Magnetic dipole moment and integration surfaces (Introductory Electromagnetics - Magnetistatics)

  • #1
Graham87
63
16
Homework Statement
See pictures
Relevant Equations
See pictures
Screenshot 2024-04-08 115126.png

A current current loop is running through this figure.
How do I design integration surfaces to find the magnetic dipole moment?

Screenshot 2024-04-08 115346.png


In the solution the three following figures were designed for integration surface, and they prove that they all give the same answer of ## m= 2IRd \hat{z} ##.

Screenshot 2024-04-08 115532.png

For a) they just did the area of the rectangle which is 2Rd for the vector area, since the sides cancel. Shouldnt one need to calculate the area for the top surface of the integration surface?

For b) they did it this way:
Screenshot 2024-04-08 123346.png

Screenshot 2024-04-08 123423.png

Screenshot 2024-04-08 123429.png


For c) they similarly
Screenshot 2024-04-08 123515.png



My problem is I do not understand how to go about finding integration surfaces. What are they for? What are the criterias? How do they work?
So the vector area of the magnetic dipole moment is not the area enclosed in the current loop?

Why do they only find the area of the top surface in b) while neglecting the top surface in a)?
Im not sure how this works mathematically. Is there anything to do with boundary surface? How come one does not calculate the boundary surface?

Thanks
 
Last edited:
Physics news on Phys.org
  • #2
The area integral formula uses any area that has the current loop as its boundary. As long as you have the correct boundary loop, they will give the same result. Ultimately, this rests upon Stoke's theorem (or Gauss' theorem if you will).
 
  • #3
To substantiate that, consider two surface integrals of ##d\vec S##
$$
I_1 = \int_{S_1} d\vec S, \quad I_2 = \int_{S_2} d\vec S
$$
where ##S_1## and ##S_2## have the same boundary. Then
$$
\Delta I = I_2 - I_1 = \oint_{S}1 \, d\vec S
$$
where ##S## is the surface of the joint surface with an outward pointing normal. Since the surfaces share the boundary, this is a closed surface. By Stokes' theorem
$$
\oint_S 1\, d\vec S = \int_\Omega (\nabla 1) dV = \int_\Omega 0\, dV = 0
$$
so ##I_1 = I_2##.
 
  • Like
Likes PhDeezNutz and nasu
  • #4
OP are you familiar with cylindrical coordinates?

The setup mentioned in your problem is a slight variation of that.
 
  • #5
To add a little bit more: You can even use different surfaces in the computation of each component of the moment. Ultimately, the component in direction ##\vec e_i## is the current multiplied by the area of the projection of the current loop on a plane perpendicular to ##\vec e_i##.

Using this idea, it is fairly easy to see that the x and y components of the magnetic moment will be zero for the loop in this problem.
 
  • Like
Likes PhDeezNutz

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
301
  • Introductory Physics Homework Help
Replies
8
Views
214
  • Introductory Physics Homework Help
Replies
25
Views
323
  • Introductory Physics Homework Help
Replies
1
Views
423
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
16
Views
1K
  • Advanced Physics Homework Help
Replies
5
Views
407
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
851
  • Introductory Physics Homework Help
Replies
7
Views
1K
Back
Top