Wire rotating inside a magnetic field

  • Thread starter Moara
  • Start date
  • #1
41
5
Homework Statement:
A conducting wire of lengh l is fixed to an insulator disk that rotates with ω. The wire is on a plane perpendicular to a uniform magnetic field B. Find the potencial difference between the middle of the wire and one of it's ends.
Relevant Equations:
V=ωr, E=BVL
15756745138187264951087213310728.jpg

For a infinitesimal wire of lengh dx, the induced potential difference in an uniform B field perpendicular to it's motion is :
dE=B.Vp.dx, where Vp is the component of the velocity perpendicular to the wire.
Looking to the big wire I tried to take an arbitrary point express dE in function of only one variable and integrate over the half of the wire. I believe that this method should work, although very long and leading to boring integrals. Would there be a way using Faraday's law, or at least some kind of simplification that could lead to the solution?
 

Answers and Replies

  • #2
phyzguy
Science Advisor
4,935
1,870
Why don't you try making a complete triangular loop by drawing lines from the center of the circle to the two ends of the wire? Then try applying Faraday's law to that loop. What does that tell you?

Or you could draw a line from the center of the loop to one end of the wire and another from the center of the loop to the center of the wire.
 
Last edited:
  • Like
Likes Moara and TSny
  • #3
rude man
Homework Helper
Insights Author
Gold Member
7,945
827
Homework Statement:: A conducting wire of lengh l is fixed to an insulator disk that rotates with ω. The wire is on a plane perpendicular to a uniform magnetic field B. Find the potencial difference between the middle of the wire and one of it's ends.
Homework Equations:: V=ωr, E=BVL

View attachment 253769
For a infinitesimal wire of lengh dx, the induced potential difference in an uniform B field perpendicular to its motion is :
dE=B.Vp.dx, where Vp is the component of the velocity perpendicular to the wire.
Looking to the big wire I tried to take an arbitrary point express dE in function of only one variable and integrate over the half of the wire. I believe that this method should work, although very long and leading to boring integrals. Would there be a way using Faraday's law, or at least some kind of simplification that could lead to the solution?
You have the right idea in trying to use the Blv law. For moving media such as here using Faraday by itself is a risky business.

Drawing a line between the wire and the loop's center as suggested in post 2 is the way to go. You can then apply the Blv law to the two lines and get the emf for the third from Faraday's law to complete the picture.
 
  • #4
41
5
I managed to solve the problem, using Faraday's law it turns out that the flux through the imaginary triangle is constant. Now, I have a question, in this case, of a triangular loop of wire, since the area of the triangle and the magnetic field is constant, the flux is constant, therefore the induced potencial difference is zero. Now, if there is another situation, where a conducting disk rotates about a axe perpendicular to it's surface passing by the center, with constant angular velocity in an uniform magnetic field perpendicular to the disks plane, although the area and the B field are constants, the flux is not constant and there would be a potencial difference induced between the edge and the center of the disk, why? Why in the case of the loop there isn't a induced potencial but in the case of a uniform conducting disk there is ?
 
  • #5
phyzguy
Science Advisor
4,935
1,870
There is in both cases a potential difference between the center of the disk and the edge, as you can see by integrating dE=B.Vp.dx along a radius, as you suggested in your first post. However, if you take a complete loop, you go from center to edge once, and then back from edge to center, so the potential cancels out. Your question makes me wonder if you have solved your original question correctly. I suggest you post your solution so we can comment.
 
  • #6
41
5
Oh, now I see, I was messing up things. Using the superposition principle to "complete" a triangle with the center, middle point and edge point, you form a triangle Wich has constant area in a constant magnetic field, so the total induced potencial difference along it's perimeter is zero. This means that the sum of eletromotive force of each side of the triangle is zero. Therefore to find the required ddp, it is enough to take the difference of the eletromotive forces between the two sides intersected in the center.
Just use dE=B.V(r).dr, and integrate over 0 to R and 0 to √4R²-l²/4 this should give the B.w.l²/8
 
  • #7
phyzguy
Science Advisor
4,935
1,870
Oh, now I see, I was messing up things. Using the superposition principle to "complete" a triangle with the center, middle point and edge point, you form a triangle Wich has constant area in a constant magnetic field, so the total induced potencial difference along it's perimeter is zero. This means that the sum of eletromotive force of each side of the triangle is zero. Therefore to find the required ddp, it is enough to take the difference of the eletromotive forces between the two sides intersected in the center.
Just use dE=B.V(r).dr, and integrate over 0 to R and 0 to √4R²-l²/4 this should give the B.w.l²/8
I think you've got it!
 
  • #8
rude man
Homework Helper
Insights Author
Gold Member
7,945
827
Now, I have a question, in this case, of a triangular loop of wire, since the area of the triangle and the magnetic field is constant, the flux is constant, therefore the induced potencial difference is zero. Now, if there is another situation, where a conducting disk rotates about a axe perpendicular to it's surface passing by the center, with constant angular velocity in an uniform magnetic field perpendicular to the disks plane, although the area and the B field are constants, the flux is not constant and there would be a potencial difference induced between the edge and the center of the disk, why? Why in the case of the loop there isn't a induced potencial but in the case of a uniform conducting disk there is ?
 
  • #9
rude man
Homework Helper
Insights Author
Gold Member
7,945
827
OK, first, don't call an emf a potential difference. A potential difference is the line integral of an electrostatic field whereas an emf is the line integral of the E field generated by a non-electric source of energy; like a battery or magnetic field.

There is no difference between the two situations. If you go around the larger loop (the radii plus the wire) the two radii emf's cancel, like 2 batteries in series & opposition, leaving the emf of the wire, which is also zero since as you correctly point out faraday says the circulation of any path is zero since dB/dt=0.. Similarly, the short line betw. the wire center and loop center also generates an emf, albeit smaller than that for the radii since it's shorter.

One way to imagine how your disc works is to imagine a very large number of radii all around the disc and closely spaced together but with a sliver of insulation between each sliver of metal. This is allowed since there is no current flow in the theta direction along the disc. So now you have a large number of "batteries" in parallel generating the same emf each. So this is again an application of what I call the Blv law i.e your d(emf) = Bv(r) dr.

I should have added that if you integrate E from the edge of one sliver down to the disc center and then back up via any other sliver you will also get zero emf.
 
Last edited:

Related Threads on Wire rotating inside a magnetic field

Replies
2
Views
7K
Replies
5
Views
5K
Replies
1
Views
2K
Replies
6
Views
3K
Replies
1
Views
2K
Replies
3
Views
7K
Replies
1
Views
2K
Replies
1
Views
3K
Replies
6
Views
1K
  • Last Post
Replies
2
Views
583
Top