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CalibratePhysics

Chapter 11

Magnetic Materials

Maharashtra HSC · 8 subtopics · textbook pages 251–264

Formula SheetEvery formula this chapter can be examined on, in one table.

Constants and conventions

  • Permeability of free space μ0=4π×10−7 T m A−1\mu_0 = 4\pi\times10^{-7}\ \text{T m A}^{-1}, also written H m−1\text{H m}^{-1} or Wb A−1m−1\text{Wb A}^{-1}\text{m}^{-1}. Numerically μ0=1.257×10−6\mu_0 = 1.257\times10^{-6}.
  • 1 gauss=10−4 T1\ \text{gauss} = 10^{-4}\ \text{T} and 1 T=1 Wb m−21\ \text{T} = 1\ \text{Wb m}^{-2}. Board numericals quote the earth's field and small fields in gauss — convert to tesla before substituting.
  • Electronic charge e=1.6×10−19 Ce = 1.6\times10^{-19}\ \text{C}, electron mass me=9.1×10−31 kgm_e = 9.1\times10^{-31}\ \text{kg}, Planck constant h=6.63×10−34 J sh = 6.63\times10^{-34}\ \text{J s}.
  • Avogadro number NA=6.023×1023 mol−1N_A = 6.023\times10^{23}\ \text{mol}^{-1}, Boltzmann constant kB=1.38×10−23 J K−1k_B = 1.38\times10^{-23}\ \text{J K}^{-1}.
  • Bohr magneton μB=9.274×10−24 A m2\mu_B = 9.274\times10^{-24}\ \text{A m}^2; gyromagnetic ratio e2me=8.8×1010 C kg−1\dfrac{e}{2m_e} = 8.8\times10^{10}\ \text{C kg}^{-1}.
  • Every temperature in a magnetic formula is absolute: T(K)=t(∘C)+273T(\text{K}) = t(^\circ\text{C}) + 273.
  • Moment of inertia given in g cm2\text{g cm}^2 converts to kg m2\text{kg m}^2 by multiplying by 10−710^{-7}; so 500 g cm2=5×10−5 kg m2500\ \text{g cm}^2 = 5\times10^{-5}\ \text{kg m}^2.

