Magnetization of a paramagnetic sphere

In summary, the magnetization of a paramagnetic sphere refers to the alignment of magnetic dipoles within the material when subjected to an external magnetic field. Unlike ferromagnetic materials, paramagnetic substances exhibit weak and temporary magnetization, which disappears once the external field is removed. The degree of magnetization is influenced by factors such as temperature and the strength of the applied magnetic field, typically described by the Curie law. The behavior of paramagnetic spheres can be analyzed through models that consider their magnetic susceptibility, leading to applications in various fields, including materials science and magnetic resonance imaging.
  • #1
AmanWithoutAscarf
22
1
Homework Statement
I'm trying to calculate the magnetic field inside and outside a uniformly distributed paramagnetic sphere. The sphere has a permeability of μ (mu) and is placed in a constant external magnetic field.
Relevant Equations
Equations related to electric and magnetic polarization
I believe there is an elementary way to solve this problem using some analogies with relevant models.

First, I consider an electric model of polarization in uniform field.

Here, there is a dielectric sphere oriented in uniform electric field ##E_0##. We can find out the electric fields inside and outside by modeling the charge distribution within the sphere as a superposition of two oppositely charged spheres. This allows us to determine the constant internal field and the external field with ##p##. Finally, using electric boundary conditions on the spherical surface, we can solve for ##p## and get the sollution.

1717417598521.png


Illustration of model​
Let's take a look at the phenomenon of conducting shell in uniform magnetic field.
1717487304414.png

The shell expells magnetic fields, so ##B_{in}=0##. Assuming the total external field is the superposition of ##B_0## and the additional field generated by a magnetic dipole ##m## (I know it is the result of some complex calculations with vector potential, but the assumption without mathematical proofs is still acceptable), we can use the boundary conditions again to find ##m## and the total external magnetic field after that.

So, in case of the initial question, I'm thinking of a model of a constant internal magnetic field (inspired by the model of electric polarization) and a superposition of the external field with a magnetic dipole (as the second model).

Is my assumption correct? And how can I get further calculation for magnetic field from that?
1717489556522.png
 
Last edited:
Physics news on Phys.org
  • #2
Where did you get the idea that the field expels the magnetic field? That’s not what your picture shows or what’s going on in the paramagnetic material. Magnetic fied lines form continuous loopds. They don't stop at magnetic charges like electric field lines.

If you wish to draw an analogy between this and the electric field case, consider the magnetic scalar potential and treat it as a boundary value problem.
 
Last edited:
  • #3
kuruman said:
Where did you get the idea that the field expels the magnetic field? That’s not what your picture shows or what’s going on in the paramagnetic material. Magnetic fied lines form continuous loopds. They don't stop at magnetic charges like electric field lines.

If you wish to draw an analogy between this and the electric field case, consider the magnetic scalar potential and treat it as a boundary value problem.
I apologize for any misunderstanding. The second picture is actually about a conducting shell (like a hollow ball) to show a possibility of modeling the external magnetic field as if it was caused by a magnetic dipole moment inside. The third one is the paramagnetic sphere we were talking about.

And rather than using advanced math (that I am not yet familiar), I would like to solve this problem in a simpler way using relevant models.
 
  • #4
AmanWithoutAscarf said:
Homework Statement: I'm trying to calculate the magnetic field inside and outside a uniformly distributed paramagnetic sphere. The sphere has a permeability of μ (mu) and is placed in a constant external magnetic field.
Relevant Equations: Equations related to electric and magnetic polarization

Here, there is a dielectric sphere oriented in uniform electric field ##E_0.Wecanfindouttheelectricfieldsinsideandoutsidebymodelingthechargedistributionwithinthesphereasasuperpositionoftwooppositelychargedspheres.Thisallowsustodeterminetheconstantinternalfieldandtheexternalfieldwithp.Finally,usingelectricboundaryconditionsonthesphericalsurface,wecansolveforp## and get the sollution.
Some special characters were lost during transmission, causing these such chaotic sentences. I'm sorry for this mistake.
This is the rewritten version:

Here, there is a dielectric sphere oriented in uniform electric field ##E_0##. We can find out the electric fields inside and outside by modeling the charge distribution within the sphere as a superposition of two oppositely charged spheres. This allows us to determine the constant internal field and the external field with p. Finally, using electric boundary conditions on the spherical surface, we can solve for ##p## and get the sollution.
 
