Mega Note: Magnetic Properties of Materials - Part 1

1. Magnetic Dipole Moment (${\mu }_m$)

Slide Concept: Consider a closed current loop with area $A$ carrying a circulating current $I$. If ${\mathbf{u}}_n$ is a unit vector coming out of the area $A$ (following the right-hand rule for the current direction), the magnetic dipole moment (${\mu }_m$) is defined as: ${\mu }_m=IA{\mathbf{u}}_n$.

Physical Significance:

• A current loop behaves just like a bar magnet, producing a magnetic field $\mathbf{B}$ around it.
• When a magnetic moment is placed in an applied external magnetic field, it experiences a rotational force (Torque, $\tau$) that attempts to align the magnetic moment’s axis with the applied magnetic field.

Relevant Past Year Questions (PYQs):

• Explain the magnetic dipole moment with relevant figure. “Magnetic moment is proportional to the orbital angular momentum through a factor that has the change to mass ratio of the electron”- justify the statement. (2016)
• Define magnetic dipole moment. Explain the procession of spin magnetic moment about external magnetic field. (2018)
• Explain the following terms: i) Magnetic dipole moment, ii) Bohr magneton… (2017)

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2. Atomic Magnetic Moment

The magnetic moment of an atom arises from its electrons, which possess both orbital and spin motions.

A. Orbital Magnetic Moment

An electron orbiting a nucleus acts as a tiny current loop. Hand-on Practice 3.a: Derive the expression for orbital magnetic moment

1. Current $I$ is charge over time: $I=\frac{-e}{T}=-ef=-e\frac{\omega }{2\pi }$ (negative sign indicates the negative charge of an electron).
2. The angular momentum $L$ of the electron is: $L=m_{e}vr=m_e\omega r^2⟹\omega r^2=\frac{L}{m_e}$.
3. The orbital magnetic moment ${\mu }_{orb}=IA{\mathbf{u}}_n=\left(-e\frac{\omega }{2\pi }\right)(\pi r^2){\mathbf{u}}_n=-\frac{1}{2}e\omega r^2{\mathbf{u}}_n$.
4. Substituting $\omega r^2$: ${\mu }_{orb}=-\frac{1}{2}\frac{e}{m_e}L{\mathbf{u}}_n$.

Assignment 3.1: State the physical significance of gyromagnetic ratio The gyromagnetic ratio ($\gamma =\frac{e}{2m_e}$) is the ratio of the magnetic moment to the orbital angular momentum. Its physical significance is that it directly links the mechanical property of the electron (angular momentum) to its electromagnetic property (magnetic moment).

B. Spin Magnetic Moment

Electrons also have an intrinsic angular momentum (Spin, $S$). The spin magnetic moment is given by: ${\mu }_{spin}=-\frac{e}{m_e}S$.

Hand-on Practice 3.b: Derive the expression for Bohr magneton

1. The spin magnetic moment along the z-axis is ${\mu }_z=-\frac{e}{m_e}{S}_z$.
2. In quantum mechanics, $S_z$ is quantized as $S_z=m_s\hslash$, where $m_s$ for an electron is $-\frac{1}{2}$.
3. ${\mu }_z=-\frac{e}{m_e}(-\frac{1}{2}\hslash )=\frac{e\hslash }{2m_e}=\beta$.
4. Therefore, $\beta =9.274\times 10^{-24}A{m}^2/J{T}^{-1}$.

Assignment 3.3: State the physical significance of Bohr magneton The Bohr magneton ($\beta$) serves as the fundamental quantum unit of magnetic moment at the atomic scale. It represents the natural magnetic moment created by the intrinsic spin of a single electron.

Assignment 3.2: “Only unfilled subshells contribute to the overall magnetic moment of an atom.” – Explain According to the Pauli exclusion principle, an atomic orbital can hold a maximum of two electrons with opposing spins (spin up and spin down). In a completely filled subshell, every positive spin or orbital quantum number is perfectly canceled by a corresponding negative value. Consequently, the net angular momentum and net magnetic moment of a closed (full) subshell are zero. Only atoms with unfilled subshells have unpaired electrons, which produce a net permanent magnetic moment.