Torque acting on a magnetic dipole in a uniform magnetic field

FormulaSymbols, units and conditions
τ⃗=m⃗×B⃗τ=mBsin⁡θ\vec{\tau} = \vec{m}\times\vec{B}\qquad \tau = mB\sin\theta
  • Torque on a magnetic dipole (a short bar magnet, or a current-carrying coil) of magnetic dipole moment mm placed in a uniform magnetic field BB; θ\theta is the angle between m⃗\vec{m} and B⃗\vec{B}. τ\tau is in N m\text{N m}, mm in A m2\text{A m}^2, BB in T\text{T}. Maximum τ=mB\tau = mB at θ=90∘\theta = 90^\circ; zero at θ=0∘\theta = 0^\circ and θ=180∘\theta = 180^\circ. Net force in a uniform field is zero, so the dipole only rotates.
m=qm×2lm=NIAm = q_m \times 2l \qquad m = NIA
  • Magnetic dipole moment of a bar magnet of pole strength qmq_m (in A m\text{A m}) and magnetic length 2l2l; and of a flat coil of NN turns of area AA carrying current II.
Um=∫τ dθ=∫mBsin⁡θ dθ⇒Um=−mBcos⁡θ=−m⃗⋅B⃗U_m = \int \tau\, d\theta = \int mB\sin\theta\, d\theta \qquad\Rightarrow\qquad U_m = -mB\cos\theta = -\vec{m}\cdot\vec{B}
  • Magnetic potential energy of a dipole at angle θ\theta to the field, with the zero taken at θ=90∘\theta = 90^\circ. In joule.
Umin=−mB (θ=0∘)Um=0 (θ=90∘)Umax=+mB (θ=180∘)U_{min} = -mB\ (\theta = 0^\circ)\qquad U_m = 0\ (\theta = 90^\circ)\qquad U_{max} = +mB\ (\theta = 180^\circ)
  • θ=0∘\theta = 0^\circ is the parallel, most stable position; θ=180∘\theta = 180^\circ is the antiparallel, most unstable position.
W=mB(cos⁡θ1−cos⁡θ2)W = mB(\cos\theta_1 - \cos\theta_2)
  • Work done by an external agent in rotating the dipole from θ1\theta_1 to θ2\theta_2; it equals the change in magnetic potential energy. From the most stable to the most unstable position, W=2mBW = 2mB.
Id2θdt2=−mBsin⁡θ≈−mBθ(θ small)I\dfrac{d^2\theta}{dt^2} = -mB\sin\theta \approx -mB\theta \qquad (\theta \text{ small})
  • Equation of the angular oscillation of a freely suspended magnet; II is the moment of inertia about the axis of suspension. Comparing with d2xdt2+ω2x=0\dfrac{d^2x}{dt^2} + \omega^2 x = 0 shows the motion is angular S.H.M.
ω=mBIT=2πImBn=1T\omega = \sqrt{\dfrac{mB}{I}}\qquad T = 2\pi\sqrt{\dfrac{I}{mB}}\qquad n = \dfrac{1}{T}
  • Angular frequency, periodic time and frequency of oscillation of a bar magnet suspended in a uniform field (the vibration-magnetometer result). If the field is the earth's horizontal component, write BHB_H in place of BB.
m=4π2IT2BB=4π2IT2mm = \dfrac{4\pi^2 I}{T^2 B}\qquad B = \dfrac{4\pi^2 I}{T^2 m}
  • The two working rearrangements — used to find the magnetic moment of a magnet, or the field (often BHB_H) in which it swings.
T′T=12\dfrac{T'}{T} = \dfrac{1}{2}
  • A bar magnet broken into two equal halves across its length: II becomes I8\dfrac{I}{8} and mm becomes m2\dfrac{m}{2}, so the new period is half the original.
BV=mg2qmB_V = \dfrac{mg}{2q_m}
  • A dip needle of pole strength qmq_m held horizontal by a weight mgmg placed on one end: the couple qmBV×2lq_m B_V \times 2l due to the vertical component of the earth's field is balanced by the moment mg×lmg \times l of the added weight about the pivot.

Origin of magnetism in materials

FormulaSymbols, units and conditions
I=eT=ef=eω2π=ev2πrI = \dfrac{e}{T} = ef = \dfrac{e\omega}{2\pi} = \dfrac{ev}{2\pi r}
  • Equivalent current of one electron of charge ee going round a circular orbit of radius rr with speed vv, period TT, frequency ff and angular speed ω=vr\omega = \dfrac{v}{r}.
morb=IA=ev2πr×πr2=evr2m_{orb} = IA = \dfrac{ev}{2\pi r}\times \pi r^2 = \dfrac{evr}{2}
  • Orbital magnetic dipole moment of a revolving electron, in A m2\text{A m}^2. For a coil of NN turns use m=NIAm = NIA instead.
L=mevrL = m_e v r
  • Orbital angular momentum of the electron, in kg m2s−1\text{kg m}^2\text{s}^{-1} (equivalently N m s\text{N m s}).
morb=eL2mem⃗orb=−(e2me)L⃗m_{orb} = \dfrac{eL}{2m_e}\qquad \vec{m}_{orb} = -\left(\dfrac{e}{2m_e}\right)\vec{L}
  • Orbital magnetic moment in terms of angular momentum. The minus sign records that m⃗orb\vec{m}_{orb} and L⃗\vec{L} are antiparallel because the electron is negatively charged; both are perpendicular to the plane of the orbit.
morbL=e2me=8.8×1010 C kg−1\dfrac{m_{orb}}{L} = \dfrac{e}{2m_e} = 8.8\times10^{10}\ \text{C kg}^{-1}
  • The gyromagnetic ratio: magnetic dipole moment per unit angular momentum of a revolving electron. Use it to convert one into the other in numericals.
L=mevr=nh2π⇒morb=neh4πmeL = m_e v r = \dfrac{nh}{2\pi}\qquad\Rightarrow\qquad m_{orb} = \dfrac{neh}{4\pi m_e}
  • Bohr's second postulate applied to the orbit; n=1,2,3,…n = 1, 2, 3, \ldots is the orbit number.
μB=eh4πme=9.274×10−24 A m2\mu_B = \dfrac{eh}{4\pi m_e} = 9.274\times10^{-24}\ \text{A m}^2
  • Bohr magneton — the orbital magnetic moment of an electron in the first orbit (n=1n=1), and the natural unit in which atomic magnetic moments are quoted.

