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Theory · Physics Circle Master Book

Magnetic Properties

01
CONCEPT

Bar Magnet & Magnetic Moment

In one line
A magnet is a dipole: two equal and opposite poles a small distance apart, with moment M from S to N.
MOLECULAR THEORY

22.1.1 What is inside a magnet

  1. Weber and Ewing: every molecule is a tiny magnet.
  2. Unmagnetised: molecular magnets are random, net moment zero.
  3. Magnetised: they line up, so a net moment appears.
UnmagnetisedMagnetised
Molecular magnets
BAR MAGNET

22.1.2 Effective and geometric length

  1. Poles are not exactly at the ends.
  2. Effective length is the distance between the poles.
Le = 2l = 56 Lg
  1. Semicircular magnet: Lg = πR, Le = 2R.
SNLe = 2lLgSN2R
Effective and geometric lengths
PROPERTIES

22.1.3 Properties of a magnet

  1. Directive: a freely suspended magnet rests in the N-S direction (magnetic meridian).
NSSNmagnetic meridian
Freely suspended magnet
  1. No monopole: break a magnet and each piece has two poles.
SNSNSNSN
Each piece is a complete magnet
  1. Field lines are closed loops: S to N inside, N to S outside.
  2. Repulsion is the sure test of magnetism: attraction also happens with plain iron.
SN
Lines form closed loops
NS
Imaginary isolated poles
PQQPattractno attraction
Only Q is a magnet
POLE STRENGTH

22.1.4 Pole strength (m)

  1. Scalar. N pole +m, S pole −m.
  2. SI unit: A m. Dimensions: [LA].
  3. Depends on material and area of cross-section.
  4. Does not depend on length.
A morem moreA less, m less
More area, more pole strength
MAGNETIC MOMENT

22.1.5 Magnetic moment (M)

M = m(2l)
  1. Vector from S to N.
  2. SI unit: A m², also J/T. Dimensions: [AL²].
SN−m+mM from S to N
Direction of M
CURRENT LOOP

22.1.6 Current loop as a magnet

M = NIA
  1. Direction: curl fingers along I; thumb gives M.
  2. Solenoid = many dipoles in a line.
  3. Toroid and straight wire: M = 0.
  4. Atoms with all electrons paired: M = 0.
current IM
Moment of a loop
=
Solenoid behaves as a bar magnet
  1. A wire of length L bent into one circle:
M = IL²4π
  1. Charge q moving in a circle of radius R at speed v:
M = qvR2
  1. Electron: L and M are opposite, angle 180°.
  2. Bohr magneton is the natural unit:
μB = eh4πm = 9.27 × 10⁻²⁴ A m²
CUTTING

22.1.7 Cutting a magnet

Cut into n partsm′M′
across the lengthmMn
along the lengthmnMn
BENDING

22.1.8 Bending a magnet

  1. m is unchanged. Use the new end-to-end distance.
M′ = M × end-to-end distancelength
SNstraight: LSN2rsemicircle
New length is the chord
ShapeM′
semicircle2Mπ
arc of 60°3Mπ
bent at the middle at 60°M2
bent at the middle at 90°M√2
JEE · NEET

Corner

Cutting a rectangular magnet into n equal parts
LbL′, b′
Cut along and across the length at the same time
  1. Applies when cut both along and across, n parts in all.
L′ = L√n  ·  b′ = b√n  ·  w′ = wn
m′ = m√n  ·  M′ = Mn
I = w L² + b²12  ·  I′ = In²
Cutting a thin magnet across its length
L′ = Ln  ·  w′ = wn  ·  M′ = Mn  ·  I′ = In³
Practise PYQs on this topicBar Magnet & Magnetic Moment · 15 questions with solutions
02
CONCEPT

B, H, I, χ and μ: The Five Quantities

In one line
The current sets H, the material answers with I, and together they give the total field B.
CurrentH = niMaterialI = χHTotalB = μH
How the quantities link up
FLUX AND B

