Axiom Physics | Class 12 - Magnetic Properties of Materials

⚡ 01. The Spark (Analogy)

Socho ek bheed bhare hall ko jisme ek speaker (external magnetic field $\vec{B}_0$) aakar lecture dena shuru karta hai:
1. Diamagnetic log stubborn hain: unke paas pehle koi opinion nahi tha (zero net intrinsic dipole), lekin speaker ke bolte hi woh unke khilaaf hokar opposite direction mein react karte hain aur hall se door bhagne lagte hain (repelled by magnetic field).
2. Paramagnetic log lazy supporters hain: sabke paas apna personal opinion (permanent dipole) hai par bheed ki garmi (thermal agitation) unhe random rakhti hai. Speaker aate hi woh halka sa uski taraf jhuk jaate hain, par garmi badhane par unka alignment toot jata hai (Curie's Law: $\chi \propto 1/T$).
3. Ferromagnetic log organized gangs (Weiss Domains) hain: lakho log pehle se hi ek direction mein disciplined khade hain! Jaise hi speaker ek halka sa ishara karta hai, poore gang ke borders shift ho jaate hain aur sab ek sath usi direction mein synchronize hokar intensely lock ho jaate hain!

📐 02. The Core Law & Formulae

Fundamental Magnetic Field Vectors

Magnetization $\vec{M}$, Magnetic Intensity $\vec{H}$, and Total Field $\vec{B}$:

$$\vec{M} = \frac{\vec{m}_{\text{net}}}{V}, \quad \vec{B} = \mu_0(\vec{H} + \vec{M})$$ $$\vec{M} = \chi \vec{H} \implies \vec{B} = \mu_0(1 + \chi)\vec{H} = \mu_r \mu_0 \vec{H}$$

Units: $\vec{M}$ and $\vec{H}$ have SI unit $\text{A/m}$. Relative permeability $\mu_r = 1 + \chi$ (dimensionless).

Curie's Law & Curie-Weiss Law

Thermal disruption of magnetic alignment:

$$\chi_{\text{para}} = \frac{C}{T} \quad (\text{Curie's Law})$$ $$\chi_{\text{ferro}} = \frac{C'}{T - T_c} \quad (T > T_c, \text{Curie-Weiss Law})$$

Where $T_c$ is the Curie Temperature. Above $T_c$, a ferromagnetic substance transitions into a simple paramagnetic substance!

03. Interactive Visualizer: Microscopic Material Response Engine

External Field ($B_0$) 1.20 Tesla

Absolute Temperature ($T$) 300 K

Material Response Class Weak Attraction

χ = +2.3 × 10⁻⁵ | μr = 1.000023

Susceptibility ($\chi$):

+2.30 × 10⁻⁵

Permeability ($\mu_r = 1 + \chi$):

1.000023

Internal Field ($B_{\text{int}}$):

1.20002 T

Thermal Law Status:

Obeys Curie's Law

📝 04. Comparative Blueprint & Mathematical Deductions

Property	Diamagnetic	Paramagnetic	Ferromagnetic
1. Physical Action in External Field $\vec{B}_0$	Feeble repulsion; moves from stronger to weaker field regions.	Feeble attraction; moves from weaker to stronger field regions.	Strong, intense attraction; pulled rapidly towards strongest field.
2. Atomic Magnetic Dipole Moment	Zero intrinsic dipole (electrons are paired with canceling spins).	Permanent intrinsic atomic dipoles exist, but randomly oriented by heat.	Permanent dipoles pre-aligned into macroscopic Weiss Domains ($10^{11}$ atoms).
3. Magnetic Susceptibility ($\chi$)	Negative & Small ($-1 \le \chi < 0$) e.g., Bismuth: $\chi \approx -1.6 \times 10^{-5}$	Positive & Small ($0 < \chi < \varepsilon$) e.g., Aluminium: $\chi \approx +2.3 \times 10^{-5}$	Positive & Extremely Large ($\chi \gg 1000$) e.g., Soft Iron: $\chi \approx 5000$
4. Relative Permeability ($\mu_r$)	Slightly less than 1 ($0 \le \mu_r < 1$)	Slightly greater than 1 ($\mu_r > 1$)	Much greater than 1 ($\mu_r \gg 1000$)
5. Behavior of Field Lines	Field lines are expelled from material ($B_{\text{in}} < B_0$).	Field lines slightly concentrated inside ($B_{\text{in}} \gtrsim B_0$).	Field lines crowd heavily inside ($B_{\text{in}} \ggg B_0$).
6. Effect of Temperature ($T$)	INDEPENDENT of $T$ (Orbital property; unchanged by heat)	Inversely proportional ($T$) $\chi = \frac{C}{T}$ (Curie's Law)	Curie-Weiss Law above $T_c$: $\chi = \frac{C'}{T - T_c}$ (Converts to Para!)
7. Classic Examples	Bismuth, Copper, Water, Lead, Nitrogen (STP), Superconductors.	Aluminium, Sodium, Platinum, Liquid Oxygen, Manganese.	Iron, Cobalt, Nickel, Gadolinium ($T_c = 16^\circ\text{C}$), Alnico.

Thermodynamic Law

Deduction of Curie's Law & Ferromagnetic-to-Paramagnetic Transition

1. In a paramagnetic material, individual atomic magnetic dipoles $\vec{m}$ experience an aligning torque from external field $\vec{B}_0$, but thermal kinetic energy ($\sim k_B T$) tends to randomize their orientations.

