Vardaan Learning Institute
Chapter 5: Magnetism and Matter
Comprehensive, board-focused class notes covering all core derivations, qualitative laws, macroscopic magnetisation parameters, dipole dynamics, and magnetic property classifications strictly according to the latest rationalized Class 12 Physics syllabus.
1. The Bar Magnet as a Magnetic Dipole
A magnetic dipole consists of two equal and opposite magnetic poles (North pole $+q_m$ and South pole $-q_m$) separated by a small finite magnetic length $2\vec{l}$.
Key Dipole Parameters
- Magnetic Dipole Moment ($\vec{m}$): Product of pole strength ($q_m$) and magnetic length ($2\vec{l}$):
$$\vec{m} = q_m (2\vec{l})$$
Vector Direction: Directed from South pole (S) to North pole (N) along the dipole axis.
SI Unit: $\text{A m}^2$ or $\text{J T}^{-1}$. Dimensions: $[M^0 L^2 T^0 A^1]$.
- Geometric Length vs Magnetic Length: Magnetic poles reside slightly inside the physical ends of a bar magnet:
$$\text{Magnetic Length } (2l) \approx 0.84 \times \text{Geometric Length}$$
- Cutting a Bar Magnet:
- Cut Transverse to length: Length halves ($l' = l/2$), pole strength unchanged $\implies m' = q_m(l/2) = m/2$.
- Cut Longitudinal along length: Length unchanged, pole strength halves ($q_m' = q_m/2$) $\implies m' = (q_m/2)(2l) = m/2$.
Figure 5.1: Magnetic dipole field lines mapped by iron filings aligning along field lines around a bar magnet.
2. Properties of Magnetic Field Lines
- Continuous Closed Loops: Outside the magnet, lines emerge from the North pole (N) and enter the South pole (S). Inside the body of the magnet, they run continuously from South pole (S) to North pole (N). (Unlike electrostatics, there are no isolated pole charges/endpoints).
- Direction of Field $\vec{B}$: Tangent drawn at any point gives the direction of net magnetic field $\vec{B}$ at that point.
- No-Intersection Rule: Two field lines never intersect. If they did, two different tangents (two field directions) would exist at the same point, which is physically impossible.
- Line Density: Relative density of field lines represents the magnitude of the field (crowded near poles $\implies$ strong field).
 a bar magnet, (b) a current-carrying finite solenoid and.png)
Figure 5.2: Continuous closed loops of: (a) Bar magnet, (b) Finite solenoid, contrasted with (c) Discontinuous electric dipole lines.
3. Magnetic Field Intensity Due to a Short Magnetic Dipole
For a short bar magnet of magnetic dipole moment $m$ ($r \gg l$):
Master Field Intensity Formulas
- Axial Field ($B_{axial}$ / End-on Position):
$$\vec{B}_{axial} = \frac{\mu_0}{4\pi} \frac{2\vec{m}}{r^3}$$
Direction: Parallel to $\vec{m}$ (along the axis from South pole to North pole).
- Equatorial Field ($B_{equatorial}$ / Broadside-on Position):
$$\vec{B}_{equatorial} = -\frac{\mu_0}{4\pi} \frac{\vec{m}}{r^3} \implies B_{equatorial} = \frac{\mu_0}{4\pi}\frac{m}{r^3}$$
Direction: Antiparallel to $\vec{m}$ (parallel to dipole axis from North pole to South pole).
- Fundamental Board Ratio: At equal distances $r$:
$$\mathbf{B_{axial} = 2 \, B_{equatorial}}$$
- General Point $(r, \theta)$ in Polar Coordinates:
$$B = \frac{\mu_0}{4\pi} \frac{m}{r^3} \sqrt{1 + 3\cos^2\theta} \quad \text{and} \quad \tan\alpha = \frac{1}{2}\tan\theta$$
4. Bar Magnet as an Equivalent Solenoid (Standard Board Derivation)
A finite current-carrying solenoid produces an identical magnetic field at large distances as a bar magnet of equivalent magnetic moment $m = N I A$.
Figure 5.3: Geometry for integrating current elements of a finite solenoid of length $2l$ and radius $a$ at axial point $P$.
Step-by-Step Derivation
Let a solenoid have length $= 2l$, radius $= a$, number of turns per unit length $= n$, and current $= I$.
Consider an elemental slice of width $dx$ located at distance $x$ from the center $O$.
Number of turns in this element $= n\,dx$.
