# Boundary conditions of linear materials

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

Homework Helper
Gold Member
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. :)

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Delta2 and Tony Hau
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? $$\nabla \cdot \vec H = -\nabla \cdot \vec M$$ $$\oint \vec H \cdot d\vec a = -\oint \vec M \cdot d\vec a$$

Homework Helper
Gold Member
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
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.

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Homework Helper
Gold Member
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.

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Tony Hau
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...

Tony Hau
The solution is finally released:

Homework Helper
Gold Member
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
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.

Tony Hau
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?

Homework Helper
Gold Member
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
Homework Helper
Gold Member
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