# Boundary conditions of linear materials

• Tony Hau
In summary: But thank you for your suggestions and advices. I really appreciate it.In summary, the conversation discusses various parts of a problem involving a uniformly magnetized sphere. The solutions provided only cover parts a) to c), and the individual attempts at solving parts d), e), and f) are described. The conversation also includes discussions on the correct equations to use and the correct approach to solving the problem. Advice and suggestions are given to help with solving the problem.
Tony Hau
Homework Statement
Relevant Equations
1) ##\vec M = \chi_m \vec H##
2)##\vec B = \mu \vec H##
Boundary conditions:
1) ##H^{\perp}_{above} - H^{\perp}_{below}= -(M^{\perp}_{above} - M^{\perp}_{below})##
2) ##H^{\parallel}_{above} - H^{\parallel}_{below} = K_{free} \times \hat n##
The question is as follows:

Solutions given only contain part a) to c), which is as follows:

So I now try to attemp d), e) and f).
d)
The magnetic field of a uniformly magnetized sphere is:
$$\vec B =\frac{2}{3}\mu_{o} \vec M = \mu_{o}\vec H$$
$$\frac{2}{3}\vec M = \vec H$$
The perpendicular component of ##\vec H## is: $$H^{\perp}_{s} = \vec H \cdot \hat r = \frac{2}{3} \vec M \cdot \hat r$$

e) ##H^{\parallel}_{above} - H^{\parallel}_{below} = \vec K_{free} \times \hat n =0##,
Therefore, $$H^{\parallel}_{l}(a) = H^{\parallel}_{s}(a)$$

f)
By $$\oint \vec H \cdot d\vec a= - \oint \vec M \cdot d\vec a$$
$$H^{\perp}_{above} - H^{\perp}_{below}= -(M^{\perp}_{above} - M^{\perp}_{below})$$
Given that:
##H^{\perp}_{below} = \vec H_{s}(a) \cdot \hat r##
##M^{\perp}_{above} = \vec M_{above} \cdot \hat r = \chi_{m} \vec H_{l}(a) \cdot \hat r##
##M^{\perp}_{below} = \vec M \cdot \hat r##
Therefore, $$H^{\perp}_{above} - \vec H_{s}(a) \cdot \hat r = - (\chi_{m} \vec H_{l}(a) \cdot \hat r - \vec M \cdot \hat r)$$
$$H^{\perp}_{l} = - (\chi_{m} \vec H_{l}(a) \cdot \hat r - \vec M \cdot \hat r)+ \vec H_{s}(a) \cdot \hat r = - [(\frac{\mu}{\mu_{o}}-1) \vec H_{l}(a) \cdot \hat r - \vec M \cdot \hat r] +\frac{2}{3}\vec M \cdot \hat r$$

I cannot proceed further, or I do not even know if I am correct. Can anyone give me any advice?

Delta2
You don't know what ## B ## is doing inside the uniformly magnetized sphere, because it will be affected by the ## H ## from contributions from the outer layer. To compute the boundary condition of the perpendicular component of ## H##, you just need ## \nabla \cdot H =-\nabla \cdot M ##, (comes from ## \nabla \cdot B=0 ##, and ## B=\mu_o (H+M) ##), which will give you ## H_{1} \cdot \hat{n}_1+H_{2} \cdot \hat{n}_2=-(M_{1} \cdot \hat{n}_1+M_2 \cdot \hat{n}_2) ##. The dot products will give you the perpendicular components. There will also be minus signs, because ## \hat{n}_1=-\hat{n}_2 ##, etc.

@vanhees71 You might find this one of interest as well, and I welcome your inputs=I think I got it right, but please confirm. :)

Last edited:
Delta2 and Tony Hau
You don't know what ## B ## is doing inside the uniformly magnetized sphere, because it will be affected by the ## H ## from contributions from the outer layer. To compute the boundary condition of the perpendicular component of ## H##, you just need ## \nabla \cdot H =-\nabla \cdot M ##, (comes from ## \nabla \cdot B=0 ##, and ## B=\mu_o (H+M) ##), which will give you ## H_{1} \cdot \hat{n}_1+H_{2} \cdot \hat{n}_2=-(M_{1} \cdot \hat{n}_1+M_2 \cdot \hat{n}_2) ##. The dot products will give you the perpendicular components. There will also be minus signs, because ## \hat{n}_1=-\hat{n}_2 ##, etc.