Relevant PYQs:

• Write down amperes law. Show that magnetic moment is proportional to the orbital angular momentum through a factor know as gyromagnetic ratio. (2023, 2019)
• Briefly explain the following terms: … ii. Bohr magneton (2018)

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3. Magnetization Vector (M)

Slide Concept: The Magnetization Vector $\mathbf{M}$ describes the extent of magnetization within a medium. It is defined as the total magnetic dipole moment per unit volume. $M=\frac{1}{\Delta V}\sum {\mu }_{mi}=N{\mu }_{av}$ (Where $N$ is the number of atoms per unit volume, and ${\mu }_{av}$ is the average magnetic moment per atom).

Surface Magnetization Current: Each magnetic moment aligned in the material acts as an elementary current loop. In the bulk of the material, internal adjacent current loops cancel each other out. However, at the surface, these un-canceled loops combine to form a macroscopic circulating surface magnetization current ($I_m$ per unit length). Total magnetic moment of a specimen = $M\times Volume=M(A\ell )$. Alternatively, total magnetic moment = $(TotalSurfaceCurrent)\times Area=(I_m\ell )A$. Equating them: $M=I_m$. Thus, the magnetization vector magnitude equals the surface magnetization current per unit length.

Relevant PYQs:

• Show that the magnetic dipole moment per unit volume is the same as the magnetization current on the surfaces per unit length of the specimen. (2015, 2016, 2017, 2018, 2020, 2022)
• What is magnetization vector? Explain how spin magnetic moment precess about the magnetic field. (2019)

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4. Magnetizing Field (H) & Magnetic Field Intensity

Slide Concept: When a material is placed in an applied external field (like inside a solenoid), the total magnetic field density $\mathbf{B}$ inside the material has two contributions: the external current and the material’s internal magnetization.

Hand-on Practice 3.c: Derive the expression for Magnetic Field Intensity (Magnetizing field)

1. The applied field from the solenoid’s free current is $B_0={\mu }_{0}nI={\mu }_0{I}^{\prime }$ (where $I^{\prime }$ is the total free current per unit length).
2. The total magnetic field inside the solenoid with the material is $B=B_0+{\mu }_0{I}_m$.
3. Since $I_m=M$ (magnetization), we write $B=B_0+{\mu }_{0}M$.
4. Rearranging for $B_0$: $B_0=B-{\mu }_{0}M⟹\frac{B_0}{{\mu }_0}=\frac{B}{{\mu }_0}-M$.
5. We define the Magnetic Field Intensity $\mathbf{H}=\frac{B_0}{{\mu }_0}$. Therefore: $H=\frac{1}{{\mu }_0}B-M$. The unit of H is $A{m}^{-1}$.

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5. Magnetic Permeability & Susceptibility

• Magnetic Permeability ($\mu$): Defined as the total magnetic field per unit magnetizing field: $\mu =\frac{B}{H}$.
• Relative Permeability (${\mu }_r$): The fractional increase in the magnetic field inside the material compared to free space: ${\mu }_r=\frac{\mu }{{\mu }_0}$.
• Magnetic Susceptibility (${\chi }_m$): Relates how responsive a material’s magnetization is to the applied field: $M={\chi }_{m}H$. Relationship: $B={\mu }_0(H+M)={\mu }_0(H+{\chi }_{m}H)={\mu }_{0}H(1+{\chi }_m)$. Thus, ${\mu }_r=1+{\chi }_m$.

Relevant PYQs:

• Explain the terms susceptibility and permeability in magnetism. (2015)
• Discuss magnetic permeability and magnetic susceptibility along with their relation. (2018)
• Numerical: A magnetizing field of $500A{m}^{-1}$ produces a magnetic flux of $2.4\times 10^{-5}$ Weber in an iron bar of $0.2c{m}^2$ cross-sectional area. Compute the permeability and susceptibility of the bar. (2018, 2020) (Solution framework: $B=Flux/Area=2.4\times 10^{-5}/0.2\times 10^{-4}=1.2T$. Permeability $\mu =B/H=1.2/500=2.4\times 10^{-3}H/m$. Relative permeability ${\mu }_r=\mu /{\mu }_0$. Susceptibility ${\chi }_m={\mu }_r-1$.)

