# Electron revolving in a circular path

1. Feb 6, 2014

### utkarshakash

1. The problem statement, all variables and given/known data
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 .............?

3. 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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2. Feb 6, 2014

### Dick

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. Feb 7, 2014

### utkarshakash

Ok. I understand that. But how can a moving charge exert magnetic force at any point?

4. Feb 7, 2014

### vanhees71

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. Feb 7, 2014

6. Feb 7, 2014

7. Feb 7, 2014

### utkarshakash

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.

8. Feb 8, 2014

### haruspex

I think you have a sign wrong.
What are the four options?

9. Feb 8, 2014

### utkarshakash

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. Feb 8, 2014

### haruspex

Yes. What does that give you?

11. Jul 4, 2014

### Vibhor

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: Jul 4, 2014
12. Jul 5, 2014

### Dick

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. Jul 6, 2014

### Vibhor

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. Jul 6, 2014

### Dick

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. Jul 7, 2014

### Vibhor

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 ?

16. Jul 7, 2014

### Delta²

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. Jul 7, 2014

### Dick

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. Jul 7, 2014

### TSny

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: Sep 25, 2014
19. Jul 7, 2014

### TSny

Although it is not clear from the wording of the problem, the line CQ lies in the plane of the circle.

20. Jul 7, 2014

### Vibhor

Thank you TSny .