22.2.1 Flux and magnetic induction (B)

  1. Flux φ: number of lines through a surface. SI: weber. CGS: maxwell.
  2. 1 Wb = 10⁸ maxwell.
  3. B is flux per unit normal area. It is a vector.
B = φA
  1. SI: tesla (Wb/m²). CGS: gauss.
  2. 1 T = 10⁴ G.
area A
Flux through area A
PERMEABILITY

22.2.2 Permeability (μ)

  1. How easily lines pass through a material.
μ = μ₀μr  ·  μr = BB₀
  1. μ₀ = 4π × 10⁻⁷ T m/A.
  2. μr has no unit.
In airIn soft iron
Lines crowd into soft iron
MAGNETISING FIELD

22.2.3 Magnetising field (H)

H = Bμ
  1. SI: A/m. CGS: oersted.
  2. 1 oersted = 80 A/m.
ROD IN A FIELD

22.2.4 Iron rod in a field

  1. Rod in H, flux φ, area A. Convert cm² to m² first.
μ = BH = φAH
MAGNETISATION

22.2.5 Intensity of magnetisation (I)

  1. Moment per unit volume, or pole strength per unit area.
  2. Vector. SI unit: A/m.
I = MV = mA
  1. Volume from mass: V = massdensity.
  2. Volume of a bar: V = length × area.
  3. N atoms, moment μ each: total M = Nμ.
SUSCEPTIBILITY

22.2.6 Susceptibility (χ)

χ = IH
  1. Scalar with no unit.
  2. How easily the material is magnetised.
RELATIONS

22.2.7 Relations between them

B = μ₀(H + I)
B = μ₀H(1 + χ)
μr = 1 + χ
μ = μ₀(1 + χ)
  1. Note: in the question, I may be printed as M. Read which one it means: moment or magnetisation.
  2. μ₀4π = 10⁻⁷ SI.
UNITS

22.2.8 Units

QuantitySI unitType
φWbscalar
BT = Wb/m²vector
HA/mvector
IA/mvector
MA m²vector
μT m/A = N/A²scalar
χ, μrnonescalar
Practise PYQs on this topicMagnetisation: B, H, M, χ & μ · 45 questions with solutions
03
CONCEPT

Solenoid & Toroid with a Core

In one line
A core multiplies the field of a solenoid by μr; the increase is simply χ as a fraction.
SOLENOID

22.3.1 Solenoid

  1. n = turns per metre, i = current.
  2. H depends only on the current, not on the core.
H = ni
B₀ = μ₀ni
B = μ₀μrni = μ₀ni(1 + χ)
core: χ, μᵣiB
Core inside the winding
  1. N turns over length L: n = NL. Convert cm to m.
  2. Diameter of the tube does not enter.
TOROID

22.3.2 Toroid

n = N2πr
r
Toroid of mean radius r
  1. Field lines stay inside the toroid; there are no poles.
MAGNETISATION OF THE CORE

22.3.3 Magnetisation of the core

I = χH = (μr − 1) ni
  1. When μr is large, I ≈ μrni.
PERCENTAGE INCREASE

22.3.4 Percentage increase in B

B − B₀B₀ = χ
% increase = χ × 100
  1. Fractional increase: χ itself.
  2. Percentage: multiply by 100.
  3. Do not add 1: the answer is not (1 + χ) × 100.
ROD WITH A WINDING

22.3.5 Rod with a winding

  1. Step 1: H = ni.
  2. Step 2: B = φA.
  3. Step 3:
μr = Bμ₀H
Practise PYQs on this topicMagnetisation: B, H, M, χ & μ · 45 questions with solutions
04
CONCEPT

Field of a Magnet & Magnet in a Uniform Field

In one line
A magnet in a uniform field feels no net force, only a torque that turns it towards the field.
FIELD OF A SHORT MAGNET

