Chapter 20: Magnetism and Matter

Introduction

Key Points

• Magnetism: Property of materials like iron, nickel, cobalt to attract certain elements.
• Earth acts as a giant magnet; magnetic field points approximately from geographic south to north.
• Bar magnet suspended freely aligns north-south; ends called north and south poles.
• Like poles repel; unlike poles attract.
• Magnetic monopoles do not exist; breaking a magnet produces smaller dipoles.
• Magnets can be made from iron and its alloys.

Solved Examples

Ex 1

• Q: What happens when two north poles of magnets are brought close?
• A: They repel each other due to like poles repulsion.

Practice Set

• Q1: Which of the following elements is attracted by a magnet?
  • A. Copper
  • B. Nickel
  • C. Aluminum
  • D. Zinc
• Q2: What is the direction of Earth's magnetic field?
  • A. Geographic north to south
  • B. Geographic south to north
  • C. East to west
  • D. West to east
• Q3: What happens if a bar magnet is broken into two halves?
  • A. Isolated north and south poles appear
  • B. Two smaller magnets with north and south poles form
  • C. Magnetism disappears
  • D. Only one pole remains

Answer Key

• Q1: B - Nickel is a ferromagnetic element attracted by magnets.
• Q2: B - Earth's magnetic field points approximately from geographic south to north.
• Q3: B - Breaking a magnet produces two smaller dipoles, each with north and south poles.

Memory Aids

• "Like poles repel, unlike poles attract" - basic magnetism rule.
• Earth's magnetic field: South to North direction.
• Magnetic monopoles do not exist; magnets always dipoles.

The Bar Magnet

Properties

• Natural dipole producing magnetic field around it.
• Attracts small pieces of iron, nickel, cobalt.
• Poles are ends where magnetic force is maximum.
• Freely suspended bar magnet aligns north-south; north pole points geographic north.
• Like poles repel; unlike poles attract.
• Magnetic poles always exist in pairs; monopoles do not exist.
• Repulsion test distinguishes magnet from iron piece.

Solved Examples

Ex 1

• Q: What is the pole called that points towards Earth's geographic north?
• A: North pole (N) of the bar magnet.

Practice Set

• Q1: What is the shape of a bar magnet?
  • A. Circular
  • B. Cylindrical
  • C. Rectangular
  • D. Bar-shaped
• Q2: Which pole of a bar magnet points towards Earth's geographic south?
  • A. North pole
  • B. South pole
  • C. Either pole
  • D. None
• Q3: What is the surest test to distinguish a magnet from a piece of iron?
  • A. Attraction
  • B. Repulsion
  • C. Weight
  • D. Color

Answer Key

• Q1: D - Bar magnet is bar-shaped.
• Q2: B - South pole points towards Earth's geographic south.
• Q3: B - Repulsion occurs only between magnets, not iron pieces.

Memory Aids

• Bar magnet poles: N and S at ends.
• Freely suspended magnet aligns N-S.
• Repulsion confirms magnet presence.

The Magnetic Field Lines

Key Concepts

• Magnetic field lines form closed continuous loops around magnets or solenoids.
• Tangent to field line at a point shows direction of magnetic field there.
• Density of field lines per unit area indicates field strength.
• Field lines never intersect; intersection would mean ambiguous field direction.
• Field lines of bar magnet and current-carrying solenoid are similar.

Solved Examples

Ex 1

• Q: What does the closeness of magnetic field lines indicate?
• A: Stronger magnetic field.

Practice Set

• Q1: Magnetic field lines around a magnet form:
  • A. Open lines
  • B. Closed loops
  • C. Straight lines
  • D. Random curves
• Q2: What does the tangent to a magnetic field line represent?
  • A. Magnitude of field
  • B. Direction of field
  • C. Force on charge
  • D. None
• Q3: Why don't magnetic field lines intersect?
  • A. They do intersect
  • B. To avoid ambiguous field direction
  • C. Because of repulsion
  • D. Due to attraction

Answer Key

• Q1: B - Magnetic field lines form closed loops.
• Q2: B - Tangent shows direction of magnetic field at that point.
• Q3: B - Intersection would cause non-unique field direction.

