Electromagnetic induction in current loops

  • I
  • Thread starter phys
  • Start date
  • #1
12
0

Main Question or Discussion Point


would be really grateful if someone could help throw some light on this - its probably really simple but I am typing myself up in loops!

I have a piece of wire which is placed in an external AC magnetic field which is varying sinusoidally. The wire is folded to make two rings of radii r1 and r2 with resistances R1 and R2. (see picture below) I have derived an equation for the current in the loop in terms of the emfs E1 and E2 induced in the 2 parts of the loops. I think that this is I = (E1+E2)/(R1+R2)

I want to work out potential difference between the two parts of the loop at their crossing point in terms of the same variables - any ideas on how I would approach this? Thank you very much

 

Attachments

Answers and Replies

  • #2
cnh1995
Homework Helper
Gold Member
3,387
1,125
I want to work out potential difference between the two parts of the loop at their crossing point
Between which two points?
 
  • #3
12
0
The bottom point of each loop
 
  • #4
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,369
6,750
I don't understand the question. You can calculate the EMF, i.e., the vorticity of the em. field (it's not a voltage!) according to Faraday's Law
$$\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B}.$$
In integrated form it's
$$\int_{\partial A} \mathrm{d} \vec{x} \cdot \vec{E}=-\frac{1}{c} \frac{\mathrm{d}}{\mathrm{d} t} \int_A \mathrm{d}^2 \vec{f} \cdot \vec{B}=-\frac{1}{c} \dot{\Phi}.$$
Here you run around two loops, i.e., the total flux is the sum of the fluxes through the small and the large loop.
 
  • #5
29
12
@vanhees71 is correct. Imagine you are sitting at the bottom point of the two loops. Let's say your potential is zero. Now you go in the direction of the electric field in the loop, around the small loop then around the big loop and you return to the point you were at. Now what is the potential you have gained. It is the closed loop integral of E.dl which is simply the time derivative of the sum of the fluxes through both loops. So if you AC field is $$Bsin(\omega t)$$ then the sum of your fluxes is $$Bsin(\omega t)\pi r_{1}^{2}+Bsin(\omega t)\pi r_{2}^{2}$$

Then $$\oint{E}{dl}=Bcos(\omega t)\omega \pi(r_{1}^{2}+r_{2}^{2})=$$ difference in potential

Your formula for current is correct, by the way.
 
  • #6
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,369
6,750
Perhaps I should have emphasized more that there is no potential here, because ##\vec{\nabla} \times \vec{E} \neq 0##. It's an EMF, not a voltage!
 
  • #7
29
12
Yes, you are right in that. There is no fixed potential assigned to every point in space as such. I meant the EMF "gained" on taking one turn around the length of the wire. Of course this is lost due the resistance.
 
  • Like
Likes vanhees71
  • #8
12
0
Thanks everyone!
The two loops are actually one loop - in the sense that it is one piece of wire.
I want to work out the pd between the two points which are at the bottom - I think that an equivalent circuit is the diagram below.
I therefore want to work out the pd between the two points that I have crossed.
I'm probably wrong but would the pd just be E1 if the current was anticlockwise or E2 if the current was clockwise?
If the field is Bsinwt would this mean that the pd changes between E1 and E2?
 
  • #10
29
12
Yes, if suppose the emf in the small loop is E1 and you go over the portion of the loop that is the small circle the pd will be E1. If you go over the portion that is the big circle the pd will be E2. It all depends on what your path is.
 

Related Threads on Electromagnetic induction in current loops

Replies
2
Views
1K
Replies
3
Views
861
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
2K
Replies
5
Views
3K
  • Last Post
Replies
7
Views
1K
Replies
3
Views
1K
Replies
78
Views
10K
Replies
6
Views
5K
Top