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Particle in an Electric and Magnetic Field

  • Thread starter kinof
  • Start date
  • #1
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Homework Statement


An electron has an initial velocity with an x component of zero, a y component of 12.4 km/s, a z component 13.5 km/s, and a constant acceleration of 2.68 x 1012 m/s2 in the positive x direction in a region in which uniform electric and magnetic fields are present. If the magnetic field has a magnitude of 463 µT and is in the positive x direction, find the (a)x, (b)y, and (c)z components of the electric field.

Homework Equations


Lorentz Force:
$$\vec{F}_B=q[\vec{E}+(\vec{v} \times \vec{B})] $$
$$\vec{F}=m\vec{a} $$

The Attempt at a Solution


I have managed to get all of them but (a), which seems like it should be the easiest.

I've found that the magnetic force along the x-axis is 2.4412*10^(-18) N. The cross product of the velocity and the magnetic field along the x-axis is zero. So we have
$$\vec{F}_B=q\vec{E}_x $$
so the answer should be roughly 15.238 V/M in the x-direction. What am I doing wrong?
 

Answers and Replies

  • #2
747
36
Consider E should be in the -x direction.
 
  • #3
rude man
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Write F = ma for x, y and z.
Solve for Ex, Ey and Ez.

(Can only be done if ay = az = 0 assumed thruout which I guess it is).

BTW I got the same number for Ex that you did.

EDIT: whoops, better take barryj's hint!
 
Last edited:
  • #4
747
36
If the electron is accelerating in the +x direction, then the E field is from + to - and has to point in the -x direction.
 
  • #5
rude man
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If the electron is accelerating in the +x direction, then the E field is from + to - and has to point in the -x direction.
Motion seconded and carried.
 

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