Magnetic field vector due to linear conductor

In summary: I know, i should get much "smaller" expression, i could do that, but the most important thing for me was to see if i solved it correctly.
  • #1
cdummie
147
5

Homework Statement


Through linear conductor flows current I, with direction shown in the picture. Axis where conductor is placed is common edge of three areas with different ferromagnetic materials. They form angles θ1, θ2, θ31 + θ2 + θ3 = 2π). If space is filled with homogeneous materials with permeabilities μ1, μ2, μ3 respectively, find intensity of magnetic field vector for arbitrary point in the space due to linear conductor.

slika za forum.jpg


Homework Equations


Generalized Ampere's law.

The Attempt at a Solution



This is my attempt:

Considering the direction of the current, it's obvious that intensity of magnetic field vector depends on the distance from the linear conductor. Since the vector B is normal to the boundary of two materials, from the boundary conditions we have that B is same for any point with same distance from the conductor regardless of the material, which means that H is different in different materials.

Using generalized Ampere's law we have:

H*dl=∫H*dl*cos(H,dl)= H11*r + H22*r + H33*r

∑I=IIf we express H2 and H3 as B/μ2 and B/μ3 respectively we get the expression for H1 and we can do similar thing for H2 and H3.

so,

H1=(I-B*r(θ22 + θ33))/θ1*r

Is this correct, and is this reasonable approach?
 
Physics news on Phys.org
  • #2
You still have H1 and B, one of them has to disappear for the final result.

The approach looks fine so far.
 
  • #3
mfb said:
You still have H1 and B, one of them has to disappear for the final result.

The approach looks fine so far.

I know what you mean, i forgot to express B in final formula, which i would do by expressing H1, H2 and H3 in first formula using B divided by corresponding permeability that way i could easily find B since it's the same for any point regardless of material.
 
  • #4
Not sure what the OP said in post 3, but the expression for Hi (or Bi) should have nothing but I, θ1, θ2, θ3 and r in it. No H's or B's. If this is what the OP meant I have no additioal comment.
 
  • Like
Likes mfb
  • #5
rude man said:
Not sure what the OP said in post 3, but the expression for Hi (or Bi) should have nothing but I, θ1, θ2, θ3 and r in it. No H's or B's. If this is what the OP meant I have no additioal comment.

I wasn't clear enough, sorry, anyway, this is what i meant:

Screenshot_3.jpg
 
  • #6
cdummie said:
I wasn't clear enough, sorry, anyway, this is what i meant:

View attachment 88740
Fine! I thought that's what you meant but wasn't sure.
 
  • #7
The last expression for H1 can be simplified significantly (e.g. starting from the expression for B).
 
  • Like
Likes cdummie
  • #8
mfb said:
The last expression for H1 can be simplified significantly (e.g. starting from the expression for B).

I know, i should get much "smaller" expression, i could do that, but the most important thing for me was to see if i solved it correctly.
 

What is a magnetic field vector?

A magnetic field vector is a mathematical representation of the direction and strength of a magnetic field. It is a vector quantity, meaning it has both magnitude and direction.

What is a linear conductor?

A linear conductor is a material, such as a wire, that allows electric current to flow through it in a straight line.

How is the magnetic field vector calculated for a linear conductor?

The magnetic field vector for a linear conductor can be calculated using the right-hand rule. The direction of the vector is perpendicular to the direction of the current and the direction of the magnetic field lines. The magnitude of the vector can be calculated using the formula B = μ0I/2πr, where μ0 is the permeability of free space, I is the current, and r is the distance from the conductor.

How does the magnetic field vector change with distance from a linear conductor?

The strength of the magnetic field vector decreases with distance from a linear conductor. It follows an inverse relationship, meaning the further away you are from the conductor, the weaker the magnetic field vector will be.

What are some real-world applications of the magnetic field vector due to a linear conductor?

The magnetic field vector due to a linear conductor is used in various technologies, such as electric motors, generators, and transformers. It is also used in medical imaging techniques like MRI machines. Additionally, the Earth's magnetic field can be considered as a magnetic field vector due to a linear conductor, as it is generated by the movement of molten iron in the Earth's core.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
2
Views
1K
  • Special and General Relativity
Replies
4
Views
146
  • Engineering and Comp Sci Homework Help
Replies
4
Views
1K
  • Electromagnetism
Replies
5
Views
321
  • Electromagnetism
Replies
7
Views
959
  • Introductory Physics Homework Help
2
Replies
40
Views
910
Replies
27
Views
1K
Replies
54
Views
3K
Replies
1
Views
44
  • Engineering and Comp Sci Homework Help
Replies
5
Views
2K
Back
Top