Electromagnetics density of Toroid

In summary: Since B is uniform around that surface, B times the area = B pi r^2. The current going through that surface is the current in the wire, which is I. So B pi r^2 = mu I.In summary, the question is about the direction of current in the Amperean loop in relation to the direction of the current in the wire. The misunderstanding is based on a different interpretation of what is meant by "going through the loop" and does not take into account the defined surface and the direction of the B field.
  • #1
baby_1
159
15
Hello
I have some problems to obtaining Electromagnetic density via Ampere's Law.

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/indtor.html

my problem is why we don't calculate and sum Inflow and outflow of Ampere's closed loop path.to clearing my question please take a look at below picture.
tor.gif

in upper side of Toroid current direction is going into center of Ampere's closed loop path however down side current direction is going out subsequently input current and output current in Ampere's closed loop path should be zero.

what is my misunderstand?

Thanks
 
Physics news on Phys.org
  • #2
baby_1 said:
Hello
I have some problems to obtaining Electromagnetic density via Ampere's Law.

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/indtor.html

my problem is why we don't calculate and sum Inflow and outflow of Ampere's closed loop path.to clearing my question please take a look at below picture.
tor.gif

in upper side of Toroid current direction is going into center of Ampere's closed loop path however down side current direction is going out subsequently input current and output current in Ampere's closed loop path should be zero.

what is my misunderstand?

Thanks

You are misunderstanding the fact that the Amperean loop in the figure is constructed BETWEEN the innner and the outer radius of the toroid and not at a radius greater than or equal to b. Here, the direction of every current element is into the plane of the paper (or according to you, to the center of the toroid).

Had the amperean loop been constructed at a radius of 'b' (outer radius of toroid) or more, the total current crossing the loop would be zero.

Hope this helps! :)
-Adithyan
 
  • #3
You're mistaken all the currents are going through the loop in the same direction. There is no upside and down side. Your misunderstanding seems to be a different (and incorrect) interpretation of what is meant by "to go through the loop".
 
  • #4
Thanks for your response.
but my question is some thing else.(sorry if i explained so weak)
as you see this picture
7913629700_1401179559.jpg

two different current direction see in Ampere's closed loop path.one of them wire is top of Toroid and we see the direction of current is going to center and the bottom current's wire that i specify with dot is going out.so sum of them should not be zero?
 
  • #5
The blue dashed line defines a round surface, area = pi r^2. The current always goes INTO that surface. The fact that the current goes OUT OF the page is irrelevant because that is not part of the surface.

Ampere's law states that the integral of the B field around a defined surface = mu times the total current through that surface.
 
  • Like
Likes 1 person

1. What is the definition of electromagnetics density of toroid?

Electromagnetics density of toroid refers to the measure of electromagnetic energy per unit volume within a toroid, which is a donut-shaped object with a circular cross section.

2. How is the electromagnetics density of toroid calculated?

The electromagnetics density of toroid is calculated by dividing the total energy stored in the toroid by its volume. The total energy can be found by multiplying the magnetic flux and current in the toroid.

3. What factors affect the electromagnetics density of toroid?

The main factors that affect the electromagnetics density of toroid are the size and shape of the toroid, the material it is made of, and the current passing through it. A larger toroid with a higher current and a material with a higher permeability will have a higher density.

4. What are the units of measurement for electromagnetics density of toroid?

The units of measurement for electromagnetics density of toroid are joules per cubic meter (J/m^3) in the SI system and ergs per cubic centimeter (erg/cm^3) in the CGS system.

5. How is the electromagnetics density of toroid used in practical applications?

The electromagnetics density of toroid is used in various practical applications, such as in the design of transformers, inductors, and electromagnets. It is also used in the study of magnetic materials and their properties, and in the development of new technologies such as wireless charging and magnetic levitation.

Similar threads

  • Introductory Physics Homework Help
2
Replies
36
Views
3K
  • Introductory Physics Homework Help
Replies
9
Views
3K
Replies
14
Views
595
  • Introductory Physics Homework Help
Replies
4
Views
2K
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
26
Views
473
  • Introductory Physics Homework Help
Replies
8
Views
13K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
6K
  • Advanced Physics Homework Help
Replies
3
Views
3K
Back
Top