CBSE Class XII • Unit 03 (Magnetism and Matter)

Chapter 05: Magnetism and Matter

Master Revision Dossier • Bar Magnet, Solenoid Analogy, Gauss's Law, Dipole Oscillations & Material Classifications

Open Access Initiative

Unit III Bracket: 17 Marks

1. Definitive Formula & Magnetic Properties Matrix

Concept	Mathematical Formulation	SI Unit / Dim	Boundary Conditions & Notes
Axial Field of Dipole	$$B_{\text{axial}} = \frac{\mu_0}{4\pi}\frac{2M}{r^3}$$	$\text{Tesla (T)}$ $[\text{M}^1\text{T}^{-2}\text{A}^{-1}]$	Valid for short magnet ($r \gg l$). Vector direction is parallel to $\vec{M}$ (from South to North).
Equatorial Field of Dipole	$$\vec{B}_{\text{eq}} = -\frac{\mu_0}{4\pi}\frac{\vec{M}}{r^3}$$	$\text{Tesla (T)}$	$B_{\text{axial}} = 2 \times B_{\text{eq}}$ at equal distance $r$. Vector direction is antiparallel to $\vec{M}$.
Dipole Torque in $\vec{B}$	$$\vec{\tau} = \vec{M} \times \vec{B} = MB\sin\theta\hat{n}$$	$\text{N}\cdot\text{m}$	Maximum at $\theta = 90^\circ$ ($\tau_{\max} = MB$). Zero at $\theta = 0^\circ$ and $\theta = 180^\circ$. Net force $\vec{F} = 0$ in uniform field.
Potential Energy in $\vec{B}$	$$U = -\vec{M}\cdot\vec{B} = -MB\cos\theta$$	$\text{Joule (J)}$	Stable Eq: $\theta = 0^\circ$ ($U = -MB$). Unstable Eq: $\theta = 180^\circ$ ($U = +MB$). Work $W_{0 \to 180^\circ} = 2MB$.
Needle Oscillation Period	$$T = 2\pi\sqrt{\frac{\mathcal{I}}{MB}} \implies B = \frac{4\pi^2\mathcal{I}}{MT^2}$$	$\text{Seconds (s)}$	Valid for small angular displacement ($\sin\theta \approx \theta$). $\mathcal{I}$ is moment of inertia about pivot.
Gauss's Law for Magnetism	$$\Phi_B = \oint \vec{B}\cdot d\vec{A} = 0$$	$\text{Weber (Wb)}$	Universal law. Confirms non-existence of isolated magnetic monopoles; field lines are closed continuous loops.
Magnetization ($\vec{M}$)	$$\vec{M} = \frac{\vec{m}_{\text{net}}}{V}$$	$\text{A/m}$ $[\text{L}^{-1}\text{A}]$	Net magnetic dipole moment per unit volume induced inside the magnetic material.
Magnetic Intensity ($\vec{H}$)	$$\vec{H} = \frac{\vec{B}}{\mu_0} - \vec{M}$$	$\text{A/m}$	Represents the external magnetizing field created by free conduction currents (e.g. $H = nI$ in solenoid).
Susceptibility & Permeability	$$\chi = \frac{M}{H}, \quad \mu_r = 1 + \chi, \quad \vec{B} = \mu_r \mu_0 \vec{H}$$	Dimensionless	Total field inside medium: $\vec{B} = \vec{B}_0 + \mu_0\vec{M}$. Independent parameter linking material response.
Curie's Law (Paramagnet)	$$\chi_{\text{para}} = \frac{C}{T} \quad (M = C\frac{B_0}{T})$$	$\text{Kelvin (K)}$	$C$ is Curie's constant. Breaks down near absolute zero or very high fields when saturation is reached.
Curie-Weiss Law (Ferro)	$$\chi_{\text{ferro}} = \frac{C'}{T - T_c} \quad (T > T_c)$$	$\text{Kelvin (K)}$	$T_c$ is Curie Temperature (Iron: $T_c = 1043\text{ K} = 770^\circ\text{C}$). Above $T_c$, ferromagnet converts to paramagnet.

