Help with a magnetic field problem

In summary, the conversation discusses the use of vector Lorentz force and its application in calculating the force on an electron in a specific problem. The participants also discuss the direction of the force and the use of the right-hand rule. The conversation also mentions the availability of math symbols and the use of equations for uniform circular motion.
  • #1
Shacking
7
0
Homework Statement
A straight wire has a current of 15 A vertically upwards, in a vacuum. An electron, presently 0.10 m from the wire moves at a speed of 5.0 x 106 m/s. Its instantaneous velocity is parallel to the wire but downward. Calculate the magnitude and direction of the force on the electron. Will this force remain constant?
Relevant Equations
F =(u I q v)/2pi(r)
Does this remain constant and what is direction
 

Attachments

  • 15752259957188600084281417179743.jpg
    15752259957188600084281417179743.jpg
    33.1 KB · Views: 105
Physics news on Phys.org
  • #2
Welcome to the PF. :smile:

I tried downloading your picture of your work so that I could try to post-process it to be able to read it, but no joy so far. It's too light to be legible for me, even after processing. Please try to attach a clearer file of your work. Better yet would be to type your work into the forum, which makes it much easier to read and comment on.

Shacking said:
F =uiqvsinthetha/2pir
Also, please make an effort to make your equations legible. Parenthesis are your friend. Also, there are math symbols available under the Sigma symbol at the top of the Edit window. You can use those to help you write basic equations.

And it's good to learn to use LaTeX to post math equations -- that makes your work even more legible. There is a LaTeX tutorial under INFO, Help at the top of the page. Thanks.
 
  • #3
Could you try to look at it now, I reposted the image
 
  • #4
Thank you, that does help.

Have you learned how to use the vector Lorentz force yet? That makes this caluculation much easier and more intuitive (especially the direction parts of it).
 
  • #5
berkeman said:
Thank you, that does help.

Have you learned how to use the vector Lorentz force yet? That makes this caluculation much easier and more intuitive (especially the direction parts of it).
No I have not learned how to use vector Lorentz force. We learned the right hand rules and cross product.
 
  • #6
Shacking said:
No I have not learned how to use vector Lorentz force. We learned the right hand rules and cross product.
Okay, that shows that you are almost there. The vector cross product and the right-hand rule are the basis for the vector Lorentz force.

https://en.wikipedia.org/wiki/Lorentz_force

1575226932539.png


Since the electric field E is zero in the problem you are working on, the force on the electron is just

[tex]F = q(V × B)[/tex]

You correctly got the B field via Ampere's Law:

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magcur.html

1575227163035.png


I think in your re-posted work that you were not sure about directions. Can you use the right-hand rule to say which direction the force is on the electron? And will that change anything about the problem over time that would change the force on the electron?
 
  • #7
berkeman said:
Also, there are math symbols available under the Sigma symbol at the top of the Edit window. You can use those to help you write basic equations.
Apologies, I just noticed that the Math symbols are under a square root symbol at the top of the Edit window ( √ ). I didn't get the memo... o0)
 
  • #8
berkeman said:
Okay, that shows that you are almost there. The vector cross product and the right-hand rule are the basis for the vector Lorentz force.

https://en.wikipedia.org/wiki/Lorentz_force

View attachment 253548

Since the electric field E is zero in the problem you are working on, the force on the electron is just

[tex]F = q(V × B)[/tex]

You correctly got the B field via Ampere's Law:

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magcur.html

View attachment 253549

I think in your re-posted work that you were not sure about directions. Can you use the right-hand rule to say which direction the force is on the electron? And will that change anything about the problem over time that would change the force on the electron?
Ya I am unsure what the direction on the force will be because the field is on a straight wire going around the wire while the electron is going down, and I am unsure of whether force is constant.
 
  • #9
Shacking said:
Ya I am unsure what the direction on the force will be because the field is on a straight wire going around the wire while the electron is going down, and I am unsure of whether force is constant.
That's fair, so let's talk about it.

The current is going up in the +Z direction, and the electron is moving down in the -Z direction. By the right-hand-rule, which way is the B-field pointing at the spot where the electron is? Assume we put x-y coordinates in the horizontal plane, and put the electron at the point (0.1m, 0m). Which way is the B-field pointing, and using the right-hand-rule, which way does the Lorentz force push the electron?

