Recent questions in Ferromagnetism

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xafeguerpf3l 2023-02-16

How to find the local extrema for $f\left(x\right)=5x-{x}^{2}$?

ElectromagnetismAnswered question

mierekaslq 2023-01-14

Why can only electrons move and get transferred? Why not protons?

ElectromagnetismAnswered question

bruinhemd3ji 2022-11-06

The exchange coupling as being responsible for ferromagnetism is not the mutual magnetic interaction between two elementary magnetic dipoles. To show this, calculate (a) the magnitude of the magnetic field a distance of 13 nm away, along a dipole axis, from an atom with magnetic dipole moment $1.2\times {10}^{-23}\text{}J/T$, and (b) the minimum energy required to turn a second identical dipole end for end in this field.

a) Number ()Units

b) Number ()Units

a) Number ()Units

b) Number ()Units

ElectromagnetismAnswered question

Alessandra Cummings 2022-10-16

6. What are the differences between magnetism in 3d and 4f electrons of periodic tables? 7. exchange interaction is a quantum mechanical effect that only occurs between identical particles mention the type of materials that can be effected by and write the exchange interaction term in energy Hamiltonian? 8. What is Stoner Criterion? Write the formula and describe its effect in ferromagnetism?

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gemauert79 2022-09-27

Describe the different types of magnetism that materials can display and account for their origins.

ElectromagnetismAnswered question

Camila Brandt 2022-09-25

What are the three principle sources of magnetic moment of a free atom? Which of them is responsible for dia, para, and ferromagnetism ?

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rialsv 2022-09-06

A few elements are naturally magnetic at room temperature. Select all that are True.

Copper is ferromagnetic

Zinc is ferromagnetic

Nickel is ferromagnetic

Iron is ferromagnetic

Cobalt is ferromagnetic

Aluminum is ferromagnetic

Copper is ferromagnetic

Zinc is ferromagnetic

Nickel is ferromagnetic

Iron is ferromagnetic

Cobalt is ferromagnetic

Aluminum is ferromagnetic

ElectromagnetismAnswered question

Janessa Benson 2022-09-06

Why is it important to know about the categories of magnatization

ElectromagnetismAnswered question

Mbalisikerc 2022-07-21

Which of the following group of elements are diamagnetic ?

A) Argon, copper, silver

B) Hydrogen, argon, copper

C) Oxygen, copper, silver

D) Hydrogen, oxygen, Argon

A) Argon, copper, silver

B) Hydrogen, argon, copper

C) Oxygen, copper, silver

D) Hydrogen, oxygen, Argon

ElectromagnetismAnswered question

asigurato7 2022-07-18

(i) What is the cause of ferromagnetism in some materials?

(ii)What is meant by a non-stoichiometric oxide?

(iii)Explain the difference between strong field and weak field.

(ii)What is meant by a non-stoichiometric oxide?

(iii)Explain the difference between strong field and weak field.

ElectromagnetismAnswered question

lnwlf1728112xo85f 2022-05-20

I have a confusion regarding how magnetic field propagate in a medium (ferromagnets or air etc..). It is knows when a ferromagnet is exposed to an external magnetic field, the magnetic domain of the ferromagnet will be aligned with the external magnetic field. My question is this alignment of domain assure the propagation of the external magnetic field in the ferromagnet? The same question for the air. Do the magnetic field propagate in the air because the magnetic domains of the air are aligned?

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Jaylene Duarte 2022-05-19

Why are there more diamagnetic substances in nature than there are paramagnetic or ferromagnetic substances?

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datomerki8a5yj 2022-05-18

I am trying to get the relation between developing a Ginzburg-Landau theory, let's say for a ferromagnet with magnetization field $\overrightarrow{m}=\overrightarrow{m}(\overrightarrow{r})$, and the formal expansion of the free energy density $\mathcal{F}=\mathcal{F}(\overrightarrow{m})$ in terms of a Taylor series.