2. Core Derivations Blueprint (Step-by-Step Board Proofs)

Blueprint 5.1: Axial Magnetic Field of a Finite Solenoid as an Equivalent Bar Magnet

1. Consider a finite solenoid of length $2l$, radius $a$, carrying current $I$ with $n$ turns per unit length ($N = 2nl$).

2. Consider a circular slice of thickness $dx$ at distance $x$ from the center $O$. Number of turns in the slice $= n\,dx$.

3. Distance of an axial observation point $P$ from this slice is $(r - x)$. Using the axial field formula of a circular ring:

$$dB = \frac{\mu_0 (n\,dx) I a^2}{2 [(r - x)^2 + a^2]^{3/2}}$$

4. For a distant point on the axis ($r \gg a$ and $r \gg x$), approximate $(r - x) \approx r$ and neglect $a^2$:

$$dB = \frac{\mu_0 n I a^2}{2 r^3} dx$$

5. Integrate over the entire length of the solenoid from $x = -l$ to $x = +l$:

$$B = \frac{\mu_0 n I a^2}{2 r^3} \int_{-l}^{+l} dx = \frac{\mu_0 n I a^2 (2l)}{2 r^3} = \frac{\mu_0 I [n(2l)] (\pi a^2)}{2\pi r^3} = \frac{\mu_0 I N A}{2\pi r^3}$$

6. Substitute total magnetic moment $M = N I A = N I (\pi a^2)$:

$$B_{\text{axial}} = \frac{\mu_0}{4\pi} \frac{2M}{r^3} \quad (\text{Identical to electrostatic dipole field } E = \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3})$$

Blueprint 5.2: Oscillation Time Period of a Magnetic Needle in a Uniform Magnetic Field

1. A magnetic needle of moment $\vec{M}$ and moment of inertia $\mathcal{I}$ is placed in uniform magnetic field $\vec{B}$.

2. When deflected by a small angle $\theta$ from equilibrium, the restoring torque tending to align $\vec{M}$ along $\vec{B}$ is:

$$\tau = -M B \sin\theta \approx -M B \theta \quad (\because \sin\theta \approx \theta \text{ for small } \theta)$$

3. By rotational dynamics, $\tau = \mathcal{I} \frac{d^2\theta}{dt^2}$:

$$\mathcal{I}\frac{d^2\theta}{dt^2} = -MB\theta \implies \frac{d^2\theta}{dt^2} = -\left(\frac{MB}{\mathcal{I}}\right)\theta$$

4. Comparing with the standard angular SHM equation $\frac{d^2\theta}{dt^2} = -\omega^2\theta$:

$$\omega = \sqrt{\frac{MB}{\mathcal{I}}} \implies T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{\mathcal{I}}{MB}} \quad \text{and} \quad B = \frac{4\pi^2\mathcal{I}}{MT^2}$$

3. High-Yield "Red Alert" Exam Trap Matrix

Master Trap 1: Comparative Classification of Magnetic Materials

Characteristic	Diamagnetic	Paramagnetic	Ferromagnetic
Behavior in External Field	Feebly repelled; moves from strong to weak field.	Feebly attracted; moves from weak to strong field.	Strongly attracted; pulled rapidly into strong field.
Atomic Magnetic Dipole	Zero intrinsic dipole (paired electrons, canceling spins).	Permanent atomic dipoles exist; randomized by heat.	Permanent dipoles pre-aligned into Weiss Domains.
Susceptibility ($\chi$)	Negative & Small ($-1 \le \chi < 0$)	Positive & Small ($0 < \chi < \varepsilon$)	Positive & Huge ($\chi \gg 1000$)
Permeability ($\mu_r = 1+\chi$)	Slightly less than 1 ($0 \le \mu_r < 1$)	Slightly greater than 1 ($\mu_r \gtrsim 1$)	Extremely large ($\mu_r \gg 1000$)
Field Line Density	Lines are expelled ($B_{\text{in}} < B_0$).	Lines slightly crowd inside ($B_{\text{in}} \gtrsim B_0$).	Lines heavily concentrated inside ($B_{\text{in}} \ggg B_0$).
Temperature Dependence	INDEPENDENT of $T$	$\chi \propto 1/T$ (Curie's Law)	$\chi = \frac{C'}{T - T_c}$ for $T > T_c$ (Curie-Weiss)
Classic Examples	Bi, Cu, H₂O, Pb, Superconductors ($\chi = -1$).	Al, Na, Pt, Liquid Oxygen, Mn.	Fe, Co, Ni, Gadolinium ($T_c = 16^\circ\text{C}$), Alnico.