@vanhees71 You might find this one of interest as well, and I welcome your inputs=I think I got it right, but please confirm. :)
But isn't it what I am doing for part f? $$\nabla \cdot \vec H = -\nabla \cdot \vec M$$ $$\oint \vec H \cdot d\vec a = -\oint \vec M \cdot d\vec a$$

Upon reading further down, you did some things very well. When I posted, I only read through part (d), and I don't agree with it. The ## M ## in that region is simply fixed. Meanwhile, where does ## B=\mu_o H ## come from? (If you meant ## B=\mu H ##, that would also be incorrect for this region). Once again, I don't think you can say ## B=(2/3) \mu_o M ##. That would be the case only if there were no outer layer.

Delta2 and Tony Hau
Upon reading further down, you did some things very well. When I posted, I only read through part (d), and I don't agree with it. The ## M ## in that region is simply fixed. Meanwhile, where does ## B=\mu_o H ## come from? (If you meant ## B=\mu H ##, that would also be incorrect for this region). Once again, I don't think you can say ## B=(2/3) \mu_o M ##. That would be the case only if there were no outer layer.
Yes, ##\vec B = \mu \vec H## is what I meant. I use this equation because there is no linear material in this inner magnetized sphere. Hence, we can assume that ##\mu## in ##\vec B = \mu \vec H## can be assumed to be ##\mu_o##. Why is it incorrect for this region?

By the way, I have come up with one solution to calculate the magnetic ##B## ## ##field inside the sphere. We can do this by calculating the magnetic field produced by the surface currents at ##r = a## and ##r=b## respectively, which is calculated in part b. After all, all the fields produced by the linear layer equals to that produced by the surface currents.

Last edited:
When you have a specified ##M ## inside the material, the ## H ## is computed as always from any and all magnetic poles, using the inverse square law, with magnetic pole density ## \rho_m=-\nabla \cdot M ##, (along with the contribution to ## H ## from any currents in conductors using Biot-Savart). The ## H ## will not affect the ##M ## in this case. You still have ## B=\mu_o(H+M) ##.
(Note: You do not have ## M=\chi H ## for this case. You cannot say ## H=M/\chi ##. )

You might find this worthwhile reading.

Last edited:
Tony Hau
When you have a specified ##M ## inside the material, the ## H ## is computed as always from any and all magnetic poles, using the inverse square law, with magnetic pole density ## \rho_m=-\nabla \cdot M ##, (along with the contribution to ## H ## from any currents in conductors using Biot-Savart). The ## H ## will not affect the ##M ## in this case. You still have ## B=\mu_o(H+M) ##.
(Note: You do not have ## M=\chi H ## for this case. You cannot say ## H=M/\chi ##. )

You might find this worthwhile reading.
Alright. But your article is a bit complicated and may take some time...

The solution is finally released:

Do you see how they did part (d)? They had to use ## H_s ##, which can have contributions from the outer layer.

If the outer layer is absent, then ## H_s ## is uniform throughout the entire interior of the sphere including ## H_s(a) ## and is then ## \vec{H}_s(a)=-\vec{M}/3 ##.

With the outer layer in place, they don't ask you to compute ## H_s ##. That would be rather difficult, and they simply want the boundary condition for now.