22.4.1 Field of a short magnet

SNaBaeBeMaxial lineequatorial line
Axial and equatorial points
axial: Ba = μ₀4π 2Mr³
equatorial: Be = μ₀4π Mr³
  1. Ba : Be = 2 : 1 at equal distance.
  2. Ba is along M; Be is opposite to M.
  3. Both vary as 1r³.
  4. Two magnets at right angles: add the fields as vectors.
TORQUE

22.4.2 Torque (τ)

BSNmBmBθ
Forces mB on the poles form a couple
τ = MB sin θ
τ = M × B
  1. Each pole feels F = mB.
  2. Uniform field: net force is zero, torque may not be.
  3. Non-uniform field: both force and torque.
  4. A current loop behaves the same, with M = NIA.
  1. θ is the angle between M and B. Plane of a coil at α to B means θ = 90° − α.
WORK

22.4.3 Work done

W = MB(cos θ₁ − cos θ₂)
  1. From the field direction, θ₁ = 0°:
W = MB(1 − cos θ)
POTENTIAL ENERGY

22.4.4 Potential energy (U)

U = −MB cos θ = −M · B
θτUW from 0°State
0°0−MB0stable
60°√32 MB−MB2MB2
90°MB0MB
180°0+MB2MBunstable
BSNθ = 0°: stableBNSθ = 180°: unstable
Two positions of equilibrium
W AND τ TOGETHER

22.4.5 W and τ together

  1. Work W to turn through θ. Torque to hold it there:
τ = W sin θ1 − cos θ
  1. θ = 60°: τ = √3 W.
  2. θ = 30°: τ = W2 − √3.
  3. Given τ at angle θ: find MB, then 2MB or U.
JEE · NEET

Corner

Coulomb’s law of magnetism
F = μ₀4π m₁m₂r²
  1. μ₀4π = 10⁻⁷ Wb per A m.
  2. In CGS, the constant is 1.
Exact field of a bar magnet of length 2l
Ba = μ₀4π 2Mr(r² − l²)²
Be = μ₀4π M(r² + l²)1.5
General position and force between dipoles
Bg = μ₀4π Mr³ √(3cos²θ + 1)
  1. Force between two short magnets, end-on: F ∝ 1x⁴.
  2. Gauss’s law: net flux through any closed surface is zero, ∮B · dS = 0.
Practise PYQs on this topicField of a Magnet & Magnet in a Uniform Field · 33 questions with solutions
05
CONCEPT

Earth’s Magnetism

In one line
The earth behaves like a huge magnet with three elements: declination, dip and BH.
THE EARTH AS A MAGNET

22.5.1 The earth as a magnet

geographic axisSNmagnetic axisequator
Magnetic and geographic axes
  1. Likely cause: currents in the molten core.
  2. Magnetic meridian: vertical plane through the magnetic axis.
  3. The field changes from place to place and with time.
THREE ELEMENTS

22.5.2 Three elements

  1. Declination θ: angle between geographic and magnetic meridians.
geographic Nmagnetic Nθdeclination θ
Declination
  1. Dip δ: angle of B with the horizontal in the magnetic meridian.
  2. BH: horizontal component of the field.
magnetic meridianBHBVBδ
Dip and the two components
BH = B cos δ
BV = B sin δ
B = √(BH² + BV²)  ·  tan δ = BVBH
POLES AND EQUATOR

22.5.3 Poles and equator

PlaceδBHBV
magnetic equator0°B0
magnetic pole90°0B
  1. At a pole, a horizontal compass needle can stay in any direction.
  2. Dip is positive in the northern hemisphere, negative in the southern.
JEE · NEET

