Electromagnetism: Moving conductor and EMF

In summary, the three questions are:1) What is the current flowing through a 0.3m long conductor when a current of 4.5A is driven through it in a perpendicular direction?2) What force is exerted on the conductor when this current is applied?3) What is the wattage of the current?
  • #1
milkism
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Homework Statement
A conductor CD with a length of 30 cm is parallel to the z-axis and rotates at a distance r= 50 cm around this axis with a rotational frequency of 120 revolutions per minute in a radial magnetic field B in the r-direction.
1) Calculate the EMF if B is 0.25 T.
2) A direct current I = 4.5 A is sent through this conductor according to the positive z-axis. Then calculate the work done W at 1 single revolution of the conductor around the z-axis.
3) Also calculate the power P required for this.
Relevant Equations
Look at solution.
For 1) I used $$V=Blv=Blwr$$, where $$w= \frac{4\pi rad}{sec}$$, $$l= 0.30m$$ and $$r=0.50m$$.
I got 0.5 V.
For 2) I used W=Vq=VIt, where $$q=It$$, where t=0.5 s, we get 1.125 J.
For 3) I used P=IV, we get 2.25 W.
Are these correct?
 
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  • #2
Work done by what force?
 
  • #3
nasu said:
Work done by what force?
It doesn't say.
 
  • #4
nasu said:
Work done by what force?
Please, no one reply and say 'magnetic force'. Or the thread is doomed. Doomed.
 
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  • #5
Steve4Physics said:
Please, no one reply and say 'magnetic force'. Or the thread is doomed. Doomed.
Not magnetic and gravity, so electric?
 
  • #6
Can someone draw a picture?
 
  • #7
malawi_glenn said:
Can someone draw a picture?
33164721a96102be631a4dea4c851929.png
 
  • #8
milkism said:
Not magnetic and gravity, so electric?
Yes. A (less-than-obvious) electric force. Arising from relativistic effects.

Whether the electromagnetic force on a charged particle is called 'magnetic' or 'electric' (or is some mixture) depends on the frame of reference used (e.g. the frame of reference of the conductor or that of the moving charges).
 
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  • #9
milkism said:
Not magnetic and gravity, so electric?
In steady state there must be external sources attached (somehow) to the rods to maintain this current. Then the devil is in the detail of how the creator of the radial field interacts with this motivator. The force will be locally electric
 
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  • #10
milkism said:
For 1) I used $$V=Blv=Blwr$$, where $$w= \frac{4\pi rad}{sec}$$, $$l= 0.30m$$ and $$r=0.50m$$.
I got 0.5 V.
Although using one significant figure is technically correct, two sig.figs. seems more appropriate here.

What is your answer to two sig. figs.?

(I get the same as you when I round my answer to one sig.fig. )

Note: the unrounded value should be used if required for subsequent calculations.

milkism said:
For 2) I used W=Vq=VIt, where $$q=It$$, where t=0.5 s, we get 1.125 J.
You have given the answer to four sig. figs.!

EDIT:
On reflection, I think my comments below are wrong and that all that 3 questions are about the system acting as a generator. So I’ve struck-through the comments.

The question is not entirely clear, but this is my interpretation.

The current of 4.5A is not due to the emf worked out in Q1).
Q1) is about the conductor acting as a generator when it is rotated, producing an emf.
Q2) is about the conductor acting as a motor when an externally provided current is driven through it.

So, for Q2), you must first answer the question: what force is exerted on a 0.3m long conductor, carrying a current of 4.5A perpendicular to a 0.25T magnetic field?

milkism said:
For 3) I used P=IV, we get 2.25 W.
Like Q2) I don't believe you are meant to assume the emf is the applied voltage. You don't know the applied voltage. You are now considering the system acting as a motor, not a generator. You should use your answer to Q2).
 
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  • #11
Sorry to be stupid here but what is the question?
The answer to Q2 is 1.125J per revolution (at 0.5 Hz) which is 2.250 J/s
The answer to Q3 is 2.250 Watts
I am assuming the 4.5V is ~exact (4.500000V)
Am I missing a subtlety here?
 
