Is it correct to say: Writing μᵣ = χ or μ = μ₀χ?

μᵣ = 1 + χ and μ = μ₀(1 + χ). For iron the difference is tiny, but for para- and diamagnetic materials it is the whole answer.

Is it correct to say: Giving H and B the same unit?

H and M are in A/m; B is in tesla (T). B = μ₀(H + M) converts between them.

Is it correct to say: Using °C in Curie's law?

χ ∝ 1/T only for absolute temperature. Add 273 before substituting.

Is it correct to say: Stating that the orbital magnetic moment is along the angular momentum?

The electron is negatively charged, so m_orb = −(e/2mₑ)L_orb: the moment is opposite to L.

Is it correct to say: Assigning a positive susceptibility to diamagnetic materials?

Diamagnetic χ is small and negative (μᵣ slightly less than 1). Such materials are weakly repelled by a magnet.

Is it correct to say: Leaving the area in cm² while finding B = Φ/A?

1 cm² = 10⁻⁴ m². Without this conversion B and every later answer come out 10⁴ times too small.

📄 Formula sheet (PDF)

Magnetic Materials

Torque on a magnetic dipole, orbital magnetic moment, Bohr magneton, magnetisation, magnetic intensity, susceptibility, permeability, dia-, para- and ferromagnetism, hysteresis — Maharashtra HSC Physics Ch 11

📐 14 formulas✏️ 4 examples🎯 6 practice🏫 Maharashtra State Board📚 Class 12✓ 2025–26 syllabus

💡

Board Exam Tips

→The derivation of the orbital magnetic moment of a revolving electron (m_orb = evr/2 = (e/2mₑ)L) and the Bohr magneton is a standard theory question. Learn it as a four-line chain: current, area, moment, angular momentum.
→Keep the three vectors straight: H (magnetic intensity, A/m) is set by the free current, M (magnetisation, A/m) is the material's response, and B = μ₀(H + M) (tesla) is the total field.
→Remember the signs of χ: small and negative for diamagnetic, small and positive for paramagnetic, large and positive for ferromagnetic. This is the quickest way to classify a material from its χ or μᵣ value.
→For Curie's law numericals, convert temperatures to kelvin before using χ₁T₁ = χ₂T₂.
→When flux and area are given for a rod in a magnetising field, find B = Φ/A first, then μ = B/H, μᵣ = μ/μ₀ and χ = μᵣ − 1 in that order.
→For hysteresis, be ready to draw and label the B–H loop: retentivity (residual magnetism) on the B-axis, coercivity on the H-axis. Then compare soft iron and steel.

📐 Formulas(14)

Torque on a Magnetic Dipole★ Board fav

τ = m \times B, τ = m B sin θ

Symbol	Meaning
$m$	Magnetic dipole moment (A m²)
$B$	Uniform magnetic field (T)
$θ$	Angle between m and B

Potential Energy of a Magnetic Dipole

U = - m \cdot B = - m B cos θ

Orbital Magnetic Moment of an Electron★ Board fav

m_{or b} = \frac{e v r}{2} = \frac{e}{2 m _{e}} L_{or b}

Symbol	Meaning
$e$	Electronic charge = 1.6×10⁻¹⁹ C
$v$	Orbital speed of the electron (m s⁻¹)
$r$	Radius of the orbit (m)
$m_{e}$	Mass of electron = 9.1×10⁻³¹ kg
$L_{or b}$	Orbital angular momentum (kg m² s⁻¹ = J s)

Bohr Magneton

μ_{B} = \frac{e h}{4 π m _{e}} \approx 9.27 \times 1 0^{- 24} A m^{2}

Symbol	Meaning
$h$	Planck's constant = 6.63×10⁻³⁴ J s

Magnetisation

M = \frac{m _{n e t}}{V}

Symbol	Meaning
$M$	Magnetisation (A m⁻¹)
$m_{n e t}$	Net magnetic dipole moment of the sample (A m²)
$V$	Volume of the sample (m³)

