Electron revolving in a circular path

In summary, an electron is circulating in an anti-clockwise direction with a constant angular speed ω along a circular path. The magnetic field at point Q, where CQ=2R, will be zero at times when the cross product of the electron's velocity and position is equal to zero. However, none of the given options satisfy this condition. A continuous ring of circulating charge is not equivalent to a single point charge moving in a circle as the fields produced by a ring of charge do not depend on time, while the fields produced by a point charge do. This can also be understood in terms of the definition of current, where a continuous ring of charge has a constant current while a point charge has a varying current.
  • #1
utkarshakash
Gold Member
854
13

Homework Statement


An electron is at P at t=0. It is circulating in anti-clockwise direction with a constant angular speed ω along the shown(see attachment) circular path. Magnetic field at Q (CQ=2R, where R is radius of circle) will be recorded as zero at times ....?


The Attempt at a Solution


The situation is equivalent to a circular ring in which current is flowing in clockwise direction. I can't really see why will the magnetic field be zero at some time. It should be constant as the current does not change.
 

Attachments

  • pf001.jpg
    pf001.jpg
    4.9 KB · Views: 566
Physics news on Phys.org
  • #2
utkarshakash said:

Homework Statement


An electron is at P at t=0. It is circulating in anti-clockwise direction with a constant angular speed ω along the shown(see attachment) circular path. Magnetic field at Q (CQ=2R, where R is radius of circle) will be recorded as zero at times ....?


The Attempt at a Solution


The situation is equivalent to a circular ring in which current is flowing in clockwise direction. I can't really see why will the magnetic field be zero at some time. It should be constant as the current does not change.

It's not equivalent to a continuous ring of circulating charge. It's a single point charge moving in a circle. That's different.
 
  • #3
Dick said:
It's not equivalent to a continuous ring of circulating charge. It's a single point charge moving in a circle. That's different.

Ok. I understand that. But how can a moving charge exert magnetic force at any point?
 
  • #4
A circulating charge of course also generates both an electric and a magnetic field, including radiation ("synchrotron radiation", "bremsstrahlung"). Thus other charged particles are subject to the Lorentz Force according to this electromagnetic field.
 
  • #5
vanhees71 said:
A circulating charge of course also generates both an electric and a magnetic field, including radiation ("synchrotron radiation", "bremsstrahlung"). Thus other charged particles are subject to the Lorentz Force according to this electromagnetic field.

Can you please give some equations to start with?
 
  • #7
berkeman said:

From the equation you gave above it is clear that B=0 when [itex]\vec{v} \times \vec{r} =0[/itex]

Let C be origin and CQ be X-axis.
Then
velocity of electron at any time t is given as [itex]\omega R (sin \omega t \vec{i} + cos \omega t \vec{j} ) \\ \vec{r} = (2R -R cos \omega t ) \vec{i} - R sin \omega t \vec{j} [/itex]

Setting cross product equal to 0 gives a relation [itex] 2 cos \omega t = cos 2 \omega t [/itex].

But from the given 4 options none of them satisfies the above relation.
 
  • #8
utkarshakash said:
From the equation you gave above it is clear that B=0 when [itex]\vec{v} \times \vec{r} =0[/itex]

Let C be origin and CQ be X-axis.
Then
velocity of electron at any time t is given as [itex]\omega R (sin \omega t \vec{i} + cos \omega t \vec{j} ) [/itex]
I think you have a sign wrong.
Setting cross product equal to 0 gives a relation [itex] 2 cos \omega t = cos 2 \omega t [/itex].

But from the given 4 options none of them satisfies the above relation.
What are the four options?
 
  • #9
haruspex said:
I think you have a sign wrong.

What are the four options?

Are you trying to say that [itex]\vec{v} = \omega R ( -sin \omega t \vec{i} + cos \omega t \vec{j} )[/itex]

The four options are
[itex] \pi / 3 \omega \\ 5 \pi / 3 \omega \\ 7 \pi / 3 \omega \\ 8 \pi / 3 \omega \\ [/itex]
 
  • #10
utkarshakash said:
Are you trying to say that [itex]\vec{v} = \omega R ( -sin \omega t \vec{i} + cos \omega t \vec{j} )[/itex]
Yes. What does that give you?
 
  • #11
Dick said:
It's not equivalent to a continuous ring of circulating charge. It's a single point charge moving in a circle. That's different.

Why is it not equivalent to a continuous ring of circulating charge ? I am not able to differentiate between the two .What is wrong with treating a rotating point charge q as equivalent to current given by i = qω/2π ?

Many Thanks
 
Last edited:
  • #12
Vibhor said:
Why is it not equivalent to a continuous ring of circulating charge ? I am not able to differentiate between the two .What is wrong with treating a rotating point charge q as equivalent to current given by i = qω/2π ?

Many Thanks

A continuous ring of circulating charge wouldn't produce a time-varying field. It always looks the same. A lump of charge moving in a circle will.
 
