Some questions about electromagnetism

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Hi,

Could you please help me with the queries below? Thank you.

Question 1:
Please have a look on this attachment. In the given attachment, the disk is mounted vertically and only a part of the disk passes thru the magnetic field at any instant. If entire surface of the disk was rotating in a uniform magnetic field then I don't think any eddy currents would be produced. Do you agree?

Question 2:
If the disk was instead mounted horizontally but still all of the disk rotating in a uniform magnetic field, would it result into production of any eddy currents?

Question 3:
It is said that an accelerating charge radiates electromagnetic waves, does this mean that the quantity of charge would reduce as a result of emanating electromagnetic waves?

Question 4:
Please have a look on this attachment. If the attachment is not clear please have a look here. I believe that "v" represents the direction of disk rotation. How is the direction of "v" is being determined? The direction of magnetic field is from left to right and disk is rotating clockwise when seen from the left side.

Question 5:
?temp_hash=93270061f59883f029fdcae9d3beec2f.gif

We can see that there is circular magnetic field around wire but where are the north and south poles of this circular magnetic field?

Thanks a lot for your time and help.
 

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  • #2
cnh1995
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If entire surface of the disk was rotating in a uniform magnetic field then I don't think any eddy currents would be produced. Do you agree?
Yes, but there will be a voltage between centre and rim of the disc.
If the disk was instead mounted horizontally but still all of the disk rotating in a uniform magnetic field, would it result into production of any eddy currents?
Yes. Emf will be indued along the thickness of the disc.
It is said that an accelerating charge radiates electromagnetic waves, does this mean that the quantity of charge would reduce as a result of emanating electromagnetic waves?
No.
How is the direction of "v" is being determined?
It is the instantaneous linear velocity of a point on the circular disc (with a direction tangential to the circle).
We can see that there is circular magnetic field around wire but where are the north and south poles of this circular magnetic field?
There cannot be a single isolated current carrying conductor, there's always at least one more conductor that acts as a return path for the current. Together they two form a single-turn coil which acts as a bar magnet and you can find its polarity.
 
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We can see that there is circular magnetic field around wire but where are the north and south poles of this circular magnetic field?
It is not a dipole field so it doesn't have a well defined pole.
 
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Thank you.

Yes, but there will be a voltage between centre and rim of the disc.
So, no eddy current but an emf will be induced. I was just trying to find some good picture to make my point clear and found out that it used to be called Faraday paradox back in the days. Please check this out: https://en.wikipedia.org/wiki/Homopolar_generator#Disk-type_generator

In the picture below I believe that the rim would have positive potential and the center negative potential. Please let me know if I'm wrong.

?temp_hash=053b97f86e3f3942ee77c4b44cb115c1.jpg


Yes. Emf will be indued along the thickness of the disc.
So, you are saying that the eddy will be produced and an emf will be also induced. I had the construction below in mind.

Using right hand rule, it shows the direction of induced current at points A and B, and it looks like that lower and upper halves of the disk will have different charge distribution.

?temp_hash=053b97f86e3f3942ee77c4b44cb115c1.jpg


No.
So, it means that an an accelerating charge won't loose its charge. As radiating electromagnetic waves is energy therefore the charge must loose something, and I think it would be mass then.

There cannot be a single isolated current carrying conductor, there's always at least one more conductor that acts as a return path for the current. Together they two form a single-turn coil which acts as a bar magnet and you can find its polarity.
It is not a dipole field so it doesn't have a well defined pole.
@cnh1995 : Let's think of an infinite wire and in such a case we can ignore the return path.
@Dale: Yes, it does look like that it's not a dipole field but a magnet cannot exist as a uni-pole which also implies no magnetic field without a 'dipole' magnet.

Thank you for your time and help.
 

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@Dale: Yes, it does look like that it's not a dipole field but a magnet cannot exist as a uni-pole which also implies no magnetic field without a 'dipole' magnet.
Just because there are no magnetic monopoles does not imply that all magnetic fields are dipoles. There are many fields that are neither.
 
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  • #6
cnh1995
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So, you are saying that the eddy will be produced and an emf will be also induced. I had the construction below in mind.
You have joined the center and the rim. I was talking about the same configuration, but not with the center and rim connected together like that, just the rotating disc.
In that case, current will flow from A to B on the upper surface and from B to A on the lower surface.
20171012_094324.jpg
 
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cnh1995
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So, it means that an an accelerating charge won't loose its charge. As radiating electromagnetic waves is energy therefore the charge must loose something, and I think it would be mass then.
I am not sure if I can comment on this since I haven't studied EM radiation in depth, but just because radiating EM waves is radiating energy doesn't mean the charge has to lose something of its own (charge or mass). The charge is accelerated by means of some energy source and this radiation energy should ultimately come from that source.
 
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  • #8
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Thank you.

Just because there are no magnetic monopoles does not imply that all magnetic fields are dipoles. There are many fields that are neither.
If you say so, I'd agree with you. But could you please tell me of some fields which are neither monopole or dipole? I'm just curious.

You have joined the center and the rim. I was talking about the same configuration, but not with the center and rim connected together like that, just the rotating disc.
Okay. But in the other configuration where the disc is horizontal, the center and rim were connected too so why is it different for this case when the disc is vertical as shown below?

vertical_disc-jpg.jpg


I am not sure if I can comment on this since I haven't studied EM radiation in depth, but just because radiating EM waves is radiating energy doesn't mean the charge has to lose something of its own (charge or mass). The charge is accelerated by means of some energy source and this radiation energy should ultimately come from that source.
No problem then. But I believe that it should be mass because I did read somewhere in the past that electrically charged body is hard to accelerate compared to an equivalent neutral body, and we can use the relation mass=E/c^2.

Thanks a lot.
 
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If you say so, I'd agree with you. But could you please tell me of some fields which are neither monopole or dipole?
The field around a straight wire for one, and all of the multipole fields greater than dipole, and a plane wave.
 
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cnh1995
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But could you please tell me of some fields which are neither monopole or dipole? I'm just curious
Induced electric field in Faraday's law is not a dipole field.
But in the other configuration where the disc is horizontal, the center and rim were connected too
That connection is necessary for a current to flow. Without that connection, there will only be voltage present between center and rim of the disc.
so why is it different for this case when the disc is vertical as shown below?
For the vertical disc in your diagram, no connection is required between center and rim as the current is flowing along the path I've shown in #6. If you connect the center and the rim, you are providing a parallel path for the current and some current will be diverted through that path.
But I believe that it should be mass because I did read somewhere in the past that electrically charged body is hard to accelerate compared to an equivalent neutral body, and we can use the relation mass=E/c^2.
I don't have enough knowledge about EM radiation and mass-energy equivalence. I can't comment on that.
 
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