Axial field due to a single circular loop at axial distance $(r - x)$:
$$dB = \frac{\mu_0 (n\,dx) I a^2}{2 [(r - x)^2 + a^2]^{3/2}}$$
For a distant point ($r \gg a$ and $r \gg l$), $[(r - x)^2 + a^2]^{3/2} \approx r^3$:
$$dB \approx \frac{\mu_0 n I a^2}{2 r^3} dx$$
Integrating along the full length of the solenoid from $x = -l$ to $x = +l$:
$$B = \frac{\mu_0 n I a^2}{2 r^3} \int_{-l}^{l} dx = \frac{\mu_0 n I a^2}{2 r^3} [2l] = \frac{\mu_0 n (2l) I (\pi a^2)}{2\pi r^3}$$
Multiplying numerator and denominator by $2$:
$$\mathbf{B = \frac{\mu_0}{4\pi} \frac{2m}{r^3}} \quad \text{where } m = N I A = [n(2l)] I (\pi a^2)$$
This confirms the exact magnetic equivalence between a finite solenoid and a bar magnet!
5. Magnetic Dipole in a Uniform Magnetic Field
Figure 5.4: Magnetic needle experiencing equal and opposite forces ($\pm q_m \vec{B}$) creating couple torque $\vec{\tau} = \vec{m} \times \vec{B}$.
Dynamics of Magnetic Dipole
- Net Translational Force: $\vec{F}_{net} = q_m \vec{B} - q_m \vec{B} = \mathbf{0}$. (No linear acceleration).
- Torque ($\vec{\tau}$):
$$\vec{\tau} = \vec{m} \times \vec{B} \implies \tau = m B \sin\theta$$
(Acts to align dipole moment $\vec{m}$ parallel to field $\vec{B}$).
- Potential Energy ($U$):
$$U = -\vec{m} \cdot \vec{B} = -m B \cos\theta$$
- Stable Equilibrium ($\theta = 0^\circ$): $\tau = 0$, $U_{min} = -mB$ (minimum potential energy).
- Unstable Equilibrium ($\theta = 180^\circ$): $\tau = 0$, $U_{max} = +mB$ (maximum potential energy).
- Zero Energy Reference ($\theta = 90^\circ$): $U = 0$, $\tau_{max} = mB$.
- Work Done in Rotating Dipole:
$$W = \int_{\theta_1}^{\theta_2} \tau \, d\theta = -m B (\cos\theta_2 - \cos\theta_1) = \mathbf{m B (\cos\theta_1 - \cos\theta_2)}$$
- Time Period of Small Oscillations:
$$I \frac{d^2\theta}{dt^2} = -m B \theta \implies \omega = \sqrt{\frac{m B}{I}} \implies \mathbf{T = 2\pi\sqrt{\frac{I}{m B}}}$$
where $I$ is the moment of inertia about the axis of rotation.
6. The Dipole Analogy (Electrostatics vs Magnetism)
| Property |
Electrostatics (Electric Dipole) |
Magnetism (Magnetic Dipole) |
| Medium Permittivity / Permeability |
$\frac{1}{\epsilon_0}$ |
$\mu_0$ |
| Dipole Moment |
$\vec{p} = q(2\vec{a})$ (Directed $-q \to +q$) |
$\vec{m} = q_m(2\vec{l})$ (Directed $S \to N$) |
| Axial Field ($r \gg a, l$) |
$\vec{E}_{axial} = \frac{1}{4\pi\epsilon_0} \frac{2\vec{p}}{r^3}$ |
$\vec{B}_{axial} = \frac{\mu_0}{4\pi} \frac{2\vec{m}}{r^3}$ |
| Equatorial Field ($r \gg a, l$) |
$\vec{E}_{eq} = -\frac{1}{4\pi\epsilon_0} \frac{\vec{p}}{r^3}$ |
$\vec{B}_{eq} = -\frac{\mu_0}{4\pi} \frac{\vec{m}}{r^3}$ |
| Torque in Uniform Field |
$\vec{\tau} = \vec{p} \times \vec{E}$ |
$\vec{\tau} = \vec{m} \times \vec{B}$ |
| Potential Energy |
$U = -\vec{p} \cdot \vec{E}$ |
$U = -\vec{m} \cdot \vec{B}$ |
7. Gauss's Law in Magnetism
Fundamental Law & Significance
Statement: "The net magnetic flux through any closed Gaussian surface is always identically zero."
$$\mathbf{\Phi_B = \oint \vec{B} \cdot d\vec{S} = 0}$$
Physical Significance:
- Magnetic field lines are continuous closed loops; number of field lines entering any closed surface equals lines exiting it.
- Isolated magnetic monopoles do NOT exist in nature. Magnetic poles always occur in equal and opposite pairs.