Memory Aids

• Field lines: Closed loops, no intersections.
• Closer lines = stronger field.
• Tangent = field direction.

Bar Magnet as an Equivalent Solenoid

Key Points

• Bar magnet behaves like a finite solenoid producing similar magnetic field.
• Magnetic field at far axial point: \( B = \frac{\mu_0}{4\pi} \frac{2m}{r^3} \)
• \( m \) is magnetic moment, \( r \) is distance from magnet.
• Experimental results confirm similarity of bar magnet and solenoid fields.

Solved Examples

Ex 1

• Q: Write the formula for magnetic field at far axial point of bar magnet.
• A: \( B = \frac{\mu_0}{4\pi} \frac{2m}{r^3} \)

Practice Set

• Q1: What does \( m \) represent in magnetic field formula of bar magnet?
  • A. Magnetic field
  • B. Magnetic moment
  • C. Distance
  • D. Permeability
• Q2: The magnetic field of a bar magnet at far point varies as?
  • A. \( 1/r \)
  • B. \( 1/r^2 \)
  • C. \( 1/r^3 \)
  • D. \( r^3 \)
• Q3: Which device produces a magnetic field similar to a bar magnet?
  • A. Straight wire
  • B. Solenoid
  • C. Capacitor
  • D. Resistor

Answer Key

• Q1: B - \( m \) is magnetic moment.
• Q2: C - Field varies as inverse cube of distance.
• Q3: B - Solenoid produces similar magnetic field as bar magnet.

Memory Aids

• Bar magnet ≈ solenoid magnetic field.
• Field \( \propto 1/r^3 \) at far point.
• Formula: \( B = \frac{\mu_0}{4\pi} \frac{2m}{r^3} \).

The Dipole in a Uniform Magnetic Field

Key Concepts

• Magnetic dipole moment \( \vec{m} \) experiences torque \( \vec{\tau} = \vec{m} \times \vec{B} \).
• Magnitude of torque: \( \tau = mB \sin \theta \), \( \theta \) is angle between \( \vec{m} \) and \( \vec{B} \).
• Torque tends to align dipole with magnetic field.
• Potential energy of dipole in field: \( U = -\vec{m} \cdot \vec{B} = -mB \cos \theta \).
• Work done to rotate dipole through angle \( \theta \) is potential energy change.
• Analogy exists between electric and magnetic dipoles in fields.

Solved Examples

Ex 1

• Q: Calculate torque on magnetic dipole with \( m=2 \mathrm{A\cdot m^2} \), \( B=0.5 \mathrm{T} \), \( \theta=30^\circ \).
• A: \( \tau = mB \sin \theta = 2 \times 0.5 \times \sin 30^\circ = 0.5 \mathrm{Nm} \).

Practice Set

• Q1: Torque on magnetic dipole is given by?
  • A. \( mB \cos \theta \)
  • B. \( mB \sin \theta \)
  • C. \( m/B \sin \theta \)
  • D. \( mB \tan \theta \)
• Q2: Potential energy of magnetic dipole in field is?
  • A. \( mB \sin \theta \)
  • B. \( -mB \cos \theta \)
  • C. \( mB \cos \theta \)
  • D. \( -mB \sin \theta \)
• Q3: What happens to dipole in uniform magnetic field?
  • A. Aligns perpendicular to field
  • B. Aligns parallel to field
  • C. No effect
  • D. Moves randomly

Answer Key

• Q1: B - Torque magnitude is \( mB \sin \theta \).
• Q2: B - Potential energy is \( -mB \cos \theta \).
• Q3: B - Dipole aligns parallel to magnetic field.

