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

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The discussion focuses on determining the directions and magnitudes of the electric and magnetic fields around a moving electron at a speed of 3*10^6 m/s. The electric field direction is identified as out of the screen using the right-hand rule, while the magnetic field direction is suggested to be east, though there is confusion regarding its calculation. The electric field magnitude is calculated using the formula E = Q/(4πε₀r²), while the magnetic field requires the use of the formula B = (μ₀/4π) * (qv × r)/r². Participants emphasize the importance of correctly applying the right-hand and left-hand rules to avoid confusion between the electric and magnetic fields. Overall, the thread highlights the complexities of electromagnetic field behavior around moving charges.
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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.
 
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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
\textbf{E}=\frac{Q}{4\pi \epsilon_0 r^2}\ \hat{\textbf{r}}​
and the B-field due to a point charge is
\textbf{B} = \frac{\mu_0}{4\pi} \frac{q\textbf{v} \times \hat{\textbf{r}}}{r^2}​
 
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