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6. Magnetostatic Energy Per Unit Volume

Hand-on Practice 3.f: Find the electric energy $\delta E$ input into the coil in time $\delta t$ and calculate the energy density.

1. From Faraday’s and Ampere’s laws, the electric energy input into a coil over time $\delta t$ is $\delta E=iv\delta t$.
2. $v=NA\frac{\delta B}{\delta t}$ and $i=\frac{H\ell }{N}$, so $\delta E=\left(\frac{H\ell }{N}\right)(NA)\left(\frac{\delta B}{\delta t}\right)\delta t=(A\ell )H\delta B$.
3. Since $A\ell$ is the volume of the toroid, the energy per unit volume is $E_{vol}={\int }_{B_1}^{B_2}HdB$.
4. Substituting $H=\frac{B}{\mu }$ and assuming $\mu$ is constant ($\mu ={\mu }_0{\mu }_r$): $E_{vol}={\int }_0^B\frac{B}{{\mu }_0{\mu }_r}dB=\frac{1}{2}\frac{B^2}{{\mu }_0{\mu }_r}=\frac{1}{2}HB$. This represents the magnetostatic energy stored per unit volume in the magnetic field.

Relevant PYQs:

• Write down Ampere’s law. Prove that magnetostatic energy density, $E_{vol}=\frac{1}{2}HB$, where symbols have their usual meaning. (2019, 2018)
• Numerical: Consider a toroidal coil with N turns that is energized from a voltage supply through a rheostat. Estimate the energy required to magnetize the toroidal coil. [Fig 5a provided] (2018)

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Mega Note: Magnetic Properties of Materials - Part 2

This section continues from the foundational concepts of magnetic moments and fields, moving into how materials are classified based on these properties, the internal mechanisms that cause ferromagnetism (domains), and how these materials behave under varying magnetic fields (hysteresis).

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7. Classifications of Magnetic Materials

Magnetic materials are classified based on how their constituent atoms respond to an applied magnetic field, which is quantified by their magnetic susceptibility (${\chi }_m$).

• Diamagnetism:

  • Definition & Concept: Occurs in substances where the constituent atoms have closed subshells and shells, meaning they possess no permanent magnetic moment in the absence of a field.
  • Mechanism: When an external field is applied, it induces a dipole moment that opposes the applied field (in accordance with Lenz’s law).
  • Properties: Typical diamagnetic materials have a negative and very small susceptibility (${\chi }_m\approx -10^{-5}$). Consequently, the relative permeability ${\mu }_r$ is slightly less than 1, and the resulting magnetic field $B$ within the material is less than the applied field ${\mu }_{0}H$.
  • Examples: Covalent crystals, many ionic crystals, and pure superconductors (which are perfect diamagnets with ${\chi }_m=-1$ and totally expel the applied field, known as the Meissner effect).

• Paramagnetism:

  • Definition & Concept: Occurs in materials where the constituent atoms/molecules possess a net permanent magnetic dipole moment.
  • Mechanism: In the absence of a field, these molecular moments are randomly oriented due to thermal collisions, resulting in zero net magnetization. When a field is applied, the moments partially align with the field.
  • Properties: Paramagnetic materials have a small positive susceptibility (${\chi }_m\approx 10^{-5}$). They are weakly attracted to strong magnets.
  • Examples: Liquid oxygen, and many metals like magnesium. In metals, this is often due to “Pauli spin paramagnetism,” where the spins of conduction electrons align with the field.

• Ferromagnetism:

  • Definition & Concept: Materials that can possess large permanent magnetizations even in the absence of an applied field.
  • Properties: Their susceptibility (${\chi }_m$) is positive, very large (can approach infinity), and depends highly on the applied field intensity. The relationship between $M$ and $H$ is highly non-linear.
  • Examples: Iron (Fe), Cobalt (Co), Nickel (Ni).

