Solenoid filled with magnetic material

Click For Summary
SUMMARY

The discussion focuses on calculating the magnetic field produced by a solenoid filled with a magnetic material, specifically addressing the effects of permeability (μ) on the field outside the solenoid. The magnetic field is derived using the formula for a circular loop of current, leading to the expression for B as a function of angle and permeability. Participants emphasize the complexity of magnetic calculations due to the non-linear relationship between magnetization and the magnetic field, advocating for numerical methods to achieve accurate results, particularly for short solenoids with magnetic cores.

PREREQUISITES
  • Understanding of solenoid physics and magnetic fields
  • Familiarity with permeability (μ) and its role in magnetism
  • Knowledge of numerical methods for solving magnetic field problems
  • Proficiency in calculus for integrating magnetic field equations
NEXT STEPS
  • Research numerical methods for calculating magnetic fields in inhomogeneous media
  • Study the effects of magnetic core materials on solenoid performance
  • Learn about iterative approaches for solving magnetization problems
  • Explore analytical solutions for long solenoids and their limitations
USEFUL FOR

Physicists, electrical engineers, and students studying electromagnetism, particularly those interested in solenoid design and magnetic field calculations.

ShayanJ
Science Advisor
Insights Author
Messages
2,802
Reaction score
605
Consider a solenoid with length L and N turns of wire wound around it, filled with a rod of permeability \mu.I want to calculate the magnetic field it produces on a point P on its axis and outside it.The angle between the axis and the lines drawn from point P to a point on the nearest and farthest loops of the solenoid are \beta and \alpha respectively.
I used the formula for the magnetic field of a circular loop of current on its axis and integrated it along the solenoid:
<br /> \vec{dH}=\frac{I R^2 \frac{N}{L} \hat{z} dz}{[(d-z)^2+R^2]^{\frac 3 2}} \Rightarrow \vec{H}=\frac{N I R^2 \hat{z}}{L} \int_\alpha^\beta \frac{R\csc^2\varphi d\varphi}{(R^2\cot^2\varphi+R^2)^{\frac 3 2}} \Rightarrow \vec B =\frac{\mu_0 N I R^2 \hat{z}}{L}(\cos\beta-\cos\alpha)<br />
I want to know how the magnetic material filling the solenoid affects the field outside it and how can I calculate its effect?Is it right to just calculate H,as I did, and then simply multiply it by \mu_0 to obtain the field outside the solenoid?
Thanks
 
Physics news on Phys.org
A good question. Hope you'll get some useful replies. I think that simply multiplying B for the air/vacuum case by μ (not μ0) would only be right if the iron surrounded the solenoid as well as filling it (that is the solenoid and the field-measuring probe are immersed in a sea of iron, as it were). Yet we know from experiment that merely filling the solenoid with an iron core does increase the B field at points on the axis outside the solenoid.

When it comes to inhomogeneous magnetic media, many textbooks handle the case of ring-shaped cores with air gaps in them, but your case tends not to be discussed. Too hard?

A nitpick: I don't think you should have the R2 (or the ^z) in your equation.
 
Last edited:
Philip Wood said:
A good question. Hope you'll get some useful replies. I think that simply multiplying B for the air/vacuum case by μ (not μ0) would only be right if the iron surrounded the solenoid as well as filling it (that is the solenoid and the field-measuring probe are immersed in a sea of iron, as it were). Yet we know from experiment that merely filling the solenoid with an iron core does increase the B field at points on the axis outside the solenoid.

When it comes to inhomogeneous magnetic media, many textbooks handle the case of ring-shaped cores with air gaps in them, but your case tends not to be discussed. Too hard?

A nitpick: I don't think you should have the R2 (or the ^z) in your equation.
I doubt that too,but I just can't find the right thing to to do!
And yeah,R^2 shouldn't be there but I think \hat{z} should be!
 
Sorry, of course ^z should be there!

The Force between electromagnets section in the Wiki article on Electromagnets purports to deal with the question we're interested in. Unfortunately the pole strength formula given is that for an air/vacuum-cored solenoid!
 
Last edited:
Philip Wood said:
Sorry, of course ^z should be there!

The Force between electromagnets section in the Wiki article on Electromagnets purports to deal with the question we're interested in. Unfortunately the pole strength formula given is that for an air/vacuum-cored solenoid!

I'm beginning to think that the magnetic field of such a solenoid has two parts.One part is given by the formula I wrote in the first post and the other part is due to the magnetization of the magnetic core.But for deriving that,we should have an expression for the magnetization of the magnetic core which can be derived from the formula for the magnetic field inside the solenoid.
I think I found how to do it,I'll tell you if I succeed!
 
I agree with this approach. The problem, of course, is finding the magnetisation of the core, or its dipole moment. I'll be interested in your progress!
 
Shyan said:
I'm beginning to think that the magnetic field of such a solenoid has two parts.One part is given by the formula I wrote in the first post and the other part is due to the magnetization of the magnetic core.But for deriving that,we should have an expression for the magnetization of the magnetic core which can be derived from the formula for the magnetic field inside the solenoid.
I think I found how to do it,I'll tell you if I succeed!
You have identified the reason that magnetic calculations are tricky. The magnetic field depends on the magnetization, but the magnetization depends on the magnetic field in a non-linear way (remember that mu depends on local H). Except for a few simple geometries that can be solved exactly, the only way to solve these problems is numerically. Iterative approaches are common; a vacuum field is assumed, magnetization is computed, the field of coil+magnetization calculated, magnetization from that field computed, and so on until convergence is reached.
 
marcusl said:
You have identified the reason that magnetic calculations are tricky. The magnetic field depends on the magnetization, but the magnetization depends on the magnetic field in a non-linear way (remember that mu depends on local H). Except for a few simple geometries that can be solved exactly, the only way to solve these problems is numerically. Iterative approaches are common; a vacuum field is assumed, magnetization is computed, the field of coil+magnetization calculated, magnetization from that field computed, and so on until convergence is reached.

Yeah but its not that tricky about this problem because \mu is constant.
Anyway,I couldn't find the magnetic field inside the solenoid!
So I think I should assume the field is consant inside the solenoid and use the formula B=\frac{\mu N I}{L}
 
If you are only interested in the field inside the solenoid and it is very long, you can certainly do it analytically.
However, in your first post you mentioned the field outside the solenoid. In this case you have to use numerical methods. I am pretty sure the same is true for a short solenoid if it is filled with a magnetic material and you are interested in the field near the edges (the core will "distort" the field).
The analytical approach isn't very viable for most real-world problems.
 
  • #10
End effects at the ends of the iron core make this a very difficult problem that can only be soved numerically.
You can get simple results for B far from the ends of the solenoid.
 
  • #11
MA:: Just to confirm: you mean INSIDE the solenoid, far from the ends?
 
  • #12
Inside or outside. Just far.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K