Memory Aids

• Torque: \( \vec{\tau} = \vec{m} \times \vec{B} \).
• Potential energy: \( U = -\vec{m} \cdot \vec{B} \).
• Dipole aligns to minimize potential energy.

Magnetism and Gauss’s Law

Magnetic Flux Concept

• Magnetic field \(\vec{B}\) passes through a surface area \(\Delta S\).
• Unit vector \(\hat{n}\) is perpendicular to the surface.
• Angle \(\theta\) between \(\vec{B}\) and \(\hat{n}\) determines flux.
• Magnetic flux \(\Phi_B = B \Delta S \cos \theta\).
• Flux measures the amount of magnetic field passing through the surface.

Gauss's Law for Magnetism

• Net magnetic flux through any closed surface is zero: \(\oint \vec{B} \cdot d\vec{S} = 0\).
• Magnetic field lines form closed loops; no beginning or end.
• No magnetic monopoles exist (no isolated magnetic charges).
• Incoming magnetic flux equals outgoing magnetic flux.
• Dipoles or current loops are simplest magnetic sources.

Solved Examples

• Ex 1: Calculate magnetic flux through a surface of area 0.5 m² with \(B = 0.2\,T\) at \(30^\circ\).
  • Use \(\Phi_B = B A \cos \theta = 0.2 \times 0.5 \times \cos 30^\circ = 0.0866\,Wb\).
• Ex 2: Verify Gauss's law for a closed cube in uniform magnetic field \(B = 0.1\,T\).
  • Flux entering one face equals flux leaving opposite face; net flux \(= 0\).

Practice Set

• Q1: Magnetic flux through a surface of area 1 m² with magnetic field 0.3 T perpendicular to surface is:
  • A) 0 Wb
  • B) 0.3 Wb
  • C) 3 Wb
  • D) 0.03 Wb
• Q2: According to Gauss's law for magnetism, the net magnetic flux through a closed surface is:
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Depends on surface shape
• Q3: A magnetic field \(B = 0.5\,T\) is at \(60^\circ\) to a surface of area 2 m². Calculate magnetic flux.
  • A) 0.5 Wb
  • B) 1 Wb
  • C) 0.5 \(\sqrt{3}\) Wb
  • D) 1 \(\sqrt{3}\) Wb

Answer Key

• Q1: B) 0.3 Wb — Flux \(= B \times A \times \cos 0^\circ = 0.3 \times 1 \times 1 = 0.3\,Wb\).
• Q2: C) Zero — Gauss's law states net magnetic flux through closed surface is zero.
• Q3: C) 0.5 \(\sqrt{3}\) Wb — Flux \(= 0.5 \times 2 \times \cos 60^\circ = 1 \times 0.5 = 0.5\,Wb\). (Note: \(\cos 60^\circ = 0.5\))

Memory Aids

• Magnetic flux \(\Phi_B = B A \cos \theta\).
• Gauss's law for magnetism: \(\oint \vec{B} \cdot d\vec{S} = 0\).
• No magnetic monopoles; field lines always loop.
• Incoming flux = Outgoing flux for closed surfaces.

Magnetisation and Magnetic Intensity

Magnetising Field Intensity (H)

• Magnetising field intensity \( \vec{H} \) is the capability of a magnetic field to magnetise a material.
• Relation: \( \vec{B} = \mu \vec{H} \), where \( \mu \) is the permeability of the medium.
• Vector quantity; SI unit: ampere per meter (A m^-1).

Intensity of Magnetisation (I)

• Defined as net magnetic moment per unit volume: \( I = \frac{M}{V} \).
• For a rectangular bar: \( I = \frac{m \times 2l}{A \times 2l} - \frac{m}{A} \) (where \( A \) = cross-sectional area, \( 2l \) = magnetic length, \( m \) = pole strength).
• Vector quantity; SI unit: A m^-1; Dimensional formula: \( [M^0 L^{-1} T^0 A] \).