Considering an isotropic ferromagnet, the lowest-order terms in our Ginzburg-Landau theory should be given by

$\mathcal{F}=\frac{r}{2}{\overrightarrow{m}}^{2}+\frac{U}{4}{\left({\overrightarrow{m}}^{2}\right)}^{2}+\frac{J}{2}[{\left({\mathrm{\partial}}_{x}\overrightarrow{m}\right)}^{2}+{\left({\mathrm{\partial}}_{y}\overrightarrow{m}\right)}^{2}+{\left({\mathrm{\partial}}_{z}\overrightarrow{m}\right)}^{2}]$

with r<0 and U,J>0.

However, when I think of a Taylor expansion of $\mathcal{F}(\overrightarrow{m})$ around the origin

$\mathcal{F}={\mathcal{F}}_{0}+{\overrightarrow{m}}^{T}\cdot D\mathcal{F}+\frac{1}{2}{\overrightarrow{m}}^{T}\cdot {D}^{2}\mathcal{F}\cdot \overrightarrow{m}+\dots $

this is giving me terms of all the individual powers in $\overrightarrow{m}$ , which are either zero or identified with the r- and U-term, but no gradient terms for the J-term? How to motivate these through a Taylor expansion?

Considering an isotropic ferromagnet, the lowest-order terms in our Ginzburg-Landau theory should be given by

$\mathcal{F}=\frac{r}{2}{\overrightarrow{m}}^{2}+\frac{U}{4}{\left({\overrightarrow{m}}^{2}\right)}^{2}+\frac{J}{2}[{\left({\mathrm{\partial}}_{x}\overrightarrow{m}\right)}^{2}+{\left({\mathrm{\partial}}_{y}\overrightarrow{m}\right)}^{2}+{\left({\mathrm{\partial}}_{z}\overrightarrow{m}\right)}^{2}]$

with r<0 and U,J>0.

However, when I think of a Taylor expansion of $\mathcal{F}(\overrightarrow{m})$ around the origin

$\mathcal{F}={\mathcal{F}}_{0}+{\overrightarrow{m}}^{T}\cdot D\mathcal{F}+\frac{1}{2}{\overrightarrow{m}}^{T}\cdot {D}^{2}\mathcal{F}\cdot \overrightarrow{m}+\dots $

this is giving me terms of all the individual powers in $\overrightarrow{m}$ , which are either zero or identified with the r- and U-term, but no gradient terms for the J-term? How to motivate these through a Taylor expansion?

ElectromagnetismAnswered question

Amappyaccon22j7e 2022-05-18

Let us suppose that there is a block of a ferromagnetic material inside a very long ideal solenoid, the magnetic field lines inside are in a straight line perpendicular to the surface of the block at both the ends.Why does the magnetic field inside the block increases? It is known form the boundary conditions of magnetic fields that the perpendicular or the normal component of the magnetic field $\overrightarrow{B}$ doesn't change when it goes from one medium to other.So, what's happening here?

ElectromagnetismAnswered question

agrejas0hxpx 2022-05-18

How to derive ferromagnet Lagrangian from the Heisenberg model: $H=-{J}_{ij}{S}_{i}{S}_{j}$ ? I understand how to obtain potential energy term, but it is not clear how to get "kinetic energy".

$L=\int {d}^{2}x\frac{\rho}{2}[\frac{({\mathrm{\partial}}_{t}\theta {)}^{2}}{{v}^{2}}-(\mathrm{\nabla}\theta {)}^{2}]$

$L=\int {d}^{2}x\frac{\rho}{2}[\frac{({\mathrm{\partial}}_{t}\theta {)}^{2}}{{v}^{2}}-(\mathrm{\nabla}\theta {)}^{2}]$

ElectromagnetismAnswered question

Amappyaccon22j7e 2022-05-17

If an electric field $\mathbf{E}$ exists at a point in free space then the energy density in that point is $\frac{1}{2}{\u03f5}_{0}|\mathbf{E}{|}^{2}$

When a field with same magnitude $|\mathbf{E}|$ exists in a material with relative permettivity k, the energy density becomes $\frac{1}{2}{\u03f5}_{0}k|\mathbf{E}{|}^{2}$, i.e. it increases k times.

