework Equations



for field direction the right hand rule is applicable

E=N/C
B=T

The Attempt at a Solution


Efield is similar to a Efield from a electric circuit, applying the right hand rule in the opposite direction of the movement of the electron gives out of the screen.

Mfield is, at a speratic guess (using a left hand rule i think) is East

|E|=1/4pi epsilon naught * chargeOfelectron/d^2

|B|=googly eyes.
 
Physics news on Phys.org
You're confusing the direction of the E-field with that of the B-field. Remember that the E-field due to a point charge is
[tex]\textbf{E}=\frac{Q}{4\pi \epsilon_0 r^2}\ \hat{\textbf{r}}[/tex]​
and the B-field due to a point charge is
[tex]\textbf{B} = \frac{\mu_0}{4\pi} \frac{q\textbf{v} \times \hat{\textbf{r}}}{r^2}[/tex]​
 

Similar threads

Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 25 ·
Replies
25
Views
6K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
3
Views
2K