Magnetic field - induced emf.

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  • #1

Homework Statement


http://www.jyu.fi/kastdk/olympiads/2005/Th2.pdf [Broken]
http://www.jyu.fi/kastdk/olympiads/2005/Th2%20Solution.pdf [Broken]
(Question 3)

My first doubt is why the magnetic field at the center of the coil is:

[tex]|\vec B|=\frac{\mu_0 NI}{2a}[/tex] and not


[tex]|\vec B|=\mu_0 nI[/tex], where n=N/l, the density of turns. The problem does not give any value for l.

Second,

They calculate the emf on each ring by doing [itex]\epsilon=\int E dr[/itex], where [tex]E=\omega Br[/tex]. Am I right?

And finally why do they sum the emfs of the rings?

----------------------
(Question 2), Why is that for the stationary regime I will calculate the mean values of the induced magnetic field? I think the magnetic needle will never stay put, there will never be a stationary regime, I think.

Thank you for the help.
 
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  • #2
nrqed
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Homework Statement


http://www.jyu.fi/kastdk/olympiads/2005/Th2.pdf [Broken]
http://www.jyu.fi/kastdk/olympiads/2005/Th2%20Solution.pdf [Broken]
(Question 3)

My first doubt is why the magnetic field at the center of the coil is:

[tex]|\vec B|=\frac{\mu_0 NI}{2a}[/tex] and not


[tex]|\vec B|=\mu_0 nI[/tex], where n=N/l, the density of turns. The problem does not give any value for l.

I did not have time to look at your second question but for the first one, the equation [tex]|\vec B|=\frac{\mu_0 NI}{2a}[/tex] is simply the B field at the center of a circular loop. That's all. The current here is an induced current, so the idea is that you can find the induced current in the loop (knowing its angular frequency of rotation and the magnitude of the external B field, etc) and then, using this induced current you can determine the B field produced by the loop at its center.
 
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  • #3
I did not have time to look at your second question but for the first one, the equation [tex]|\vec B|=\frac{\mu_0 NI}{2a}[/tex] is simply the B field at the center of a circular loop. That's all. The current here is an induced current, so the idea is that you can find the induced current in the loop (knowing its angular frequency of rotation and the magnitude of the external B field, etc) and then, using this induced current you can determine the B field produced by the loop at its center.

Ah ok, I was calculating the magnetic field inside a solenoid. My bad. :frown:
Thank you so much for replying.

This is easily prooved with Biot-Savart Law, which is not in IPhO syllabus.
 
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  • #4
nrqed
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Homework Statement


http://www.jyu.fi/kastdk/olympiads/2005/Th2.pdf [Broken]
http://www.jyu.fi/kastdk/olympiads/2005/Th2%20Solution.pdf [Broken]
(Question 3)

My first doubt is why the magnetic field at the center of the coil is:

[tex]|\vec B|=\frac{\mu_0 NI}{2a}[/tex] and not


[tex]|\vec B|=\mu_0 nI[/tex], where n=N/l, the density of turns. The problem does not give any value for l.

Second,

They calculate the emf on each ring by doing [itex]\epsilon=\int E dr[/itex], where [tex]E=\omega Br[/tex]. Am I right?

And finally why do they sum the emfs of the rings?

Thank you for the help.

About your second question, what is your E?? There is no E field here. The emf is the potential difference and the potential difference can is related to the force by [itex] e \Delta V \simeq Work \simeq \int {\vec F} \cdot d{\vec r} [/itex] (where I have not been careful with any minus signs). So [itex] \Delta V = \int \frac{{\vec F}}{e} \cdot d{\vec r} [/itex] and this is what they are using. They have to do an integral because the magnetif force varies with "r" in the disks. So the total emf has to be the integral over the entire disk.

Hope this helps
 
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  • #5
My E is an electric field, which is force per unit charge. Which gives exactly your expression, since F/e is equal to my E. This integral, [itex] \delta V = \int \frac{{\vec F}}{e} \cdot d{\vec r} [/itex], is done radially. But it could be done over the perimeter of each ring of the disk. Since there will be an emf too in each one. Isn't this idea right?
 
  • #6
nrqed
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My E is an electric field, which is force per unit charge. Which gives exactly your expression, since F/e is equal to my E. This integral, [itex] \delta V = \int \frac{{\vec F}}{e} \cdot d{\vec r} [/itex], is done radially. But it could be done over the perimeter of each ring of the disk. Since there will be an emf too in each one. Isn't this idea right?

My point is that there is no electric field here so I don't see why it should be introduced. The force is a magnetic force. So one should go directly from F/e to vB, without mentioning any E field.

And the magnetic force is acting everywhere on the disk, not just on the perimeter. This is why the integral must be done over all the "rings" . Maybe a better way to explain this is to think of a charged moving through one of the disk (from the shaft to the wire touching the perimeter of the disk or the other way around). That charge will feel a force that will depend on its distance from the axis of rotation so one must integrate over the radial distance.
 
  • #7
Ok, I got it.
I just don't agree with you in what concerns the first statement. But I understand your point of view. There is an induced electric field if I don't distinguish magnetic and electric forces. The magnetic force is an relativistic consequence of the the electric force.

Thank you for your help. :) We can go to the last question. :D
 
  • #8
nrqed
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Ok, I got it.
I just don't agree with you in what concerns the first statement. But I understand your point of view. There is an induced electric field if I don't distinguish magnetic and electric forces. The magnetic force is an relativistic consequence of the the electric force.

Thank you for your help. :) We can go to the last question. :D

I agree that the electric and magnetic forces are aspects of a single force, the electromagnetic force. And that changing frame of reference will mix E and B fields. But if you start calling the field that creates a force proprotional to the speed and perpendicualr to both the velocity and the field itself an "electric field", you will be using a notation that nobody will understand.

regards
 
  • #9
Oh, I see, if you say so, I will be careful, next time. Thanks again for your patience and time you are dispending to help me.
 
  • #10
nrqed
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Oh, I see, if you say so, I will be careful, next time. Thanks again for your patience and time you are dispending to help me.

No problem. Again, it's just a question of semantics. But in the frame of the apparatus presented in the problem, there is no E field acting on the disks. So if you solve that and talk about an E field in the solution, it will confuse everybody.


I am glad if I could help a little. Those are very good problems, by the way.

best luck!
 
  • #11
Yes, they are very interesting. :) Questions 6 and 7 are very interesting too.

Would you help me in my other doubts?
 
  • #12
nrqed
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Yes, they are very interesting. :) Questions 6 and 7 are very interesting too.

Would you help me in my other doubts?

Of course...Post any questions you may have. If I am not around, many others very knowledgeable helpers can give you a hand.
 
  • #13
I think you did not read my refreshed opening post. :redface:
 
  • #14
Forget the other questions. I could figure them out. :) Thanks.
 

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