Corner

Apparent dip
BHBH cos ββ
Plane at angle β to the magnetic meridian
  1. In another vertical plane, BV stays the same; BH becomes BH cos β.
tan δ′ = tan δcos β
  1. Apparent dip is greater than or equal to true dip.
  2. Two vertical planes at right angles:
cot²δ = cot²δ₁ + cot²δ₂
Earth as a dipole
Bequator = μ₀4π MR³
  1. Magnetic axis makes about 11.5° with the geographic axis.
  2. At magnetic latitude λ:
tan δ = 2 tan λ
  1. Southern half of the earth’s magnet lies in the northern hemisphere.
  2. Total intensity: pole 0.66 oersted, equator 0.33 oersted.
Neutral points
  1. Resultant field is zero: the magnet’s field cancels BH.
  2. N pole towards north: neutral points on the equatorial line.
μ₀4π Mr³ = BH
  1. N pole towards south: neutral points on the axial line.
μ₀4π 2Mr³ = BH
Magnetic maps
LineJoins places of equal
isogonicdeclination
agoniczero declination
isoclinicdip
acliniczero dip (magnetic equator)
isodynamicBH
Tangent law and tangent galvanometer
SNmBmBHmBmBHθ
Needle rests under B and BH at right angles
B = BH tan θ
coilcompass box
Circular coil with a compass at its centre
  1. Coil is kept in the magnetic meridian.
μ₀Ni2r = BH tan θ
i = k tan θ  ·  k = 2rBHμ₀N
  1. k is the reduction factor. It measures current.
Deflection magnetometer
Works on the tangent law
Tan A: BH tan θ = μ₀4π 2Mr³
Tan B: BH tan θ = μ₀4π Mr³
M₁M₂ = tan θ₁tan θ₂  ·  null method: M₁M₂ = (d₁d₂)³
Practise PYQs on this topicEarth's Magnetism · 16 questions with solutions
06
CONCEPT

Vibration Magnetometer

In one line
A magnet displaced from the field direction performs S.H.M.: stronger M or B, faster oscillation.
TIME PERIOD

22.6.1 Time period

SNBHtorsion head
Magnet oscillating in BH
T = 2π √(IMB)
ν = 12π √(MBI)
I = wL²12
  1. Here I is moment of inertia, not magnetisation.
  2. Small angle: τ = −MBθ gives S.H.M.
  3. Frequency from a count: ν = oscillations per second.
FINDING M OR B

22.6.2 Finding M or B

M = 4π²IBT²  ·  B = 4π²IMT²
COMPARING

22.6.3 Comparing fields and magnets

Same magnet: B₁B₂ = T₂²T₁² = ν₁²ν₂²
Same size, same B: M₁M₂ = T₂²T₁²
Same B: M₁M₂ = I₁I₂ × T₂²T₁²
  1. An added field opposite to BH gives net B = BH − B′; along it, BH + B′.
  2. Field of a magnet at a new place: use the total B, not the increase.
SUM AND DIFFERENCE

22.6.4 Sum and difference positions

SNSNSum: M₁ + M₂SNNSDifference: M₁ − M₂
Like poles together and opposed
Ts = 2π √(I₁ + I₂(M₁ + M₂)B)
Td = 2π √(I₁ + I₂(M₁ − M₂)B)
M₁M₂ = Td² + Ts²Td² − Ts²
  1. Moment of inertia always adds.
  2. Td is greater than Ts.
  3. Equal moments, difference position: Mnet = 0, so Td = ∞.
CHANGES TO THE MAGNET

22.6.5 Changes to the magnet

ChangeIMT′
mass × 4 (same size)4IM2T
thin, cut across into nIn³MnTn
cut along and across, n partsIn²MnT√n
  1. Heating: M falls, so T rises.
  2. At the Curie temperature it stops oscillating.
JEE · NEET

Corner

Compass needle at two places of different dip
  1. Only BH = B cos δ makes the needle oscillate.
ν₁²ν₂² = B₁ cos δ₁B₂ cos δ₂
Ratio of a magnet’s field to BH
SNSNdBHprimarysecondary
Secondary magnet placed near the primary
BBH = (ν′ν)² − 1
Practise PYQs on this topicVibration Magnetometer · 20 questions with solutions
07
CONCEPT