A similar analysis for energy density in magnetic field $\frac{1}{2{\mu}_{0}}|\mathbf{B}{|}^{2}$ states that: if the field $\mathbf{B}$ exists in a material with relative permeability k (consider a ferromagnetic material, for example) then $\mu =k{\mu}_{0}$ and thus the energy density decreases k times (?). But this does not seem right because inductors with ferromagnetic cores store more energy.

When a field with same magnitude $|\mathbf{E}|$ exists in a material with relative permettivity k, the energy density becomes $\frac{1}{2}{\u03f5}_{0}k|\mathbf{E}{|}^{2}$, i.e. it increases k times.

A similar analysis for energy density in magnetic field $\frac{1}{2{\mu}_{0}}|\mathbf{B}{|}^{2}$ states that: if the field $\mathbf{B}$ exists in a material with relative permeability k (consider a ferromagnetic material, for example) then $\mu =k{\mu}_{0}$ and thus the energy density decreases k times (?). But this does not seem right because inductors with ferromagnetic cores store more energy.

ElectromagnetismAnswered question

William Santiago 2022-05-17

I was wondering what is the difference between a spin-fluid Heisenberg Magnet and spin-glass Heisenberg Magnet. As far as I understand in a spin-glass the spins are randomly oriented compared to a ferromagnet material but I don't quite understand what we mean by spin-fluid phase. And also in the article I am reading its mentioned that spin-fluid phase is found to be generically "gapless", but I am not sure what we mean by phase being gapless and how is it related to the spin phase of the material?

ElectromagnetismAnswered question

hetriamhageh6k20 2022-05-17

Consider one-dimensional ferromagnet namely N spin-1/2 objects placed around a circle with the Hamiltonian

$\mathcal{H}=-\mathcal{J}\sum _{n=1}^{N}{\overrightarrow{\mathcal{S}}}_{n}\cdot {\overrightarrow{\mathcal{S}}}_{n+1}$

where we assume the periodic boundary condition ${\overrightarrow{\mathcal{S}}}_{N+1}\equiv {\overrightarrow{\mathcal{S}}}_{1}$ and $\mathcal{J}>0$

I'm trying to show that total spin ket is a good quantum number that is they commute with $\mathcal{H}$ and finding out the energy corresponding to

$|{\psi}_{0}\u27e9=|\uparrow {\u27e9}_{1}\otimes |\uparrow {\u27e9}_{2}\otimes \cdots \otimes |\uparrow {\u27e9}_{N}$

By definition:

${\mathcal{S}}^{2}={\left(\sum _{n}{\overrightarrow{\mathcal{S}}}_{n}\right)}^{2}=\sum _{n}{\mathcal{S}}_{n}^{2}+\sum _{i,j}{\overrightarrow{\mathcal{S}}}_{i}\cdot {\overrightarrow{\mathcal{S}}}_{j}$

The second term has our Hamiltonian but there are other terms also. I don't understand, How do I proceed from here?

$\mathcal{H}=-\mathcal{J}\sum _{n=1}^{N}{\overrightarrow{\mathcal{S}}}_{n}\cdot {\overrightarrow{\mathcal{S}}}_{n+1}$

where we assume the periodic boundary condition ${\overrightarrow{\mathcal{S}}}_{N+1}\equiv {\overrightarrow{\mathcal{S}}}_{1}$ and $\mathcal{J}>0$

I'm trying to show that total spin ket is a good quantum number that is they commute with $\mathcal{H}$ and finding out the energy corresponding to