Magnetic Susceptibility (\( x_m \))

• Ratio of intensity of magnetisation to magnetising field: \( x_m = \frac{I}{H} \).
• Scalar quantity; dimensionless and unitless.
• Measures response of magnetic material to external field.

Magnetic Permeability (\( \mu \))

• Ratio of magnetic induction to magnetising field intensity: \( \mu = \frac{B}{H} \).
• Scalar quantity; SI unit: henry per meter (H m^-1); Dimensional formula: \( [M L T^{-2} A^{-2}] \).
• Indicates how much a magnetic field penetrates a material.

Relative Permeability (\( \mu_r \))

• Ratio of permeability of medium to free space: \( \mu_r = \frac{\mu}{\mu_0} \).
• Dimensionless and unitless.

Relation Between Magnetic Permeability and Susceptibility

• For linear magnetic materials: \( B = \mu_0 (H + I) = \mu H \).
• Therefore, \( \mu = \mu_0 (1 + x_m) \) or \( \mu_r = 1 + x_m \).

Solved Examples

• Ex 1: Calculate magnetic susceptibility if \( I = 0.02 \) A/m and \( H = 0.1 \) A/m.
  • \( x_m = \frac{I}{H} = \frac{0.02}{0.1} = 0.2 \).
• Ex 2: Find relative permeability if magnetic susceptibility \( x_m = 0.5 \).
  • \( \mu_r = 1 + x_m = 1 + 0.5 = 1.5 \).

Practice Set

• Q1: What is the SI unit of magnetising field intensity \( H \)?
  • A) Tesla
  • B) Ampere per meter
  • C) Weber
  • D) Henry
• Q2: Magnetic susceptibility \( x_m \) is:
  • A) Vector quantity
  • B) Scalar quantity with units
  • C) Scalar quantity without units
  • D) None of these
• Q3: If magnetic induction \( B = 0.3 \) T and magnetising field intensity \( H = 200 \) A/m, find magnetic permeability \( \mu \).
  • A) 0.0015 H/m
  • B) 0.6 H/m
  • C) 66 H/m
  • D) 150 H/m

Answer Key

• Q1: B) Ampere per meter
  Reason: SI unit of \( H \) is A m^-1.
• Q2: C) Scalar quantity without units
  Reason: Magnetic susceptibility is dimensionless scalar.
• Q3: A) 0.0015 H/m
  Reason: \( \mu = \frac{B}{H} = \frac{0.3}{200} = 0.0015 \) H/m.

Memory Aids

• \( B = \mu H \) relates magnetic induction and field intensity.
• \( x_m = \frac{I}{H} \) is dimensionless susceptibility.
• \( \mu_r = 1 + x_m \) links relative permeability and susceptibility.
• SI unit of \( H \) and \( I \) is A m^-1.

Magnetic Properties of Materials

Classification and Properties

• Materials classified by magnetisation intensity (I), susceptibility (\( \chi_m \)), and relative permeability (\( \mu_r \)).
• Diamagnetic: Caused by orbital electron motion; weak magnetism opposite to field; \( \chi_m < 0 \), \( \mu_r < 1 \); no permanent magnetic moment; examples: bismuth, copper, lead, water.
• Paramagnetic: Caused by electron spin; weak magnetism along field; \( \chi_m > 0 \), slightly \( \mu_r > 1 \); permanent moments randomly oriented; examples: aluminium, sodium, calcium, oxygen.
• Ferromagnetic: Due to magnetic domains; strong magnetism along field; large positive \( \chi_m \), \( \mu_r \gg 1 \); permanent moments aligned; examples: iron, cobalt, nickel.

Temperature Dependence of Susceptibility

• Diamagnetic: \( \chi_m \) small, negative, temperature independent.
• Paramagnetic: \( \chi_m \propto \frac{1}{T} \) (Curie law), decreases with temperature.
• Ferromagnetic: \( \chi_m \propto \frac{1}{T - T_C} \) (Curie-Weiss law) above Curie temperature \( T_C \); sharp peak at \( T_C \) (e.g., iron \( T_C = 770^\circ C \)).