Dia, Para & Ferromagnetic Materials

In one line
Materials differ by what the atoms do: dia opposes the field, para weakly follows it, ferro strongly follows it.
ORIGIN

22.7.1 Origin of magnetism

  1. Diamagnetic: paired electrons, atom has no net moment. Moment is induced by orbital motion, opposite to the field. Every material has this property.
  2. Paramagnetic: unpaired spins, each atom has a moment. Without a field the moments are random; a field lines them up a little.
  3. Ferromagnetic: atoms align together in groups called domains.
no fieldfield applied: B →
Paramagnetic atoms
DOMAINS

22.7.2 Domains

  1. A domain is a region fully magnetised to saturation.
  2. Unmagnetised: domains point in different directions.
  3. In a field: favourable domains grow, others shrink, and domains rotate.
  4. Atoms interact and align by themselves.
unmagnetisedmagnetised
Domains without and with a field
BEHAVIOUR

22.7.3 Behaviour in a field

diaparaferro
Field lines near the three materials
  1. Dia: lines bend away. Para: lines come in slightly. Ferro: lines crowd in strongly.
  2. Rod in a uniform field: dia sets perpendicular; para and ferro set parallel.
SNdia: fallsSNpara: risesSNferro: rises a lot
Liquid in a U-tube with one arm between poles
COMPARISON

22.7.4 Comparison

PropertyDiaParaFerro
χsmall, negativesmall, positivelarge, positive
range−1 ≤ χ < 00 < χ < εχ ≫ 1
μr0 ≤ μr < 1slightly above 1μr ≫ 1
B insideB < B₀B > B₀B ≫ B₀
I and Hoppositesame waysame way, large
magnetweakly repelledweakly attractedstrongly attracted
movesstrong to weak fieldweak to strongweak to strong
atom’s momentzeronon-zeronon-zero
gas between polesexpands across fieldexpands along fieldrapidly along field
examplesCu, Ag, Au, Zn, Bi, Sb, NaCl, H₂O, air, diamondAl, Mn, Pt, Na, CaCl₂, O₂Fe, Co, Ni, Fe₃O₄
HIdiaHIparaHIferro
I against H
SUPERCONDUCTOR

22.7.5 Superconductor: a perfect diamagnet

  1. χ = −1 and μr = 0.
  2. Field inside is zero: lines are fully expelled.
  3. The relative permittivity of any material stays above 1.
Practise PYQs on this topicMagnetic Materials & Curie Law · 46 questions with solutions
08
CONCEPT

Curie Law & Curie Temperature

In one line
Heat disturbs alignment: χ of para falls as T rises, ferro turns para at Tc, and dia does not care.
CURIE LAW

22.8.1 Curie law

χ = CT  ·  χT = constant
I = C BT
  1. C is the Curie constant; T is in kelvin.
  2. Convert °C: add 273.
  3. Magnetisation rises with B and falls with T.
χ₂χ₁ = T₁T₂  ·  I₂I₁ = B₂B₁ × T₁T₂
χ AGAINST T

22.8.2 χ against T

TχdiaTχparaTχTcferro
Dia is flat and negative; para falls smoothly; ferro turns para at Tc
  1. χ against 1T for a paramagnet: straight line through the origin.
  2. Dia: χ does not depend on T (except Bi at low temperature).
CURIE TEMPERATURE

22.8.3 Curie temperature (Tc)

  1. Above Tc a ferromagnet becomes paramagnetic.
  2. Heating makes the domains random; atoms keep their own moments.
  3. It does not become diamagnetic.
MaterialTc
Ni358°C
Fe770°C
Co1120°C
CURIE-WEISS LAW

