Magnetism and Matter

What this chapter is about

This chapter explores how materials respond to magnetic fields and introduces the idea that magnetism, at its core, arises from circulating currents at the atomic level. You have already studied how moving charges and currents produce magnetic fields; here you learn that every magnetic phenomenon in matter can be traced to tiny current loops within atoms.

The chapter develops the analogy between electric dipoles and magnetic dipoles. A bar magnet behaves like a magnetic dipole, and the Earth itself acts as a giant magnet with its own dipole field. You will learn how to describe the magnetic properties of materials using quantities such as magnetisation, magnetic intensity and susceptibility, and how these ideas classify substances into diamagnetic, paramagnetic and ferromagnetic categories.

After studying this chapter, you should be able to describe a bar magnet in dipole terms, solve problems on torque and potential energy of a magnetic dipole in an external field, explain the Earth's magnetism, and distinguish the magnetic behaviour of different materials using measurable quantities.

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Key ideas

• A bar magnet is equivalent to a magnetic dipole; its dipole moment m = (pole strength) × (magnetic length), direction from S-pole to N-pole inside the magnet.
• When a magnetic dipole of moment m sits in a uniform field B, it experiences a torque τ = m × B and has potential energy U = −m · B.
• Gauss's law for magnetism states that the net magnetic flux through any closed surface is zero: ∮B · dA = 0. This implies isolated magnetic poles (monopoles) do not exist.
• The axial and equatorial fields of a short bar magnet mirror those of an electric dipole: B_axial ≈ (μ₀/4π)(2m/r³), B_equatorial ≈ (μ₀/4π)(m/r³).
• The Earth's magnetic field at any location is described by three elements: declination (angle between geographic and magnetic meridians), dip or inclination (angle the field makes with horizontal), and horizontal component B_H.
• Magnetisation M is the magnetic moment per unit volume of a material. Magnetic intensity H is defined by B = μ₀(H + M).
• Magnetic susceptibility χ = M/H tells how easily a material is magnetised; permeability μ = B/H and relative permeability μ_r = 1 + χ connect these quantities.
• Materials are classified as diamagnetic (χ small and negative), paramagnetic (χ small and positive), or ferromagnetic (χ large and positive, with hysteresis).

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Formulas and facts to remember

1. Magnetic dipole moment: m = n I A for a coil of n turns, area A carrying current I.
2. Torque on dipole: τ = m B sin θ, maximum when θ = 90°.
3. Potential energy of dipole: U = −m B cos θ; minimum when m is along B.
4. Axial field of a short magnet: B = (μ₀/4π)(2m/r³).
5. Equatorial field of a short magnet: B = (μ₀/4π)(m/r³), direction opposite to m.
6. Relation between B, H and M: B = μ₀(H + M) = μ₀ μ_r H.
7. Susceptibility and permeability: χ = M/H; μ_r = 1 + χ.
8. Curie law (paramagnets): χ ∝ 1/T.

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Worked examples

Example 1 – Torque on a bar magnet

A bar magnet of dipole moment 0.40 A m² is placed in a uniform field of 0.25 T with its axis making 30° with the field. Find the torque acting on it.

Solution

Torque τ = m B sin θ

τ = 0.40 × 0.25 × sin 30° = 0.40 × 0.25 × 0.5 = 0.050 N m

The magnet experiences a torque of 5.0 × 10⁻² N m, tending to align it with the field.

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Example 2 – Finding Earth's field components

At a place in southern India the total magnetic field of the Earth is 4.0 × 10⁻⁵ T and the angle of dip is 30°. Calculate the horizontal and vertical components of the field.

Solution

Horizontal component B_H = B cos δ, vertical component B_V = B sin δ

B_H = 4.0 × 10⁻⁵ × cos 30° = 4.0 × 10⁻⁵ × 0.866 ≈ 3.5 × 10⁻⁵ T

B_V = 4.0 × 10⁻⁵ × sin 30° = 4.0 × 10⁻⁵ × 0.5 = 2.0 × 10⁻⁵ T

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Example 3 – Susceptibility and magnetisation

A paramagnetic sample is placed in a magnetic intensity H = 1.2 × 10⁴ A/m. Its susceptibility at that temperature is 2.5 × 10⁻⁴. Find (a) the magnetisation and (b) the magnetic field inside the sample.

Solution

(a) Magnetisation M = χ H = 2.5 × 10⁻⁴ × 1.2 × 10⁴ = 3.0 A/m

(b) B = μ₀ (H + M) = 4π × 10⁻⁷ × (1.2 × 10⁴ + 3.0)

Since M is much smaller than H, H + M ≈ 1.2 × 10⁴ A/m

B ≈ 4π × 10⁻⁷ × 1.2 × 10⁴ ≈ 1.51 × 10⁻² T ≈ 15 mT

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Common mistakes

• Confusing magnetic field B with magnetic intensity H → remember B includes the contribution of the material (M), while H relates to free currents alone.
• Treating the angle of dip as the same everywhere → dip varies with latitude; it is 0° at the magnetic equator and 90° at the magnetic poles.
• Assuming isolated north or south poles can exist → Gauss's law for magnetism forbids monopoles; every magnet has both poles.
• Using the electric-dipole energy formula with a wrong sign → U = −m B cos θ; minimum energy (stable) is when θ = 0.
• Forgetting that ferromagnets show hysteresis → their magnetisation depends on magnetic history, not just the current field.

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Quick revision

1. A bar magnet is a magnetic dipole with moment m; its field resembles that of an electric dipole at large distances.
2. Torque τ = m B sin θ; potential energy U = −m B cos θ.
3. Earth's field is described by declination, dip and horizontal component.
4. B = μ₀(H + M); χ = M/H; μ_r = 1 + χ.
5. Diamagnets have χ < 0; paramagnets have small χ > 0; ferromagnets have large χ > 0 and exhibit hysteresis.
6. Gauss's law for magnetism: ∮B · dA = 0 — no magnetic monopoles.

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