Magnetic Intensity (Magnetising Field)

H = \frac{B _{0}}{μ _{0}}, H = n I (inside a solenoid)

Symbol	Meaning
$H$	Magnetic intensity (A m⁻¹)
$B_{0}$	Applied magnetic field in vacuum (T)
$μ_{0}$	Permeability of free space = 4π×10⁻⁷ T m A⁻¹
$n$	Number of turns per unit length (m⁻¹)

Total Magnetic Field in a Material★ Board fav

B = μ_{0} (H + M)

Magnetic Susceptibility

χ = \frac{M}{H}

Symbol	Meaning
$χ$	Magnetic susceptibility (no unit)

Permeability

B = μ H, μ = μ_{0} (1 + χ)

Symbol	Meaning
$μ$	Permeability of the material (T m A⁻¹)

Relative Permeability

μ_{r} = \frac{μ}{μ _{0}} = 1 + χ

Symbol	Meaning
$μ_{r}$	Relative permeability (no unit)

Fractional Change in Field Due to a Material

\frac{B - B _{0}}{B _{0}} = χ

Flux Density in a Specimen

B = \frac{Φ}{A}

Symbol	Meaning
$Φ$	Magnetic flux (Wb)
$A$	Area of cross-section (m²)

Curie's Law (Paramagnetic Materials)★ Board fav

M = C \frac{B}{T}, χ = \frac{C μ _{0}}{T} \propto \frac{1}{T} \Rightarrow χ_{1} T_{1} = χ_{2} T_{2}

Symbol	Meaning
$C$	Curie constant (depends on the material)
$B$	Applied magnetic field, B = μ₀H (T)
$T$	Absolute temperature (K)

Saturation Magnetisation

M_{s a t} = N m_{a t o m}

Symbol	Meaning
$N$	Number of atomic dipoles per unit volume (m⁻³)
$m_{a t o m}$	Dipole moment of each atom (A m²)

✏️ Solved Examples

1Solved Exampleeasy2 steps

A bar magnet of magnetic moment 2.5 A m² is placed in a uniform magnetic field of 0.2 T with its axis at 30° to the field. Find the torque acting on it and its potential energy.

Torque on a magnetic dipole.

τ = m B sin θ = 2.5 \times 0.2 \times sin 3 0^{\circ} = 0.25 N m

2Solved Exampleboard4 steps

A magnetising field of 2000 A/m produces a magnetic flux of 3.0×10⁻⁵ Wb in an iron rod of cross-sectional area 0.25 cm². Find the permeability, relative permeability and susceptibility of the rod. (μ₀ = 4π×10⁻⁷ T m/A)

Convert area to m² and find B.

A = 0.25 \times 1 0^{- 4} m^{2}, B = \frac{Φ}{A} = \frac{3.0 \times 1 0 ^{- 5}}{0.25 \times 1 0 ^{- 4}} = 1.2 T

3Solved Exampleboard4 steps

An electron revolves in a circular orbit of radius 2.1×10⁻¹⁰ m with a speed of 1.1×10⁶ m/s. Find its orbital magnetic moment and orbital angular momentum. (e = 1.6×10⁻¹⁹ C, mₑ = 9.1×10⁻³¹ kg)

Orbital magnetic moment.

m_{or b} = \frac{e v r}{2} = \frac{1.6 \times 1 0 ^{- 19} \times 1.1 \times 1 0 ^{6} \times 2.1 \times 1 0 ^{- 10}}{2}

4Solved ExampleHOTS3 steps

A paramagnetic salt has susceptibility 3.0×10⁻⁴ at 27 °C. Find its susceptibility at −73 °C. If a toroid is filled with the salt at −73 °C, by what percentage does the magnetic field inside increase?

Convert to kelvin.