  • #12
Steve4Physics said:
Although using one significant figure is technically correct, two sig.figs. seems more appropriate here.

What is your answer to two sig. figs.?

(I get the same as you when I round my answer to one sig.fig. )

Note: the unrounded value should be used if required for subsequent calculations.You have given the answer to four sig. figs.!

EDIT:
On reflection, I think my comments below are wrong and that all that 3 questions are about the system acting as a generator. So I’ve struck-through the comments.

The question is not entirely clear, but this is my interpretation.

The current of 4.5A is not due to the emf worked out in Q1).
Q1) is about the conductor acting as a generator when it is rotated, producing an emf.
Q2) is about the conductor acting as a motor when an externally provided current is driven through it.

So, for Q2), you must first answer the question: what force is exerted on a 0.3m long conductor, carrying a current of 4.5A perpendicular to a 0.25T magnetic field?Like Q2) I don't believe you are meant to assume the emf is the applied voltage. You don't know the applied voltage. You are now considering the system acting as a motor, not a generator. You should use your answer to Q2).
Haha sorry, didn't really put much attention at significant figures, if it was at the good ol days when I was at school, I would have put attention!
 
  • #13
hutchphd said:
Sorry to be stupid here but what is the question?
The answer to Q2 is 1.125J per revolution (at 0.5 Hz) which is 2.250 J/s
The answer to Q3 is 2.250 Watts
I am assuming the 4.5V is ~exact (4.500000V)
Am I missing a subtlety here?
My question was if my answers to the three questions were correct, I've got same for Q2 and Q3. Where did you get 4.5 V from? Did you mean 4.5 A? :)
 
  • #14
Yes of course. Apologies.
I believe the answers are correct. So what are your concerns here? I don't understand some of the dialog.
And the question is sort of artificial because it is difficult to have a current without a complete circuit and a radial B field is also an odd duck.
 
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  • #15
milkism said:
Haha sorry, didn't really put much attention at significant figures, if it was at the good ol days when I was at school, I would have put attention!
If you evaluate ##\mathscr E = Bl\omega r## it gives ##\mathscr E =## 0.47124V.

You’ve rounded to 0.5V (ok) but then used this rounded value in Q2) and Q3).

It's worth noting that using the unrounded value for emf:
- the work in Q2) is 1.06J (not 1.125J);
- the power in Q3) is 2.12J (not 2.25J).
It makes a significant (pun intended) difference!
 
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  • #16
Yes good catch. But I dislike this problem because the topology is very wierd: in particular how does one build such a scenario in a real world. Radial B fields and currents in disconnected segments.
 
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  • #17
Yep, that's very weird.
 
  • #18
hutchphd said:
Yes good catch. But I dislike this problem because the topology is very wierd: in particular how does one build such a scenario in a real world. Radial B fields and currents in disconnected segments.
I can see the disconnected segment as a piece of a very big closed loop, say the Earth's circumference, and the question essentially asking to find the work per unit length. It's the radial field that bothers me because it violates the "no magnetic monopoles" Maxwell equation ##~\mathbf{\nabla}\cdot\mathbf{B}=0.##
 
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  • #19
hutchphd said:
But I dislike this problem because the topology is very wierd: in particular how does one build such a scenario in a real world. Radial B fields and currents in disconnected segments.
kuruman said:
It's the radial field that bothers me because it violates the "no magnetic monopoles" Maxwell equation ##~\mathbf{\nabla}\cdot\mathbf{B}=0.##
A little convoluted but…

For a pretty good radial field, use the gap in a loud-speaker magnet as shown in cross-section here:

https://mynewmicrophone.com/wp-cont..._Moving-Coil_Loudspeaker_Driver_Diagram-1.jpg

Put a vertical conducting rod in the gap. Have a flying lead connected to the top of the rod and some rotating mechanical support to keep the rod aligned as it moves around the gap.. Have a ring of mercury at the bottom of the gap providing the lower electrical contact to the rod.

Voilà!
 
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