8. Magnetisation and Magnetic Intensity Parameters
5 Essential Parameters (Definitions & Formulas)
- 1. Magnetisation Vector ($\vec{M}$): Net magnetic dipole moment developed per unit volume of the material:
$$\vec{M} = \frac{\vec{m}_{net}}{V} \quad \left[\text{Unit: } \text{A m}^{-1}\text{, Dimensions: } [L^{-1} A]\right]$$
- 2. Magnetic Intensity ($\vec{H}$): The capability of an external magnetising field to magnetise a material:
$$\vec{H} = \frac{\vec{B}}{\mu_0} - \vec{M} \implies \mathbf{\vec{B} = \mu_0(\vec{H} + \vec{M})} \quad \left[\text{Unit: } \text{A m}^{-1}\right]$$
For a long solenoid: $H = n I$.
- 3. Magnetic Susceptibility ($\chi$): Ratio of induced magnetisation $M$ to applied intensity $H$:
$$\mathbf{\chi = \frac{M}{H}} \quad \text{(Dimensionless scalar)}$$
Measures how easily a substance gets magnetised when placed in an external field.
- 4. Relative Magnetic Permeability ($\mu_r$): Ratio of permeability of medium ($\mu$) to that of free space ($\mu_0$):
$$\mathbf{\mu_r = 1 + \chi = \frac{\mu}{\mu_0}} \quad \text{(Dimensionless scalar)}$$
- 5. Magnetic Permeability ($\mu$): Total magnetic field produced per unit magnetic intensity:
$$\mathbf{\mu = \mu_0 \mu_r = \mu_0 (1 + \chi)} \quad \left[\text{Unit: } \text{T m A}^{-1} \text{ or } \text{H m}^{-1}\right]$$
9. Classification of Magnetic Materials (Dia, Para, Ferro)
Figure 5.5: Magnetic field line repulsion in Diamagnetic material ($B < B_0$) vs attraction in Paramagnetic material ($B > B_0$).
| Property |
Diamagnetic Substances |
Paramagnetic Substances |
Ferromagnetic Substances |
| Origin / Atomic Dipoles |
Paired electrons (net atomic dipole moment $= 0$). Induced moment opposes external field. |
Unpaired electrons (each atom has permanent dipole moment; randomly oriented by thermal agitation). |
Permanent atomic dipoles spontaneously aligned in macroscopic domains via exchange coupling. |
| In Non-Uniform Field |
Moves feebly from stronger to weaker field regions (repelled). |
Moves feebly from weaker to stronger field regions (attracted). |
Moves strongly from weaker to stronger field regions (strongly attracted). |
| Susceptibility ($\chi$) |
Small & negative ($-1 \le \chi < 0$) |
Small & positive ($0 < \chi < \epsilon$) |
Very large & positive ($\chi \gg 1000$) |
| Relative Permeability ($\mu_r$) |
Slightly less than $1$ ($0 \le \mu_r < 1$) |
Slightly greater than $1$ ($\mu_r > 1$) |
Very large ($\mu_r \gg 1000$) |
| Field Line Pattern |
Expels field lines ($B_{in} < B_0$) |
Concentrates field lines feebly ($B_{in} > B_0$) |
Crowds field lines intensely inside ($B_{in} \gg B_0$) |
| Temperature Dependence |
Independent of Temperature |
Inversely proportional to $T$:
Curie's Law: $\chi = \frac{C\mu_0}{T}$ |
Decreases with $T$. Above Curie temperature ($T_c$), turns Paramagnetic:
Curie-Weiss Law: $\chi = \frac{C'}{T - T_c}$ |
| Key Examples |
$\text{Bi, Cu, Pb, Si, Water, NaCl, } \text{N}_2 \text{ (STP)}$ |
$\text{Al, Na, Ca, } \text{O}_2 \text{ (STP), } \text{CuCl}_2$ |
$\text{Fe, Co, Ni, Gd, Alnico, Lodestone}$ |
Figure 5.6: Ferromagnetic domains: (a) Random unmagnetized state ($\sum \vec{M} = 0$), (b) Domain alignment in external field $\vec{B}_0$.
Important Special Phenomena
- Superconductors & The Meissner Effect: Superconductors exhibit perfect diamagnetism ($\chi = -1, \mu_r = 0$). They completely expel all internal magnetic field lines, creating magnetic levitation.
- Curie Temperature ($T_c$): The critical temperature at which a ferromagnetic substance loses its spontaneous domain magnetization and becomes purely paramagnetic. (For Iron: $T_c \approx 1043\text{ K} = 770^\circ\text{C}$).
10. Selected Core Board Exam Practice Problems
Board Practice 1: Dipole Torque & Work
Question: A bar magnet of magnetic moment $1.5\text{ J T}^{-1}$ lies aligned with a uniform magnetic field of $0.22\text{ T}$. Calculate: (a) the work required to rotate the magnet to $90^\circ$ and $180^\circ$, (b) the torque on the magnet at $\theta = 90^\circ$ and $\theta = 180^\circ$.