Also, are you familiar with the equations for uniform circular motion yet? like when a mass is on a string and is swung in a horizontal circle?
 
  • #10
berkeman said:
That's fair, so let's talk about it.

The current is going up in the +Z direction, and the electron is moving down in the -Z direction. By the right-hand-rule, which way is the B-field pointing at the spot where the electron is? Assume we put x-y coordinates in the horizontal plane, and put the electron at the point (0.1m, 0m). Which way is the B-field pointing, and using the right-hand-rule, which way does the Lorentz force push the electron?

Also, are you familiar with the equations for uniform circular motion yet? like when a mass is on a string and is swung in a horizontal circle?
I understand how the centripetal field exists around the wire does that mean the force is towards the wire. How would this affect the force being constant(acceleration in centripetal stays same but for this it's going from down to centripetal).
 
  • #11
Shacking said:
I understand how the centripetal field exists around the wire does that mean the force is towards the wire.
Yes, good.
Shacking said:
How would this affect the force being constant(acceleration in centripetal stays same but for this it's going from down to centripetal).
Can you compare the inward (centripital) force with the force it takes to keep the electron circulating at a constant radius? What do you find? And can you say how this might be different from the situation where the vertical B-field is uniform, instead of getting stronger closer to the source/wire?
 
  • #12
berkeman said:
Yes, good.

Can you compare the inward (centripital) force with the force it takes to keep the electron circulating at a constant radius? What do you find? And can you say how this might be different from the situation where the vertical B-field is uniform, instead of getting stronger closer to the source/wire?
I still don't understand.
 
  • #13
Shacking said:
Ya I am unsure what the direction on the force will be because the field is on a straight wire going around the wire while the electron is going down, and I am unsure of whether force is constant.
It seems to me you are making this overly complicated. There is an expression for the ##\vec B## field at the position of the electron and you have an expression for the Lorentz force ##\vec F##. The exercise asks or the instantaneous force and isn't concerned with the trajectory of the electron..., except that it asks whether the force remains the same if the electron moves in the direction of the force. That's an easy one.
 
  • Like
Likes berkeman
  • #14
BvU said:
It seems to me you are making this overly complicated. There is an expression for the ##\vec B## field at the position of the electron and you have an expression for the Lorentz force ##\vec F##. The exercise asks or the instantaneous force and isn't concerned with the trajectory of the electron..., except that it asks whether the force remains the same if the electron moves in the direction of the force. That's an easy one.
So from what I understand instanteous force has direction towards/centripetal to the wire, what I fail to understand is if this force is constant and if not why.
 
  • #15
So far, so good. What about the field closer to the wire ? Is its strength equal, bigger or smaller ?
 
  • Like
Likes berkeman and mfb
  • #16
Fortunately, this problem has a qualitative answer ("yes" or "no") only. To solve it quantitatively would involve solving a system of non-linear differential equations whih is beyond the scope of an introductory physics course.
 
  • Like
Likes berkeman

Related to Help with a magnetic field problem

1. What is a magnetic field?

A magnetic field is a region of space around a magnet or a moving electric charge where the force of magnetism can be detected.

2. How do I calculate the strength of a magnetic field?

The strength of a magnetic field can be calculated using the formula B = μ0(I / 2πr), where B is the magnetic field strength, μ0 is the permeability of free space, I is the current, and r is the distance from the current.

3. What factors affect the strength of a magnetic field?

The strength of a magnetic field is affected by the current, the distance from the current, and the permeability of the material the field is passing through.

4. How can I determine the direction of a magnetic field?

The direction of a magnetic field can be determined by using the right-hand rule, where the thumb points in the direction of the current and the curled fingers indicate the direction of the field.

5. How can I use a magnetic field to move objects?

A magnetic field can be used to move objects by creating a magnetic force on the object. This can be done by using an electromagnet or by manipulating the magnetic field of a permanent magnet.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
392
  • Introductory Physics Homework Help
Replies
9
Views
301
  • Introductory Physics Homework Help
Replies
16
Views
440
  • Introductory Physics Homework Help
Replies
12
Views
367
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
263
  • Introductory Physics Homework Help
Replies
2
Views
300
  • Introductory Physics Homework Help
Replies
7
Views
904
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
1K
Back
Top