Magnetization and magnetic intensity

FormulaSymbols, units and conditions
M⃗=mnetV\vec{M} = \dfrac{m_{net}}{V}
  • Magnetization: net magnetic dipole moment per unit volume of the specimen. SI unit A m−1\text{A m}^{-1}, dimensions [M0L−1T0I1][M^0L^{-1}T^0I^1]. It is a vector.
B0=μ0nIH=nIB_0 = \mu_0 nI \qquad H = nI
  • Field of an empty solenoid of nn turns per unit length carrying current II, and the magnetic intensity (magnetising field) HH. HH is in A m−1\text{A m}^{-1} and does not depend on the material placed inside.
Bm=μ0MB=B0+Bm=μ0(H+M)B_m = \mu_0 M \qquad B = B_0 + B_m = \mu_0(H + M)
  • BmB_m is the contribution of the magnetised material; BB is the total field inside the rod, in tesla.
H=Bμ0−MH = \dfrac{B}{\mu_0} - M
  • The same relation rearranged — the form used when BB and MM are known.
χ=MHM=χH\chi = \dfrac{M}{H} \qquad M = \chi H
  • Magnetic susceptibility: a pure number (no unit, no dimensions), because MM and HH have the same unit. Negative for diamagnetics, small and positive for paramagnetics, large and positive for ferromagnetics.
B=μ0(1+χ)H=μ0μrH=μHB = \mu_0(1+\chi)H = \mu_0\mu_r H = \mu H—
μr=1+χ=μμ0μ=μ0(1+χ)\mu_r = 1 + \chi = \dfrac{\mu}{\mu_0}\qquad \mu = \mu_0(1+\chi)
  • μ\mu is the permeability of the material in T m A−1\text{T m A}^{-1} (or H m−1\text{H m}^{-1}, or Wb A−1m−1\text{Wb A}^{-1}\text{m}^{-1}) and μr\mu_r its relative permeability, a pure number. For a superconductor χ=−1\chi = -1, so μ=0\mu = 0 and no lines pass through it.
B−B0B0=μ−μ0μ0=χpercentage increase=χ×100\dfrac{B - B_0}{B_0} = \dfrac{\mu - \mu_0}{\mu_0} = \chi \qquad \text{percentage increase} = \chi \times 100
  • Fractional (and percentage) increase in the field inside a solenoid or toroid when its core is filled with a material of susceptibility χ\chi.
Φ=BAB=ΦA\Phi = BA \qquad B = \dfrac{\Phi}{A}
  • Magnetic flux through a rod of cross-sectional area AA, in weber. Use it to get BB first, then μ=BH\mu = \dfrac{B}{H}, then μr\mu_r, χ\chi and MM.

Magnetic properties of materials

FormulaSymbols, units and conditions
M=CBTχ=μ0CTχ∝1TM = \dfrac{CB}{T}\qquad \chi = \dfrac{\mu_0 C}{T}\qquad \chi \propto \dfrac{1}{T}
  • Curie law for a paramagnetic material: magnetization is directly proportional to the applied field and inversely proportional to the absolute temperature. CC is the Curie constant.
χ1T1=χ2T2\chi_1 T_1 = \chi_2 T_2
  • The working form of Curie's law for every board numerical on magnesium and other paramagnetics: χT\chi T is a constant.
χ=CT−TC(T>TC)\chi = \dfrac{C}{T - T_C}\qquad (T > T_C)
  • Susceptibility of a ferromagnetic material above its Curie temperature TCT_C, where it has become paramagnetic.
Msat=NmM_{sat} = N m
  • Maximum (saturation) magnetization of a sample containing NN atoms per unit volume, each of dipole moment mm; reached only when every atomic moment is perfectly aligned.
Ethermal=32kBTcompare withUmax=mBE_{thermal} = \dfrac{3}{2}k_B T \qquad\text{compare with}\qquad U_{max} = mB
  • Saturation test: if the thermal energy of an atom is far larger than its maximum magnetic potential energy in the field, thermal agitation destroys the alignment and saturation cannot be reached.
N=VρNAmolar massN = \dfrac{V \rho N_A}{\text{molar mass}}
  • Number of atoms in a domain (or specimen) of volume VV and density ρ\rho; then mmax=N×m_{max} = N \times (moment per atom) and M=mmaxVM = \dfrac{m_{max}}{V}.