  • #13
Dick said:
A continuous ring of circulating charge wouldn't produce a time-varying field. It always looks the same. A lump of charge moving in a circle will.


Sorry . I didn't understand much .

Could you explain in a little simpler terms why a ring of charge is a current but not a point charge ?

Thanks
 
  • #14
Vibhor said:
Sorry . I didn't understand much .

Could you explain in a little simpler terms why a ring of charge is a current but not a point charge ?

Thanks

If you have a single charge moving around a circle when you look at the circle at different times then you will see the charge in a different place at each time. This means the fields will be different at different times. If it's a continuous ring of charge smeared around the circle then it looks the same at all times. So the fields don't depend on time. There's just a circulating current.
 
  • #15
Dick said:
If you have a single charge moving around a circle when you look at the circle at different times then you will see the charge in a different place at each time. This means the fields will be different at different times. If it's a continuous ring of charge smeared around the circle then it looks the same at all times. So the fields don't depend on time. There's just a circulating current.

Thanks .

I have understood the difference between a ring of charge and point charge in terms of magnetic field .

But how do I understand it in terms of the definition of current i.e - rate of charge flow through a cross section ? Don't you think if we go by this definition ,then a point charge is crossing a cross section after a fixed time t = 2π/ω ,hence a current ?

Thanks for your patience :)
 
  • #16
Vibhor said:
Thanks .

I have understood the difference between a ring of charge and point charge in terms of magnetic field .

But how do I understand it in terms of the definition of current i.e - rate of charge flow through a cross section ? Don't you think if we go by this definition ,then a point charge is crossing a cross section after a fixed time t = 2π/ω ,hence a current ?

Thanks for your patience :)

I think (not sure though) that it is about the instanteneous rate of charge flow versus the average rate of charge flow. If q is the charge of the point particle, the average rate of charge flow for Δt=2π/ω will be the same for all the points in circle and equal to qω/2π , however the instanteneous rate of charge flow for a point of the circle at angle phi will be qω/2π only for discrete times t_n=(2nπ+phi)/ω, and will be zero for any other time t.
 
  • #17
Vibhor said:
Thanks .

I have understood the difference between a ring of charge and point charge in terms of magnetic field .

But how do I understand it in terms of the definition of current i.e - rate of charge flow through a cross section ? Don't you think if we go by this definition ,then a point charge is crossing a cross section after a fixed time t = 2π/ω ,hence a current ?

Thanks for your patience :)

Yes, it's a current. It's not a continuous current. You can approximate it as a continuous current if you want the average values of the fields over a whole rotation and not instantaneous values.
 
  • #18
Vibhor,

You have indicated elsewhere that you are a high school student. Here is a video that reviews how to find the magnetic field of a moving point charge.



It is at the high school level (AP Physics). See if you can apply the concepts discussed there to the problem you are trying to work. The formula discussed here is the same as given in Berkeman's link. (Post #6)

[If the charge is moving very fast, then the field becomes much more complicated. But, from the given choices of answers, I don't think you are expected to worry about that. Assume that the equation given in the video is applicable.]
 
Last edited by a moderator:
  • #19
Although it is not clear from the wording of the problem, the line CQ lies in the plane of the circle.
 
  • #20
Thank you TSny .
 

1. What is an electron revolving in a circular path?

An electron revolving in a circular path is a concept in physics where an electron, which is a subatomic particle with a negative charge, moves in a circular motion around an atomic nucleus. This motion is known as an orbit, and it is caused by the attraction between the negatively charged electron and the positively charged nucleus.

2. What keeps an electron in a circular path?

The centripetal force is what keeps an electron in a circular path. This force is directed towards the center of the circular motion and is caused by the electric force between the electron and the nucleus. Without this force, the electron would move in a straight line and not in a circular path around the nucleus.

3. Is the electron's path truly circular?

No, the electron's path around the nucleus is not truly circular. It is actually an elliptical shape, with the nucleus at one of the foci. However, for simplicity, we often refer to it as a circular path since the difference between a circle and an ellipse at the atomic scale is very small.

4. How fast does an electron move in its circular path?

The speed of an electron in its circular path can vary depending on the type of atom and the energy level of the electron. Generally, the speed is around 2.2 million meters per second, which is about 1% of the speed of light.

5. Can the electron's circular path change?

Yes, the electron's circular path can change. The energy level of the electron can change, causing it to move into a different orbit around the nucleus. This change can occur through interactions with other particles or by absorbing or releasing energy. Additionally, the electron's path can also be affected by external forces, such as a magnetic field.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
213
  • Introductory Physics Homework Help
Replies
3
Views
304
  • Introductory Physics Homework Help
Replies
7
Views
73
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
6
Replies
196
Views
19K
  • Introductory Physics Homework Help
Replies
7
Views
859
  • Introductory Physics Homework Help
Replies
20
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
678
  • Introductory Physics Homework Help
Replies
5
Views
736
  • Introductory Physics Homework Help
Replies
4
Views
1K
Back
Top