Solution:
Given: $m = 1.5\text{ J T}^{-1}$, $B = 0.22\text{ T}$, $\theta_1 = 0^\circ$.
(a) Work Done $W = m B (\cos\theta_1 - \cos\theta_2)$:
• At $\theta_2 = 90^\circ$:
$$\begin{aligned}
W &= (1.5)(0.22)(\cos 0^\circ - \cos 90^\circ) = 0.33 \times (1 - 0) = \mathbf{0.33 \text{ J}}
\end{aligned}$$
• At $\theta_2 = 180^\circ$:
$$\begin{aligned}
W &= (1.5)(0.22)(\cos 0^\circ - \cos 180^\circ) = 0.33 \times [1 - (-1)] = \mathbf{0.66 \text{ J}}
\end{aligned}$$
(b) Torque $\tau = m B \sin\theta$:
• At $\theta = 90^\circ$: $\tau = (1.5)(0.22)\sin 90^\circ = \mathbf{0.33 \text{ N m}}$
• At $\theta = 180^\circ$: $\tau = (1.5)(0.22)\sin 180^\circ = \mathbf{0 \text{ N m}}$
Board Practice 2: Solenoid Magnetic Moment & Torque
Question: A closely wound solenoid of $2000\text{ turns}$ and area $1.6 \times 10^{-4}\text{ m}^2$ carries a current of $4.0\text{ A}$. Calculate: (a) magnetic moment $m$, (b) torque in a uniform magnetic field of $7.5 \times 10^{-2}\text{ T}$ at an angle of $30^\circ$.
Solution:
(a) Magnetic Dipole Moment:
$$m = N I A = 2000 \times 4.0 \times (1.6 \times 10^{-4}) = \mathbf{1.28 \text{ A m}^2}$$
(b) Torque $\tau$:
$$\begin{aligned}
\tau &= m B \sin\theta \\
&= 1.28 \times (7.5 \times 10^{-2}) \times \sin 30^\circ \\
&= 0.096 \times 0.5 = \mathbf{0.048 \text{ N m}} = \mathbf{4.8 \times 10^{-2} \text{ N m}}
\end{aligned}$$
Board Practice 3: Axial & Equatorial Field
Question: A short bar magnet has a magnetic moment of $0.48\text{ J T}^{-1}$. Find the magnitude and direction of magnetic field at a distance of $10\text{ cm}$ from its center: (a) on the axis, (b) on the equatorial line.
Solution:
Given: $m = 0.48\text{ J T}^{-1}$, $r = 0.1\text{ m}$.
(a) Axial Field ($B_A$):
$$\begin{aligned}
B_A &= \frac{\mu_0}{4\pi} \frac{2m}{r^3} = 10^{-7} \times \frac{2 \times 0.48}{(0.1)^3} \\
&= 10^{-7} \times \frac{0.96}{10^{-3}} = 0.96 \times 10^{-4}\text{ T} = \mathbf{0.96 \text{ G}} \text{ (along } \vec{m} \text{, South to North)}
\end{aligned}$$
(b) Equatorial Field ($B_E$):
$$\begin{aligned}
B_E &= \frac{B_A}{2} = \frac{0.96 \times 10^{-4}}{2} = \mathbf{0.48 \times 10^{-4} \text{ T}} = \mathbf{0.48 \text{ G}} \text{ (opposite to } \vec{m} \text{, North to South)}
\end{aligned}$$
Board Practice 4: Magnetisation Parameters
Question: A solenoid with $1000\text{ turns/m}$ carries a current of $2.0\text{ A}$ with an iron core of relative permeability $\mu_r = 400$. Calculate: (a) Magnetic Intensity $H$, (b) Susceptibility $\chi$, (c) Magnetisation $M$, (d) Total magnetic field $B$.
Solution:
Given: $n = 1000\text{ m}^{-1}$, $I = 2.0\text{ A}$, $\mu_r = 400$.
(a) Magnetic Intensity $H$:
$$H = n I = 1000 \times 2.0 = \mathbf{2 \times 10^3 \text{ A m}^{-1}}$$
(b) Magnetic Susceptibility $\chi$:
$$\chi = \mu_r - 1 = 400 - 1 = \mathbf{399}$$
(c) Magnetisation $M$:
$$M = \chi H = 399 \times (2 \times 10^3) = \mathbf{7.98 \times 10^5 \text{ A m}^{-1}}$$
(d) Total Magnetic Field $B$:
$$B = \mu_r \mu_0 H = 400 \times (4\pi \times 10^{-7}) \times (2 \times 10^3) = 32\pi \times 10^{-2} \approx \mathbf{1.0 \text{ T}}$$
VARDAAN COMET