Solved Examples

Ex 1

• Find relative permeability of a diamagnetic material with susceptibility \( \chi_m = -1.2 \times 10^{-5} \).
• \( \mu_r = 1 + \chi_m = 1 - 1.2 \times 10^{-5} = 0.999988 \).

Practice Set

• Q1: Which material has \( \mu_r \gg 1 \)?
  A) Copper
  B) Aluminium
  C) Iron
  D) Water
• Q2: Magnetic susceptibility of paramagnetic materials varies as:
  A) Constant
  B) \( \propto T \)
  C) \( \propto \frac{1}{T} \)
  D) \( \propto T^2 \)
• Q3: Which of the following is NOT ferromagnetic?
  A) Nickel
  B) Cobalt
  C) Sodium
  D) Iron

Answer Key

• Q1: C) Iron — Ferromagnetic materials have very high relative permeability.
• Q2: C) \( \propto \frac{1}{T} \) — Paramagnetic susceptibility follows Curie law.
• Q3: C) Sodium — Sodium is paramagnetic, not ferromagnetic.

Memory Aids

• Diamagnetic: "Dia" = opposite magnetism, \( \mu_r < 1 \).
• Paramagnetic: "Para" = parallel but weak, \( \chi_m \propto 1/T \).
• Ferromagnetic: "Ferro" = strong, domains aligned, \( \mu_r \gg 1 \).

Magnetic Field of a Bar Magnet

Field at Axial Position

• Magnetic field at point P due to north pole: \( B_1 = \frac{\mu_0}{4\pi} \frac{m}{(r - l)^2} \) (away from north pole)
• Due to south pole: \( B_2 = \frac{\mu_0}{4\pi} \frac{m}{(r + l)^2} \) (towards north pole)
• Net field: \( B_{\text{axis}} = B_1 - B_2 = \frac{\mu_0 m}{4\pi} \left[ \frac{1}{(r - l)^2} - \frac{1}{(r + l)^2} \right] = \frac{\mu_0 m}{4\pi} \frac{4 r l}{(r^2 - l^2)^2} \)
• Define magnetic moment \( M = m (2l) \), then \( B_{\text{axis}} = \frac{\mu_0}{4\pi} \frac{2 M r}{(r^2 - l^2)^2} \)
• For short magnet \( (r \gg l) \), \( B_{\text{axis}} \approx \frac{\mu_0}{4\pi} \frac{2 M}{r^3} \) (like electric dipole)

Field at Equatorial Position

• Field due to each pole at P: \( B_1 = B_2 = \frac{\mu_0}{4\pi} \frac{m}{(r^2 + l^2)} \)
• Net field: \( B_{\text{eq}} = 2 B \cos \theta = 2 \frac{\mu_0}{4\pi} \frac{m}{r^2 + l^2} \cos \theta \)
• With \( \cos \theta = \frac{r}{\sqrt{r^2 + l^2}} \), \( B_{\text{eq}} = \frac{\mu_0}{4\pi} \frac{M}{(r^2 + l^2)^{3/2}} \)
• For short magnet \( (r \gg l) \), \( B_{\text{eq}} \approx \frac{\mu_0}{4\pi} \frac{M}{r^3} \)

Field at General Position

• Radial component: \( B_r = 2 \frac{\mu_0}{4\pi} \frac{M \cos \theta}{r^3} \)
• Normal component: \( B_n = \frac{\mu_0}{4\pi} \frac{M \sin \theta}{r^3} \)
• Resultant field magnitude: \( B_{res} = \frac{\mu_0 M}{4\pi r^3} \sqrt{1 + 3 \cos^2 \theta} \)
• Angle between \( B_{res} \) and \( B_r \): \( \tan \phi = \frac{B_n}{B_r} = \frac{\tan \theta}{2} \)