• Antiferromagnetism:

  • Definition: Materials where adjacent atomic magnetic moments are equal in magnitude but perfectly opposed to each other, resulting in a net macroscopic magnetization of zero. Example: Chromium (Cr).

• Ferrimagnetism:

  • Definition: Materials containing two sets of opposing atomic magnetic moments that are unequal in strength. This imbalance leaves a net spontaneous magnetization. They behave very similarly to ferromagnets macroscopically. Example: Ferrites like $F{e}_3{O}_4$.

Explicit PYQ Answers for this Section:

• Write down the differences between paramagnetism and ferromagnetism. Answer: Paramagnetic materials have small, positive susceptibility and no permanent magnetization; their moments randomly orient without a field. Ferromagnetic materials have very large, positive susceptibility and can retain spontaneous permanent magnetization without a field due to strong quantum exchange interactions aligning the moments.
• Differentiate between Ferrimagnetic and Ferromagnetic materials. Answer: In ferromagnetic materials, all magnetic moments in a domain align parallel to each other, creating a strong net magnetization. In ferrimagnetic materials, there are two distinct sublattices of magnetic moments aligned anti-parallel to each other, but because one sublattice has stronger moments than the other, a net macroscopic magnetization still exists.
• Derive the expression for Langevin’s function. Write down the significance of it. Explicit Answer (Beyond Source Text): The Langevin function $L(x)=\coth (x)-\frac{1}{x}$ is derived from classical statistical mechanics to describe the net alignment of magnetic dipoles in a paramagnetic material under an applied field $B$ at temperature $T$. By integrating the energy of a dipole $U=-\mu \cdot B$ over all solid angles weighted by the Boltzmann factor $\exp (-U/kT)$, we find the average magnetization $M=N\mu L(\frac{\mu B}{kT})$. Significance: It mathematically proves that orientational polarization is inversely proportional to temperature (Curie’s Law) at low fields, and it shows that magnetization saturates at very high fields or very low temperatures when all dipoles are perfectly aligned.

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8. Temperature Dependence & Curie Temperature

Concept: The alignment of magnetic moments in ferromagnetic and ferrimagnetic materials is heavily dependent on temperature.

• At absolute zero, spins are perfectly aligned.
• As temperature increases, thermal energy causes random vibrations that disrupt this alignment, reducing the overall saturation magnetization ($M_{sat}$).
• Curie Temperature ($T_c$): The critical temperature at which the thermal energy completely overcomes the exchange interaction aligning the spins. Above $T_c$, the material loses its ferromagnetism and becomes paramagnetic.
• Curie-Weiss Law: Above the Curie temperature, the paramagnetic susceptibility follows the rule: ${\chi }_m=\frac{C}{T-T_c}$, where C is the material-specific Curie constant.

Explicit PYQ Answers for this Section:

• Explain the variation of saturation magnetization of a ferromagnetic material with respect to temperature. Answer: Saturation magnetization is maximum at 0 K. As temperature rises, thermal agitation increasingly disrupts the parallel alignment of the spin magnetic moments. This causes the saturation magnetization to decrease gradually at first, and then drop sharply to zero as the temperature reaches the Curie temperature ($T_c$).
• Show that at room temperature, normalized, saturated magnetization of iron is very close to 1. Explicit Answer (Beyond Source Text): Iron has a very high Curie temperature ($T_c\approx 1043K$). Room temperature ($\approx 300K$) is much lower than this $T_c$. Because thermal energy at 300 K ($kT$) is exceedingly small compared to the strong exchange interaction energy holding the domains aligned in Iron, the thermal disruption is minimal. Thus, the ratio of magnetization at room temperature to magnetization at absolute zero ($M/M_0$) remains very close to 1.