2. At moderate fields and temperatures where magnetic potential energy is much smaller than thermal energy ($m B_0 \ll k_B T$), statistical Boltzmann mechanics gives net magnetization proportional to applied field and inversely to absolute temperature:

$$M = C \frac{B_0}{T} \implies \frac{M}{H} = \frac{\mu_0 C}{T} \implies \chi = \frac{C}{T}$$

3. Where $C$ is Curie's Constant ($C = \frac{\mu_0 n m^2}{3 k_B}$). As $T \to 0\text{ K}$, magnetization approaches maximum saturation value $M_{\text{sat}} = n m$.

4. In a ferromagnetic substance, atomic exchange coupling creates spontaneous domain alignment below the Curie Temperature $T_c$ (e.g., for Iron $T_c = 1043\text{ K}$ or $770^\circ\text{C}$). Above $T_c$, thermal agitation completely overpowers the exchange coupling, breaking the domain structure into randomized atomic dipoles. The susceptibility then obeys the Curie-Weiss Law:

$$\chi = \frac{C'}{T - T_c} \quad (T > T_c)$$

⚠️ 05. Red Alert Exam Trap Zone

Trap 1: Perfect Diamagnetism & The Superconductor Identity

*Board Question:* What is the magnetic susceptibility of a perfect superconductor?
A superconductor below its critical temperature exhibits the Meissner Effect: it completely expels all magnetic flux from its interior ($\vec{B} = 0$). $$\vec{B} = \mu_0(\vec{H} + \vec{M}) = 0 \implies \vec{M} = -\vec{H}$$ $$\chi = \frac{M}{H} = \mathbf{-1}, \quad \mu_r = 1 + \chi = \mathbf{0}$$ A superconductor is a perfect diamagnet ($\chi = -1, \mu_r = 0$).

Trap 2: Does Diamagnetism depend on Temperature?

*Exam Trap:* "If a piece of bismuth is heated from $300\text{ K}$ to $600\text{ K}$, how does its susceptibility change?"
*Correct Answer:* IT REMAINS COMPLETELY UNCHANGED! Diamagnetism arises from Lenz's law modification of paired electron orbital speeds, which does not depend on thermal agitation. (Only paramagnetism and ferromagnetism depend on temperature).

Trap 3: Liquid in a U-Tube Between Magnetic Poles

• Diamagnetic Liquid (Water/Bismuth): The liquid meniscus in the limb between strong magnetic poles DEPRESSES (goes down) because it is repelled from the strong field.
• Paramagnetic Liquid (Liquid Oxygen/FeSO₄ sol): The meniscus RISES UP because it is attracted towards the region of stronger field.

Trap 4: Electromagnet Core vs Permanent Magnet Materials

• Electromagnet / Transformer Core (Soft Iron): Requires High Permeability $\mu_r$, High Retentivity, but Low Coercivity and thin hysteresis loop to minimize heat dissipation during AC cycling.
• Permanent Magnet (Steel / Alnico): Requires High Retentivity and HIGH COERCIVITY so that stray fields or mechanical drops do not demagnetize it easily.

💡 06. Handcrafted Exemplars

Board Classic Level

Problem: A solenoid has a core of material with relative permeability $\mu_r = 400$. The windings of the solenoid are insulated from the core and carry a current of $2.0\text{ A}$. If the number of turns is $1000\text{ per metre}$, calculate (i) Magnetic Intensity $H$, (ii) Magnetic Field $B$, and (iii) Magnetization $M$ inside the core.

Solution:

1. Magnetic Intensity $H$ depends strictly on the winding free current:

$$H = n I = (1000\text{ m}^{-1}) \times (2.0\text{ A}) = 2.0 \times 10^3\text{ A/m}$$

2. Magnetic field $B$ inside the material core:

$$B = \mu_r \mu_0 H = 400 \times (4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}) \times (2.0 \times 10^3\text{ A/m})$$

$$B = 3200\pi \times 10^{-4}\text{ T} \approx 1.005\text{ Tesla}$$

3. Magnetization $M$:

$$M = \chi H = (\mu_r - 1) H = (400 - 1) \times (2.0 \times 10^3) = 399 \times 2.0 \times 10^3$$

$$M = 7.98 \times 10^5\text{ A/m}$$

JEE/NEET Competitive Trap

Problem: The magnetic susceptibility of a paramagnetic salt is $1.2 \times 10^{-5}$ at $300\text{ K}$. At what temperature will its susceptibility increase to $1.8 \times 10^{-5}$? If the sample is placed in a field of $0.6\text{ T}$ and cooled to $4\text{ K}$, it reaches 15% of its saturation magnetization. If the field is increased to $2.0\text{ T}$, can Curie's law still be applied?

Solution:

1. By Curie's Law, $\chi \propto 1/T \implies \chi_1 T_1 = \chi_2 T_2$:

$$T_2 = T_1 \times \left(\frac{\chi_1}{\chi_2}\right) = 300\text{ K} \times \left(\frac{1.2 \times 10^{-5}}{1.8 \times 10^{-5}}\right) = 300 \times \frac{2}{3} = 200\text{ K}$$

2. At $T = 4\text{ K}$ and higher field ($2.0\text{ T}$), the ratio $B/T$ becomes very large ($B/T = 2.0 / 4 = 0.5\text{ T/K}$). Atomic dipoles become almost fully aligned, approaching complete magnetic saturation ($M \to M_{\text{sat}}$). In this regime, magnetization stops increasing linearly with field; therefore, Curie's Law ceases to be valid!

🗂️ 07. Quick Revision Flashcard Core Formulae

Diamagnetic

$-1 \le \chi < 0$, $\mu_r < 1$

Paramagnetic

$\chi \propto 1/T$, $\mu_r > 1$

Ferromagnetic

$\chi \gg 1000$, $\mu_r \gg 1000$

Curie-Weiss Law

$\chi = \frac{C'}{T - T_c}$ ($T > T_c$)