  • #5
AmanWithoutAscarf said:
Some special characters were lost during transmission, causing these such chaotic sentences. I'm sorry for this mistake.
This is the rewritten version:

Here, there is a dielectric sphere oriented in uniform electric field ##E_0##. We can find out the electric fields inside and outside by modeling the charge distribution within the sphere as a superposition of two oppositely charged spheres. This allows us to determine the constant internal field and the external field with p. Finally, using electric boundary conditions on the spherical surface, we can solve for ##p## and get the sollution.
I think you are confusing your examples.

The offset positively and negatively conducting shells argument is used in the case of a conducting sphere or shell in a uniform electric field where charges are free to move on the surface and the conductor is an equipotential. See for example



I don't think that you can carry this idea to a a dielectric sphere where there are no free charges. I think you should buckle down and learn some intermediate (not advanced) math like Legendre polynomials and how they can be used with this kind of problem. The electric potential serves as an analogue of the magnetic scalar potential that I mentioned earlier. Once you have solved the former, you can carry the solutions to the latter.
 
  • Like
Likes AmanWithoutAscarf
  • #6
AmanWithoutAscarf said:
So, in case of the initial question, I'm thinking of a model of a constant internal magnetic field (inspired by the model of electric polarization) and a superposition of the external field with a magnetic dipole (as the second model).

Is my assumption correct? And how can I get further calculation for magnetic field from that?
View attachment 346423
Are you making an "ansatz" that the solution inside the sphere will be a uniform field ##B_i## and the solution outside will be the superposition of the external field ##B_0## and the field of a magnetic dipole ##m##? Then, with this assumption, do you want to know how to determine the values of ##B_i## and ##m##?

If so, that seems like a decent exercise. Are you familiar with the boundary conditions that ##\mathbf B## must satisfy at the surface of the sphere?
 
  • Like
Likes AmanWithoutAscarf
  • #7
TSny said:
Are you making an "ansatz" that the solution inside the sphere will be a uniform field ##B_i## and the solution outside will be the superposition of the external field ##B_0## and the field of a magnetic dipole ##m##?
It is not clear what the ansatz is.
AmanWithoutAscarf said:
The shell expells magnetic fields, so ##B_{in}=0##.
The third figure in post #1 shows a field outside but only a dipole inside with no field. It looks like OP thinks that the magnetic field is zero inside in sphere and dipolar outside in a manner analogous to the conducting sphere in a uniform electric field.
 
  • #8
kuruman said:
It is not clear what the ansatz is.

The third figure in post #1 shows a field outside but only a dipole inside with no field. It looks like OP thinks that the magnetic field is zero inside in sphere and dipolar outside in a manner analogous to the conducting sphere in a uniform electric field.
Ok. Thanks. Hopefully the OP will clarify.
 
  • #9
kuruman said:
It is not clear what the ansatz is.

The third figure in post #1 shows a field outside but only a dipole inside with no field. It looks like OP thinks that the magnetic field is zero inside in sphere and dipolar outside in a manner analogous to the conducting sphere in a uniform electric field.
I'm sorry for the confusion. I assume that the internal magnetic field is uniform, similar to the model of a dielectric sphere in a uniform electric field. This is because the magnetic permeability (##\mu##) of a paramagnetic sphere and the electric permittivity (##\varepsilon##) of a dielectric sphere seem to have similar roles in their polarization effects
1717560316863.png

kuruman said:
I don't think that you can carry this idea to a a dielectric sphere where there are no free charges. I think you should buckle down and learn some intermediate (not advanced) math like Legendre polynomials and how they can be used with this kind of problem. The electric potential serves as an analogue of the magnetic scalar potential that I mentioned earlier. Once you have solved the former, you can carry the solutions to the latter.
I think it is still adaptable to apply the electric dipole model to the case of a dielectric sphere.
It is easy to point out the internal field is uniform:
1717561684921.png