T_{1} = 27 + 273 = 300 K, T_{2} = - 73 + 273 = 200 K

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Writing μᵣ = χ or μ = μ₀χ
✓μᵣ = 1 + χ and μ = μ₀(1 + χ). For iron the difference is tiny, but for para- and diamagnetic materials it is the whole answer.
2
Giving H and B the same unit
✓H and M are in A/m; B is in tesla (T). B = μ₀(H + M) converts between them.
3
Using °C in Curie's law
✓χ ∝ 1/T only for absolute temperature. Add 273 before substituting.
4
Stating that the orbital magnetic moment is along the angular momentum
✓The electron is negatively charged, so m_orb = −(e/2mₑ)L_orb: the moment is opposite to L.
5
Assigning a positive susceptibility to diamagnetic materials
✓Diamagnetic χ is small and negative (μᵣ slightly less than 1). Such materials are weakly repelled by a magnet.
6
Leaving the area in cm² while finding B = Φ/A
✓1 cm² = 10⁻⁴ m². Without this conversion B and every later answer come out 10⁴ times too small.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
A specimen of volume 1.6×10⁻⁵ m³ has a net magnetic dipole moment of 0.8 A m². Find its magnetisation.
Q2
A solenoid with 1000 turns per metre carries 2 A. Its core has susceptibility 299. Find H, M and B inside the core. (μ₀ = 4π×10⁻⁷ T m/A)
Q3
A toroid is filled with a diamagnetic material of susceptibility −1.5×10⁻⁵. Find the percentage change in the magnetic field inside it.
Q4
A paramagnetic gas has 2.5×10²⁶ atoms per m³, each with a magnetic dipole moment of 1.2×10⁻²³ A m². Find the maximum (saturation) magnetisation possible.
Q5
A magnet of moment 1.5 A m² lies along a uniform field of 0.4 T. How much work is needed to turn it through (a) 90°, (b) 180°?
Q6
Using e = 1.6×10⁻¹⁹ C, h = 6.63×10⁻³⁴ J s and mₑ = 9.1×10⁻³¹ kg, calculate the Bohr magneton.

📝 Notes

Magnetic Materials — Maharashtra HSC Overview

Chapter 11 of the Maharashtra Board Std XII Physics textbook explains why materials respond to a magnetic field, and how to describe that response with a few quantities.

Where magnetism comes from

An orbiting electron is a tiny current loop with moment $m_{orb} = evr/2 = (e/2m_e)L$ . Quantising $L$ gives the smallest possible value, the Bohr magneton $\mu_B = eh/(4\pi m_e)$ . Electron spin adds its own moment. A material's behaviour depends on whether these atomic moments cancel (diamagnetic), are present but randomly oriented (paramagnetic), or line up in domains (ferromagnetic).

The working set of quantities

$H$ : what the free current supplies (A/m).
$M$ : how strongly the material magnetises (A/m).
$B = \mu_0(H + M)$ : the resulting field (T).
$\chi = M/H$ , $\mu_r = 1 + \chi$ , $\mu = \mu_0\mu_r$ .

Most numericals are a chain through these: $\Phi \rightarrow B \rightarrow \mu \rightarrow \mu_r \rightarrow \chi \rightarrow M$ .

Comparing the three classes

Diamagnetic: $\chi$ small and negative, $\mu_r$ slightly less than 1, almost independent of temperature.
Paramagnetic: $\chi$ small and positive, $\mu_r$ slightly more than 1, $\chi \propto 1/T$ (Curie's law).
Ferromagnetic: $\chi$ large and positive, $\mu_r \gg 1$ , becomes paramagnetic above the Curie temperature.

Hysteresis and applications

The B–H loop of a ferromagnet shows retentivity and coercivity. Soft iron (narrow loop, low coercivity, small energy loss per cycle) suits electromagnets and transformer cores. Steel has a large coercivity, so it keeps its magnetism against stray fields and suits permanent magnets. Magnetic shielding works because field lines prefer to pass through a high-permeability shell rather than the space it encloses.