Electron revolving in a circular path

AI Thread Summary
An electron is circulating in an anti-clockwise direction along a circular path, and the discussion revolves around determining when the magnetic field at a specific point will be zero. Participants clarify that the situation involves a single point charge rather than a continuous ring of charge, which affects the magnetic field's behavior over time. The magnetic field can be zero when the cross product of the velocity and position vector equals zero, leading to a specific mathematical relation that none of the provided options satisfy. The conversation also touches on the distinction between instantaneous and average current flow, emphasizing that while a point charge can be considered a current, it does not produce a continuous current like a ring of charge. Understanding these concepts is crucial for solving the problem effectively.
utkarshakash
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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.
 

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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.
 
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?
 
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.
 
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?
 
berkeman said:

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

Let C be origin and CQ be X-axis.
Then
velocity of electron at any time t is given as \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}

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

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

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

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

What are the four options?

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

The four options are
\pi / 3 \omega \\ 5 \pi / 3 \omega \\ 7 \pi / 3 \omega \\ 8 \pi / 3 \omega \\
 
  • #10
utkarshakash said:
Are you trying to say that \vec{v} = \omega R ( -sin \omega t \vec{i} + cos \omega t \vec{j} )
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
 
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  • #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.]
 
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  • #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 .
 
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