22.8.4 Curie-Weiss law

χ = CT − Tc
TχTc
Valid for T greater than Tc
Practise PYQs on this topicMagnetic Materials & Curie Law · 46 questions with solutions
09
CONCEPT

Hysteresis, Soft Iron vs Steel & Shielding

In one line
A ferromagnet remembers: B lags behind H, and the loop shape decides what the material is used for.
HYSTERESIS CURVE

22.9.1 Hysteresis curve

  1. Only ferromagnetic materials show hysteresis.
HBACDEFGO
O to A: first magnetisation. A C D E F G A: one full cycle
  1. OA: first magnetisation up to saturation at A.
  2. OC: B left at H = 0: retentivity (residual magnetism).
  3. OD: reverse H that makes B zero: coercivity.
  4. E: saturation in the reverse direction.
  5. Point on the B axis = retentivity. Point on the H axis = coercivity.
DEMAGNETISING

22.9.2 Demagnetising current

  1. Put the magnet inside a solenoid, with H opposite to its magnetisation.
Hc = ni  ·  i = Hcn
ENERGY LOSS

22.9.3 Energy loss

  1. Area of the loop is the energy lost per unit volume per cycle.
Energy loss = VAft
  1. V: volume, A: loop area, f: frequency, t: time.
SOFT IRON VS STEEL

22.9.4 Soft iron vs steel

HIsoft ironHIsteel
Narrow and wide loops
PropertySoft ironSteel
loop areasmalllarge
energy losslowhigh
retentivitylowhigh
coercivitylowhigh
permeabilityhighlower
magnetise and demagnetiseeasilynot easily
used forelectromagnet, transformer, dynamo core, tape recorderpermanent magnet
  1. Electromagnet: high μ, low retentivity, low coercivity.
  2. Permanent magnet: high retentivity, high coercivity.
  3. Transformer core: narrow loop, low loss.
MAGNETIC SHIELDING

22.9.5 Magnetic shielding

  1. Put the instrument in a soft iron enclosure.
  2. High μ draws the lines into the wall.
  3. Field inside is nearly zero.
  4. Copper, aluminium or plain insulation will not do it.
B = 0
Lines go through the iron wall
Practise PYQs on this topicHysteresis, Soft Iron vs Steel & Shielding · 20 questions with solutions
EXAM-DAY CARD

Just see this before the exam

Magnetism
M = m(2l) = NIA
moment
B = μ₀(H + I)
total field
μr = 1 + χ
permeability
U = −MB cos θ
energy
T = 2π√(IMB)
oscillation
χ = CT
Curie law
FORMULA SHEET

Whole chapter on one page

Bar magnet

M = m(2l), Le = 56 Lg

loop: M = NIA

semicircle: M′ = 2Mπ

cut across: m same, M′ = Mn

B, H, I, χ, μ

B = φA, μ = BH

I = MV, χ = IH

B = μ₀(H + I)

μr = 1 + χ

With a core

H = ni

B = μ₀ni(1 + χ)

I = (μr − 1)ni

% increase = χ × 100

Magnet in a field

Ba = μ₀4π 2Mr³, Be = μ₀4π Mr³

τ = MB sin θ

W = MB(1 − cos θ)

U = −MB cos θ

Earth & oscillation

BH = B cos δ, BV = B sin δ

T = 2π√(IMB)

B₁B₂ = T₂²T₁²

M₁M₂ = Td² + Ts²Td² − Ts²

Materials

dia: χ < 0; para: small χ > 0

ferro: χ ≫ 1

χ = CT, χ = CT − Tc

i = Hcn

TOP MISTAKES

Avoid these = easy marks

MISTAKE ✗

Using χ + 1 as the % increase

Curie law with T in °C

Above Tc: ferro becomes dia

Electromagnet from steel

CORRECT ✓

% increase = χ × 100

T always in kelvin

Above Tc: ferro becomes para

Soft iron: low retentivity and coercivity