Solved Examples

Ex 1

• Calculate axial field at \( r = 10 \text{ cm} \), \( l = 1 \text{ cm} \), \( M = 0.05 \text{ A·m}^2 \).
• \( B_{axis} = \frac{\mu_0}{4\pi} \frac{2 M r}{(r^2 - l^2)^2} \approx 10^{-7} \times \frac{2 \times 0.05 \times 0.1}{(0.01)^4} = 10^{-7} \times 10 = 1 \times 10^{-6} \text{ T} \)

Practice Set

• Q1: Magnetic field at equatorial point for short magnet varies as:
  A) \( \frac{1}{r^2} \)
  B) \( \frac{1}{r^3} \)
  C) \( \frac{1}{r} \)
  D) Constant
• Q2: At axial point, net magnetic field is:
  A) Sum of fields from poles
  B) Difference of fields from poles
  C) Zero
  D) Product of fields from poles
• Q3: For a bar magnet with magnetic moment \( M \), field at distance \( r \) on axis is proportional to:
  A) \( \frac{M}{r^2} \)
  B) \( \frac{M}{r^3} \)
  C) \( \frac{M}{r} \)
  D) \( M r^3 \)

Answer Key

• Q1: B) \( \frac{1}{r^3} \) — Equatorial field decreases as inverse cube of distance.
• Q2: B) Difference of fields from poles — Axial field is \( B_1 - B_2 \).
• Q3: B) \( \frac{M}{r^3} \) — Axial field varies inversely with cube of distance.

Memory Aids

• Axial field: \( B_{axis} \propto \frac{2 M}{r^3} \)
• Equatorial field: \( B_{eq} \propto \frac{M}{r^3} \)
• Resultant field angle: \( \tan \phi = \frac{\tan \theta}{2} \)

Magnetic Hysteresis

Concept and Characteristics

• Hysteresis: lag of magnetic flux density (B) behind magnetic field strength (H) in ferromagnetic materials.
• Hysteresis loop shows B vs. H during magnetisation and demagnetisation.
• Key points on loop:
  • \( B_r \) (remanence): residual magnetism when \( H = 0 \).
  • \( H_c \) (coercivity): reverse field needed to reduce \( B \) to zero.
  • Saturation: maximum magnetisation.
• Only ferromagnetic materials show hysteresis.
• Cause: magnetic domains remain aligned after removing external field.
• Retentivity: ability to retain magnetism (equal to \( B_r \)).
• Coercivity: measure of field needed to demagnetise.

Soft vs Hard Magnetic Materials

• Soft magnetic materials: Low retentivity and coercivity; easy to magnetise/demagnetise; used in transformer cores, electromagnets.
• Hard magnetic materials: High retentivity and coercivity; retain magnetism; used for permanent magnets.

Solved Examples

Ex 1

• Identify coercivity if \( B \) reduces to zero at \( H = -50 \text{ A/m} \).
• Coercivity \( H_c = 50 \text{ A/m} \).

Practice Set

• Q1: What does remanence represent?
  A) Maximum magnetisation
  B) Residual magnetism at zero field
  C) Coercive field
  D) Magnetic susceptibility
• Q2: Which material shows wide hysteresis loop?
  A) Soft iron
  B) Steel
  C) Copper
  D) Aluminium
• Q3: Coercivity is the measure of:
  A) Magnetic field to saturate
  B) Magnetic field to reduce magnetism to zero
  C) Magnetic susceptibility
  D) Magnetic moment

Answer Key

• Q1: B) Residual magnetism at zero field — Remanence is residual B when H=0.
• Q2: B) Steel — Hard magnetic materials like steel have wide loops.
• Q3: B) Magnetic field to reduce magnetism to zero — Definition of coercivity.

Memory Aids

• Hysteresis loop shows magnetic memory.
• Retentivity = remanence = residual magnetism.
• Coercivity = field to demagnetise.
• Soft materials: narrow loop; hard materials: wide loop.