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9. Magnetic Domains, Domain Walls, and Anisotropy

Concept: A magnetic domain is a microscopic region within a ferromagnetic crystal where all the spin magnetic moments are spontaneously aligned in the same direction, producing a strong net magnetization in that specific region. An unmagnetized specimen contains many domains pointing in different directions, yielding zero net external field.

A. Magnetocrystalline Anisotropy

• Definition: The excess energy required to magnetize a unit volume of a crystal in a particular “hard” direction compared to an “easy” direction.
• Significance: In a single iron crystal, it is easiest to magnetize along the direction and hardest along the direction. Rotating magnetization away from easy directions consumes substantial energy.

B. Domain Walls (Bloch Walls)

• A domain wall is the boundary between two adjacent domains with different magnetization directions.

• Assignment 3.5: Explain the reason for changing the Bloch wall.

  • Answer: A domain wall cannot be simply one atomic spacing wide because forcing two neighboring spins to instantly flip 180° to each other would require a massive amount of exchange interaction energy. Instead, the spin magnetic moments rotate gradually over several hundred atomic spacings.

• Hand-on Practice 2.g: Prove that the minimum energy occurs at optimum domain wall thickness.

  • Derivation: The total potential energy per unit area of a domain wall ($U_{wall}$) is the sum of Exchange Energy ($E_{ex}$) and Anisotropy Energy ($K$).
  • As wall thickness $\delta$ increases, the angle between adjacent spins decreases, which lowers the exchange energy ($U_{exchange}\propto 1/\delta$).
  • However, a thicker wall means more spins are pointing in “hard” crystallographic directions, which increases the anisotropy energy ($U_{anisotropy}\propto \delta$).
  • The total energy is $U_{wall}=\frac{{\pi }^2{E}_{ex}}{2a\delta }+K\delta$ (where $a$ is interatomic spacing). To find the minimum energy, we take the derivative with respect to $\delta$ and set it to zero: $\frac{d{U}_{wall}}{d\delta }=-\frac{{\pi }^2{E}_{ex}}{2a{\delta }^2}+K=0$
  • Solving for $\delta$ gives the optimum thickness: ${\delta }^{\prime }=\sqrt{\frac{{\pi }^2{E}_{ex}}{2aK}}$.
  • Substituting ${\delta }^{\prime }$ back into the two energy components shows that at this optimum thickness, the exchange energy contribution equals the anisotropy energy contribution. This balances the conflicting requirements of both forces.

Explicit PYQ Answers for this Section:

• Explain the different contributions for the formation of domains in a ferromagnetic materials… Answer: Domains form to minimize the total magnetostatic energy of the specimen. A single large domain creates a large external magnetic field, storing high potential energy. By splitting into multiple domains pointing in opposite directions (and forming closure domains at the ends), the external magnetic field lines are kept entirely within the material, drastically reducing the external magnetostatic potential energy.

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10. Polycrystalline Materials & M vs H Behavior (Hysteresis)

When an unmagnetized polycrystalline ferromagnetic material (initially heated above $T_c$ and cooled) is subjected to an external magnetic field ($H$), its magnetization ($M$) changes non-linearly. This tracks the Hysteresis Loop.

1. Initial Magnetization Curve (O to d):
   • Small Field (O to a): Domains that are aligned favorably with the applied field grow slightly at the expense of unfavorably aligned domains through reversible boundary (domain wall) motion.
   • Medium Field (a to c): As $H$ increases, domain walls move larger distances and encounter crystal imperfections (voids, dislocations). Overcoming these obstacles causes sudden, jerky, irreversible jumps in wall motion. These discrete jumps in magnetization are called the Barkhausen effect.
   • High Field (c to d): Once domain walls have moved as far as possible, the magnetization vectors within the domains must rotate away from their “easy” crystallographic axes to strictly align with the applied field $H$.
   • Saturation ($M_{sat}$): At point d, all domain magnetizations are perfectly aligned with $H$. No further magnetization is possible.
2. Removing the Field (d to e):
   • When $H$ is reduced to 0, the specimen does not demagnetize completely. It retains a permanent magnetization called the Remanent Magnetization (or Residual Magnetization, $M_r$).
3. Reverse Field (e to f):
   • To bring the magnetization back to zero, a reverse magnetic field must be applied. The specific magnitude of the reverse field required to completely demagnetize the sample is called the Coercive Field (or Coercivity, $H_c$).
4. Hysteresis Loss:
   • The complete closed curve generated by one cycle of magnetization and demagnetization is the Hysteresis Loop.
   • The area inside this loop represents the energy dissipated per unit volume per cycle as heat, primarily due to the energy lost overcoming pinning sites during domain wall motion.

