Electron Moving in a Uniform Magnetic Field

It's not mentioned in the problem, but I assume the electron is moving in the xy plane, parallel to the magnetic field. In that case, the electric field (if there is one) will be perpendicular to both the velocity and the magnetic field, so the cross product will be zero. In summary, the conversation discusses finding the magnitude of the magnetic force on an electron and proton moving through a uniform magnetic field with given velocity and magnetic field components. The solution involves combining the components and using the equation F = q(v x B). The correct angle between the two vectors is calculated to be 76.723 degrees and the magnitude of the magnetic force is found to be -6.27 x 10^-14 N. It is also
  • #1
iceman2048
6
0

Homework Statement


An electron that has a velocity with x component 2.3 x 106 m/s and y component 2.9 x 106 m/s moves through a uniform magnetic field with x component 0.040 T and y component -0.12 T. (a) Find the magnitude of the magnetic force on the electron. (b) Repeat your calculation for a proton having the same velocity.


Homework Equations


F = q vXB


The Attempt at a Solution



Since both magnetic field and velocit were given in component form I figured they could be combined into one (meaning one for velocity and one for magnetic field). After doing

a = sqrt[ (ax)^2 + (ay)^2] for both V and B I got V =3.7 x10 ^6 m/s, angle = 5.158 degrees; B = 1.2649 x 10^-1, angle = -71.565.

Then I added the two angles (I figured their sum is the total difference b/w the B and V vectors?) and plugged in:

F = q v X B
F= (1.6 x 10^-19)(3.7 x 10^6)(1.2649 x 10^-1)sin (-66.407)
F = -6.862 x 10^-14 N

For part b) it should be the same # as the charge of the proton and electron is the same (only sign is different and we use absolute values in that equation).

I hope that my approach is ok, though, assuming my approach is ok I think I might have messed up the combining of the two angles. Any help/guidance would be greatly appreciated. Thank you in advance.

P.S. I thought I should mention that no diagram is provided. Thanks again.
 
Physics news on Phys.org
  • #2
You got the angle wrong. Also, this is all much easier to do using the components. Both v and B are in the xy plane, so their cross product will have only a z-component. The z component of q(v x B) is just q(vxBy - vyBx).
 
  • #3
dx said:
You got the angle wrong. Also, this is all much easier to do using the components. Both v and B are in the xy plane, so their cross product will have only a z-component. The z component of q(v x B) is just q(vxBy - vyBx).

Hmm by using that, I get an answer of -6.27 x 10^-14, which is incorrect...Just out of curiosity, what would be the correct angle, had I not realized that components were easier? Thanks again, especially for that godly-quick response.
 
  • #4
They asked for the magnitude, so you should remove the minus sign.
 
  • #5
Assuming you calculated the angle of each of the vectors correctly, the angle between them would be 71.565 + 5.158 = 76.723. Just draw a picture and it should be obvious why.
 
  • #6
dx said:
Assuming you calculated the angle of each of the vectors correctly, the angle between them would be 71.565 + 5.158 = 76.723. Just draw a picture and it should be obvious why.

Duh! I can't believe I didn't see that! Also, the answer yielded with your method is correct. Thank you very much!
 
  • #7
No problem.
 
  • #8
what about the component qE of the lorentz force?
 
  • #9
E is zero here.
 

1. What is an electron moving in a uniform magnetic field?

An electron moving in a uniform magnetic field refers to the motion of an electron when it is subjected to a constant magnetic field. This motion is characterized by the electron moving in a circular path perpendicular to the direction of the magnetic field.

2. How is the motion of an electron affected by a uniform magnetic field?

The motion of an electron is affected by a uniform magnetic field because the magnetic field exerts a force on the electron. This force, known as the Lorentz force, causes the electron to move in a circular path.

3. What is the direction of the force on an electron moving in a uniform magnetic field?

The direction of the force on an electron moving in a uniform magnetic field is perpendicular to both the direction of the magnetic field and the direction of the electron's velocity. This means that the force always acts in a direction that is at a right angle to the electron's motion.

4. How does the strength of the magnetic field affect the motion of an electron?

The strength of the magnetic field directly affects the motion of an electron. A stronger magnetic field will result in a larger force on the electron and therefore a tighter circular path. A weaker magnetic field will result in a smaller force and a wider circular path.

5. What is the relationship between the speed of an electron and the radius of its circular path in a uniform magnetic field?

The speed of an electron and the radius of its circular path in a uniform magnetic field are inversely proportional. This means that as the speed of the electron increases, the radius of its circular path decreases and vice versa.

Similar threads

  • Introductory Physics Homework Help
Replies
13
Views
938
  • Introductory Physics Homework Help
Replies
31
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
336
  • Introductory Physics Homework Help
Replies
6
Views
744
  • Introductory Physics Homework Help
Replies
1
Views
350
  • Introductory Physics Homework Help
Replies
25
Views
279
  • Introductory Physics Homework Help
Replies
12
Views
203
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
809
  • Introductory Physics Homework Help
Replies
4
Views
355
Back
Top