⚠️ Trap 2: Cutting a Bar Magnet Rules

Initial moment $M = m \times 2l$.
• Transverse cut (across length): Length halves ($l$), pole strength unchanged $\implies M' = m \times l = \mathbf{M/2}$.
• Longitudinal cut (along length): Length unchanged ($2l$), pole strength halves ($m/2$) $\implies M' = (m/2) \times 2l = \mathbf{M/2}$.
Each cut piece becomes a complete, functional dipole with both N and S poles.

⚠️ Trap 3: Uniform vs Non-Uniform Field Forces

• Uniform $\vec{B}$: Equal and opposite forces act on poles ($\vec{F}_N = +m\vec{B}, \vec{F}_S = -m\vec{B}$). Net translational force is strictly ZERO ($\vec{F}_{\text{net}} = 0, \tau \ne 0$).
• Non-Uniform $\vec{B}$: Field strengths at poles differ ($B_N \ne B_S$). Dipole experiences BOTH a net force and a torque ($\vec{F}_{\text{net}} \ne 0, \tau \ne 0$).

4. Handcrafted Exemplars (Analytical Numerical Solutions)

Problem 1: Oscillation Magnetometer Determination of Magnetic Moment Exemplar 01

A magnetic needle completes 30 oscillations in 1 minute in a horizontal field of $B_H = 0.36\text{ G}$. If its moment of inertia is $\mathcal{I} = 7.2 \times 10^{-6}\text{ kg}\cdot\text{m}^2$, calculate its magnetic dipole moment $M$.

Solution: Time period $T = \frac{60\text{ s}}{30} = 2.0\text{ s}$. Field $B = 0.36 \times 10^{-4}\text{ T} = 3.6 \times 10^{-5}\text{ T}$.

$$T = 2\pi\sqrt{\frac{\mathcal{I}}{MB}} \implies M = \frac{4\pi^2\mathcal{I}}{B T^2} = \frac{4 \times (9.87) \times (7.2 \times 10^{-6})}{(3.6 \times 10^{-5}) \times (2.0)^2}$$

$$M = \frac{2.842 \times 10^{-4}}{1.44 \times 10^{-4}} \approx 1.97\text{ A}\cdot\text{m}^2 \quad (\text{or J/T})$$

Problem 2: Solenoid with Ferromagnetic Core Parameters Exemplar 02

A solenoid of $1000\text{ turns/m}$ carries current $2.0\text{ A}$ and contains an iron core of relative permeability $\mu_r = 400$. Find (i) Magnetic Intensity $H$, (ii) Magnetic Field $B$, and (iii) Magnetization $M$ inside the core.

Solution:

1. $H = nI = 1000 \times 2.0 = 2.0 \times 10^3\text{ A/m}$.

2. $B = \mu_r \mu_0 H = 400 \times (4\pi \times 10^{-7}) \times (2.0 \times 10^3) = 3200\pi \times 10^{-4} \approx 1.005\text{ Tesla}$.

$$M = (\mu_r - 1)H = (400 - 1) \times (2.0 \times 10^3) = 399 \times 2000 = 7.98 \times 10^5\text{ A/m}$$

5. Printable Pocket Flashcard Grid

Axial Field

$B = \frac{\mu_0 2M}{4\pi r^3}$

Needle Period

$T = 2\pi\sqrt{\frac{\mathcal{I}}{MB}}$

Gauss Law

$\oint \vec{B}\cdot d\vec{A} = 0$

Curie's Law

$\chi \propto \frac{1}{T}$