Tony Hau
However I have come up with another way for d) and f):
For d), $$\nabla \cdot \vec H = - \nabla \cdot \vec M$$, $$H^{\perp}_{l} - H^{\perp}_{s} = -(M^{\perp}_{l} - M^{\perp}_{s})$$
Given that:
##M^{\perp}_{l} = \vec M_{l} \cdot \hat r##

For ##\vec M_{l}##, because ## \vec B_{l} = \mu_o (\vec H_{l} + \vec M_{l})##,
##\mu \vec H_{l} = \mu_o (\vec H_{l} +\vec M_{l})##,
##(\frac{\mu}{\mu_o}-1)\vec H_{l} = \vec M_{l}##,
##M^{\perp}_{l} = (\frac{\mu}{\mu_o}-1)\vec H_{l} \cdot \hat r = \vec M_{l} \cdot \hat r##,

Therefore, $$H^{\perp}_{l} - H^{\perp}_{s} = -((\frac{\mu}{\mu_o}-1)H^{\perp}_{l} - \vec M_{l} \cdot \hat r)$$
$$\mu H^{\perp}_{l} = \mu_o \vec M_{l} \cdot \hat r + \mu_o H^{\perp}_{s}$$

For f),
By $$H^{\perp}_{v} - H^{\perp}_{l} = -(M^{\perp}_{v} - M^{\perp}_{l} )$$,
Because $$M^{\perp}_{v} =0$$,
therefore, $$H^{\perp}_{v}= \frac{\mu}{\mu_o} H^{\perp}_{l}$$

After reading a passage from Electromagnetics by John Kraus, I finally understand when to apply ##\vec B = \mu_o(\vec H + \vec M)## and ##\vec B = \mu \vec H##. The paragraph reads as follows:

##\dots## In isotropic media ##\vec M## and ##\vec H## are in the same direction, so that their quotient is a scalar and hence ##\mu## is a scalar. In nonisotropic media, such as crystals, ##\vec M## and ##\vec H## are, in general, not in the same direction, and ##\mu## is not a scalar. Hence, ##\vec B = \mu_o(\vec H + \vec M)## is a general relation, while ##\vec B = \mu \vec H## is a more consise expression, which, however, has a simple significance only for isotropic media or certain special cases in nonisotropic media.

Do you see how they did part (d)? They had to use ## H_s ##, which can have contributions from the outer layer.

If the outer layer is absent, then ## H_s ## is uniform throughout the entire interior of the sphere including ## H_s(a) ## and is then ## \vec{H}_s(a)=-\vec{M}/3 ##.

With the outer layer in place, they don't ask you to compute ## H_s ##. That would be rather difficult, and they simply want the boundary condition for now.
The outer layer is magnetized and radiates H field, which alters the H field inside the sphere, am I right?

Tony Hau said:
Therefore,
Right after "Therefore", you should have ## M_s ## in that expression, not ## M_l ##. Otherwise, part (d) that you have there looks correct to me. :)

Tony Hau
Tony Hau said:
The outer layer is magnetized and radiates H field, which alters the H field inside the sphere, am I right?
Yes, and it is not uniform and not simply in the z-direction, like the ## H ## field is inside the sphere for the case of no outer layer. With no outer layer, the ## H=-M/3 ## field inside the sphere is much simpler.

Tony Hau

## 1. What are boundary conditions of linear materials?

Boundary conditions of linear materials refer to the constraints or limitations placed on the behavior of a material at its boundaries or interfaces with other materials. They determine how a material will respond to external forces or stimuli.

## 2. How do boundary conditions affect the behavior of linear materials?

Boundary conditions can significantly influence the mechanical, thermal, and electrical properties of linear materials. They can affect the material's stiffness, strength, and ability to conduct heat or electricity.

## 3. What are some common types of boundary conditions for linear materials?

Some common boundary conditions for linear materials include fixed or clamped boundaries, free or unconstrained boundaries, and various types of applied loads or displacements at the boundaries.

## 4. How do boundary conditions differ for different types of linear materials?

Boundary conditions can vary depending on the type of linear material. For example, the boundary conditions for a metal may differ from those for a polymer due to differences in their mechanical and thermal properties.

## 5. Why are boundary conditions important to consider in material testing and analysis?

Boundary conditions are crucial to accurately predicting and understanding the behavior of linear materials. They can affect the validity of experimental results and the accuracy of computational models, making them essential to consider in material testing and analysis.

Replies
1
Views
478
Replies
6
Views
451
• Introductory Physics Homework Help
Replies
3
Views
892
Replies
2
Views
774
Replies
3
Views
1K
Replies
11
Views
1K
Replies
1
Views
1K
Replies
2
Views
1K