$|{\psi}_{0}\u27e9=|\uparrow {\u27e9}_{1}\otimes |\uparrow {\u27e9}_{2}\otimes \cdots \otimes |\uparrow {\u27e9}_{N}$

By definition:

${\mathcal{S}}^{2}={\left(\sum _{n}{\overrightarrow{\mathcal{S}}}_{n}\right)}^{2}=\sum _{n}{\mathcal{S}}_{n}^{2}+\sum _{i,j}{\overrightarrow{\mathcal{S}}}_{i}\cdot {\overrightarrow{\mathcal{S}}}_{j}$

The second term has our Hamiltonian but there are other terms also. I don't understand, How do I proceed from here?

ElectromagnetismAnswered question

Alisa Durham 2022-05-15

What are the specific electronic properties that make an atom ferromagnetic versus simply paramagnetic?

ElectromagnetismAnswered question

Jayden Mckay 2022-05-14

Given the quantum Heisenberg model with Hamiltonian

$\hat{H}=-\frac{1}{2}\sum _{i,j}{J}_{ij}{\hat{\mathbf{\text{S}}}}_{i}\cdot {\hat{\mathbf{\text{S}}}}_{j}$,

the uniform mean-field approximation ${\hat{\mathbf{\text{S}}}}_{i}=\u27e8\hat{\mathbf{\text{S}}}\u27e9+({\hat{\mathbf{\text{S}}}}_{i}-\u27e8\hat{\mathbf{\text{S}}}\u27e9)$ allows to rewrite it as

${\hat{H}}_{MF}=\sum _{i}{\mathbf{\text{B}}}_{eff}\cdot {\hat{\mathbf{\text{S}}}}_{i}+\text{const.}$,

in order to perform a diagonalization by means of a Fourier transform. To do so, I am told to choose the z-axis so to align with the effective field ${\mathbf{\text{B}}}_{eff}$, which I find to be ${\mathbf{\text{B}}}_{eff}=-\u27e8\hat{\mathbf{\text{S}}}\u27e9\sum _{j}({J}_{ij}+{J}_{ji})$. That's where I remain stuck. First of all, how should the Fourier transform which allows me to diagonalize ${\hat{H}}_{MF}$ look like? And how is this related to the alignment of the effective field?

$\hat{H}=-\frac{1}{2}\sum _{i,j}{J}_{ij}{\hat{\mathbf{\text{S}}}}_{i}\cdot {\hat{\mathbf{\text{S}}}}_{j}$,

the uniform mean-field approximation ${\hat{\mathbf{\text{S}}}}_{i}=\u27e8\hat{\mathbf{\text{S}}}\u27e9+({\hat{\mathbf{\text{S}}}}_{i}-\u27e8\hat{\mathbf{\text{S}}}\u27e9)$ allows to rewrite it as

${\hat{H}}_{MF}=\sum _{i}{\mathbf{\text{B}}}_{eff}\cdot {\hat{\mathbf{\text{S}}}}_{i}+\text{const.}$,

in order to perform a diagonalization by means of a Fourier transform. To do so, I am told to choose the z-axis so to align with the effective field ${\mathbf{\text{B}}}_{eff}$, which I find to be ${\mathbf{\text{B}}}_{eff}=-\u27e8\hat{\mathbf{\text{S}}}\u27e9\sum _{j}({J}_{ij}+{J}_{ji})$. That's where I remain stuck. First of all, how should the Fourier transform which allows me to diagonalize ${\hat{H}}_{MF}$ look like? And how is this related to the alignment of the effective field?

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If you’re looking for a simple ferromagnetism definition, it can be summed up as the magnetism related to iron, cobalt, nickel, and other related compounds where these elements can be met. The examples for this field include questions on ferromagnetism Class 12 and permanent magnetization. You can take a look at the list of questions and answers on the topic that will help you get ferromagnetism explained. Since this subject will include several samples for every task, we have collected the most efficient solutions to relevant challenges and lab experiments that help to see the methods that are most applicable.