(Here ##E_e## is the extra field generated by the polarization)
We can describe the total internal field as: $$\mathbf{E}_{in} =\mathbf{E}_{0} +\mathbf{E}_{e} =\mathbf{E}_{0} +\frac{-\mathbf{p}}{4\pi \varepsilon _{0} R^{3}}$$
and the external field:
$$\mathbf{E}_{exr} =\mathbf{E}_{0} +\frac{1}{4\pi \varepsilon _{0} R^{3}}( 3(\mathbf{p} .\hat{r}) .\hat{r} -\mathbf{p})$$
Using boundary conditions, we can solve for ##p##:
$$\mathbf{p} =4\pi \varepsilon _{0} R^{3}\frac{\varepsilon -1}{\varepsilon +2} .\mathbf{E}_{0}$$
TSny said:
Are you making an "ansatz" that the solution inside the sphere will be a uniform field Bi and the solution outside will be the superposition of the external field B0 and the field of a magnetic dipole m? Then, with this assumption, do you want to know how to determine the values of Bi and m?
Yes, I do assume the internal and external magnetic field as you described. However, I cannot think of a "superposition model" to support this internal uniform magnetic field. In the case of the dielectric sphere, the uniform electric field is generated by the superposition of two oppositely charged spheres, but a similar model for the paramagnetic sphere isn't coming to mind.

These illustrations are hand-drawn, so there may be some errors. Please point out, and I will correct them.
 
  • #10
AmanWithoutAscarf said:
but a similar model for the paramagnetic sphere isn't coming to mind.
That's because, strictly speaking, positive and negative magnetic "charge" distributions are not known to exist. The way to get around this is by using the "advanced math" that you are trying to avoid. In electrostatics you can prove the Uniqueness Theorem which basically says that if you can find a solution to Laplace's equation that satisfies the boundary conditions, it is the solution for the electrostatic potential to within a constant.

This seemingly naive statement is very powerful because it allows you to solve Laplace's equation in it most general form and then, for particular problem, filter out the solutions that do not satisfy the boundary conditions. What's left is, by the Uniqueness Theorem, your solution to within a constant. This constant, by the way, is the zero of potential which is usually taken at infinity.

You see then that if someone tells you what the filtered out part of the general solution is, all you have to do is make sure that the boundary conditions are satisfied. That's exactly what you did here without knowing it. You crafted the dielectric-sphere-in-a-uniform-field solution, by making an assumption (where did it come from?) about the two offset charge distributions and matched the boundary conditions.

Carrying this over to the paramagnetic sphere case is trivial when you use vector calculus. I

In electrostatics, when free charges are absent, the relevant Maxwell's equations are
$$\mathbf{\nabla}\cdot \mathbf{D}=0~;~~\mathbf{\nabla}\times \mathbf E=0$$ where ##\mathbf D## is the electric displacement field. This leads to Laplace's equation $$\nabla ^2\varphi=0$$ where ##\varphi## is the electrostatic potential, the negative gradient of which is the electric field ##\mathbf E.##

In magnetostatics, when transport currents are absent, Maxwell's equations are
$$\mathbf{\nabla}\cdot \mathbf{B}=0~;~~\mathbf{\nabla}\times \mathbf H=0$$where ##\mathbf H## is the magnetic intensity field. This leads to Laplace's equation $$\nabla ^2\varphi^*=0$$ where ##\varphi^*## is the magnetic scalar potential, the negative gradient of which times ##\mu_0## is the magnetic induction field ##\mathbf B.##

You can see now that the magnetostatic case is mathematically identical to the electrostatic except for the symbols representing the fields. Furthermore, the proof of the Uniqueness Theorem in electrostatics is valid in magnetostatics as well. So once you have the electrostatic solution, you can change the symbols and write down the solution to the corresponding magnetostatic problem.

Doesn't all this make you eager to learn some vector calculus and expand your understanding?
 
  • Informative
Likes AmanWithoutAscarf
  • #11
kuruman said:
Doesn't all this make you eager to learn some vector calculus and expand your understanding?
Well, honestly, these Nabla symbols always seem scary to me, as a high school student. But if this is the only solution, I will try to learn it. Thank you so much for your sharing!
 
  • #12
AmanWithoutAscarf said:
Well, honestly, these Nabla symbols always seem scary to me, as a high school student. But if this is the only solution, I will try to learn it. Thank you so much for your sharing!
Yes, they seemed scary to me too, but once I took a formal course in vector calculus, they became my friend. Now high school may be a bit early for these, however if you wish to tackle intermediate-level problems like this one, you have to use intermediate-level maths.
 
  • Love
Likes AmanWithoutAscarf
Back
Top