Explicit PYQ Answers for this Section:

• Define the terms (i) Barkhausen effect (ii) Residual magnetization (iii) Coercive field (iv) Hysteresis loop. (See definitions precisely outlined in the numbered points above).
• Explain how the hysteresis curve is explained on the basis of the domain theory. Answer: Domain theory explains hysteresis through the irreversibility of domain wall motion. Initially, an applied field causes favorably aligned domains to grow by wall motion. Because walls get “pinned” by crystal defects, they require extra energy to break free (causing Barkhausen jumps). When the external field is removed, these domain walls do not simply snap back to their original positions; they remain pinned, leaving the material with a net Remanent Magnetization ($M_r$). To force the walls back and rotate the domains to a net-zero state, a forcible reverse field (Coercive field, $H_c$) must be applied, creating the lagging B-H curve known as hysteresis.

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11. Demagnetization

Slide Concept & PYQ Answer:

• Briefly explain the process of demagnetization. Answer: If a specimen has a remanent magnetization ($M_r$), it can be completely demagnetized (depermed) by subjecting it to an alternating magnetic field (AC field) whose amplitude is gradually decreased to zero. As the alternating field cycles, the B-H operating point traces out smaller and smaller hysteresis loops, eventually spiraling inward to the origin ($H=0,B=0$), leaving the material with zero net magnetization.

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12. Soft & Hard Magnetic Materials

Engineering magnetic materials are classified based on the shape of their B-H hysteresis loops.

A. Soft Magnetic Materials

• Properties:
  • Easy to magnetize and demagnetize.
  • Require very low magnetic field intensities ($H_c$ is very small).
  • Narrow B-H loops, resulting in very small hysteresis energy losses per cycle.
  • High initial and maximum permeability (${\mu }_{ri}$ and ${\mu }_{r,max}$).
• Applications: Used where repeated, rapid cycles of magnetization and demagnetization occur. Examples include AC electric motors, power transformers, inductors, and sensitive relays.
• Examples: Silicon-steels (adding Si to Fe increases resistivity, reducing unwanted eddy current losses), Nickel-Iron alloys (like Supermalloy, which have extremely low magnetocrystalline anisotropy).

B. Hard Magnetic Materials

• Properties:
  • Difficult to magnetize and demagnetize.
  • Require very large magnetic field intensities ($H_c$ is exceptionally large—sometimes millions of times greater than soft materials).
  • Broad, almost rectangular B-H curves with high Remanent Magnetization ($B_r$).
  • High maximum energy product $(BH{)}_{max}$, meaning they can store a large amount of magnetic energy in their external field (air gap).
• Applications: Used as permanent magnets and in magnetic data storage media (like hard drives, where the strong $+B_r$ and $-B_r$ states stably represent binary 1s and 0s).
• Examples: Alnico alloys, Rare-earth cobalt alloys, Neodymium-Iron-Boron (NdFeB).

Explicit PYQ Answers for this Section:

• Explain clearly the differences between hard and soft magnetic materials. Mention their uses. Answer: The primary difference lies in their hysteresis loops. Soft materials have narrow loops, low coercivity, and low hysteresis loss, making them easy to flip magnetically; they are used in AC applications like transformers and motors. Hard materials have broad rectangular loops, massive coercivity, and high remanence, making them highly resistant to being demagnetized; they are used as permanent magnets and in data storage.

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This concludes the complete study materials for Magnetic Properties of Materials based on your slides and the reference text.