Summary & Master Memory Capsule

Master Formula Sheet

Everything computable in this chapter, on one card:

#	Result	Formula
1	Magnetic moment of coil/solenoid	$m = NIA$ (unit: A m$^2$ = J/T)
2	Far axial field	$B_A = \dfrac{\mu_0}{4\pi}\dfrac{2m}{r^3}$ (parallel to $\vec{m}$)
3	Far equatorial field	$\vec{B}_E = -\dfrac{\mu_0}{4\pi}\dfrac{\vec{m}}{r^3}$, so $|\vec{B}_E| = \dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}$
4	General point (JEE)	$B = \dfrac{\mu_0 m}{4\pi r^3}\sqrt{1+3\cos^2\theta}$, where $\theta$ is measured from the magnetic axis
5	Torque	$\vec{\tau} = \vec{m}\times\vec{B}$, $\tau = mB\sin\theta$
6	Potential energy	$U = -\vec{m}\cdot\vec{B} = -mB\cos\theta$ (zero at 90 degrees)
7	Work in slow rotation by an external agent	$W_{\text{ext}} = \Delta U = mB(\cos\theta_1 - \cos\theta_2)$; flip from stable: $2mB$
8	Oscillating needle (JEE)	$T = 2\pi\sqrt{\mathcal{I}/(mB)}$
9	Gauss's law	$\displaystyle \oint_S \vec{B}\cdot d\vec{S} = 0$ (any closed surface)
10	Magnetisation	$M = m_{\text{net}}/V$ (A/m)
11	Magnetic intensity	$H = B/\mu_0 - M$; long solenoid: $H = nI$ (A/m)
12	Master relation	$B = \mu_0(H + M) = \mu_0(1+\chi)H = \mu_0\mu_r H = \mu H$
13	Response chain	$M = \chi H$; $\mu_r = 1+\chi$; $\mu = \mu_0\mu_r$

Constants: $\mu_0/4\pi = 10^{-7}$ T m/A; 1 gauss $= 10^{-4}$ T.

The Electrostatic Analog Dictionary

Translate any Chapter 1 dipole result into magnetism:

Electrostatics	Magnetism
$\vec{E}$	$\vec{B}$
$\vec{p}$	$\vec{m}$
$1/(4\pi\varepsilon_0)$	$\mu_0/(4\pi)$
$-\dfrac{\vec{p}}{4\pi\varepsilon_0 r^3}$ (equatorial)	$-\dfrac{\mu_0\vec{m}}{4\pi r^3}$
$\dfrac{2\vec{p}}{4\pi\varepsilon_0 r^3}$ (axial)	$\dfrac{\mu_0}{4\pi}\dfrac{2\vec{m}}{r^3}$
$\vec{p}\times\vec{E}$ (torque)	$\vec{m}\times\vec{B}$
$-\vec{p}\cdot\vec{E}$ (energy)	$-\vec{m}\cdot\vec{B}$

And the deep difference the dictionary cannot translate: electric charges exist in isolation; magnetic monopoles do not. Hence electrostatic field lines start on positive charges and end on negative charges (or at infinity), while magnetic field lines are closed loops — and Gauss's law for magnetism has a zero on the right-hand side.

Materials at a Glance

Property	Diamagnetic	Paramagnetic	Ferromagnetic
$\chi$	ordinary: small negative (~$-10^{-5}$); ideal superconductor: $-1$	small + (~$+10^{-5}$)	$\gg 1$ (can be $10^3$ or more)
$\mu_r$	ordinary: just less than 1; ideal superconductor: 0	just greater than 1	$\gg 1$
Field lines	expelled	weakly concentrated	strongly concentrated
In non-uniform field	strong $\to$ weak	weak $\to$ strong (weakly)	weak $\to$ strong (strongly)
Atomic origin	zero net moment; induced opposite moment (Lenz)	permanent moments vs thermal chaos	permanent moments aligned in domains (typical microscopic size ~$1\,\mu\text{m}$; about $10^{11}$ atoms)
Temperature	nearly independent	$M$ rises on cooling (saturates at very low temperature/high field)	ferromagnetism disappears above the Curie temperature; material becomes paramagnetic
Examples	Bi, Cu, Pb, Si, N$_2$(STP), H$_2$O, NaCl	Al, Na, Ca, O$_2$(STP), CuCl$_2$	Fe, Co, Ni, Gd

Special cases to quote: superconductor = perfect diamagnet ($\chi = -1$, $\mu_r = 0$, $B = 0$ inside, Meissner effect, maglev); hard ferromagnets (Alnico, lodestone) keep magnetisation — permanent magnets; soft iron loses it — switchable cores.

One-Glance Revision Flow

The story of the chapter in six steps:

1. Magnets behave like dipoles — two poles, N-S alignment, no monopoles ever (cutting gives smaller magnets).
2. A bar magnet is equivalent to a solenoid (Ampere's hypothesis): $m = NIA$; both have the same magnetic dipole far field, and on the axial line $B = \dfrac{\mu_0}{4\pi}\dfrac{2m}{r^3}$.
3. In a uniform field: no net force, torque magnitude $mB\sin\theta$, energy $-mB\cos\theta$; stable at 0 degrees, unstable at 180 degrees; flip costs $2mB$.
4. The electrostatic analog hands you every dipole-field formula free of charge.
5. Gauss's law: net magnetic flux through any closed surface is zero — the formal 'no monopoles'. Use it to reject impossible field-line diagrams with isolated starting/ending points. Circulation questions involving currents are handled by Ampere's law.
6. Matter responds: $H = nI$ is the cause in a long solenoid, $M = \chi H$ the response, $B = \mu_0(H+M)$ the total. The sign and size of $\chi$ sorts all materials into dia ($<0$), para (small $+$), ferro ($\gg 1$).

Morning-of-exam checklist: equatorial field is antiparallel to $\vec{m}$ … axial = 2 x equatorial … U is zero at 90 degrees, not at 0 … $T \propto 1/\sqrt{B}$ … for a long solenoid with fixed current, $H=nI$ is set by free current and a linear core changes $B$ by the factor $\mu_r$ … $\chi$ and $\mu_r$ are dimensionless … oxygen para, nitrogen dia … flux through ANY closed surface = 0 … superconductor: $\chi = -1$ exactly.

Now go score. This chapter rewards memory with marks faster than any other in Class 12 Physics.