Hysteresis

FormulaSymbols, units and conditions
B=μ0(H+M)M=χHB = \mu_0 (H + M) \qquad M = \chi H
  • The two relations used to plot the loop: pick H=nIH = nI from the solenoid current, get MM, then get BB.
μ=BH\mu = \dfrac{B}{H}
  • Permeability read off any point of the BB–HH curve. A magnetising field of 360 A m−1360\ \text{A m}^{-1} giving B=0.6 TB = 0.6\ \text{T} means μ=1600 T m A−1\mu = \dfrac{1}{600}\ \text{T m A}^{-1}.
  • Area enclosed by one BB–HH loop == energy dissipated as heat per unit volume of the material per cycle of magnetisation, in J m−3\text{J m}^{-3}.

Permanent magnet and electromagnet

  • Soft iron core: μr>1000\mu_r > 1000, so the field of a solenoid rises about a thousandfold when a soft iron rod is inserted. No new formula — the numerical work is B=μH=μ0μrnIB = \mu H = \mu_0\mu_r nI.

Magnetic shielding

  • No formula is set on this section. It is examined as theory: a closed shell of soft ferromagnetic material (μr\mu_r very large) crowds the lines of force into its own walls and leaves the enclosed space almost field-free.
Short NotesThe chapter subtopic by subtopic, in the textbook's own order.8 subtopics
  1. 1Introduction
  2. 2Torque Acting on a Magnetic Dipole in a Uniform Magnetic Field1 board question
  3. 3Origin of Magnetism in Materials10 board questions
  4. 4Magnetization and Magnetic Intensity11 board questions
  5. 5Magnetic Properties of Materials13 board questions
  6. 6Hysteresis
  7. 7Permanent Magnet and Electromagnet
  8. 8Magnetic Shielding
The whole chapter in brief

11.1 Introduction

  • Recall from Std XI: a short bar magnet has a magnetic dipole moment mm, sets up a field along its axial and equatorial lines, and when suspended freely comes to rest along the geographic north–south direction (the basis of the compass and of navigation).
  • Magnetic monopoles do not exist: cut a magnet anywhere and each piece is again a complete dipole with a north and a south pole. Magnetic lines of force are therefore always closed loops.
  • This chapter covers: a bar magnet in a uniform field (torque, potential energy, oscillations), the atomic origin of magnetism, magnetization and magnetic intensity, the three classes of magnetic material, hysteresis, and three applications — permanent magnets, electromagnets and magnetic shielding.

11.2 Torque acting on a magnetic dipole in a uniform magnetic field

  • A bar magnet in a uniform field feels equal and opposite forces on its two poles along different lines of action. These form a couple, so there is pure rotation and no translation.
  • τ⃗=m⃗×B⃗\vec{\tau} = \vec{m}\times\vec{B}, magnitude τ=mBsin⁡θ\tau = mB\sin\theta — the same expression already obtained for a rectangular current-carrying coil in Chapter 10.
  • Magnetic potential energy Um=−mBcos⁡θ=−m⃗⋅B⃗U_m = -mB\cos\theta = -\vec{m}\cdot\vec{B}, obtained by integrating the torque through the angular displacement.
Positionθ\thetaUmU_mState
m⃗\vec{m} parallel to B⃗\vec{B}0∘0^\circ−mB-mBMinimum energy, most stable
m⃗\vec{m} perpendicular to B⃗\vec{B}90∘90^\circ00Reference position
m⃗\vec{m} antiparallel to B⃗\vec{B}180∘180^\circ+mB+mBMaximum energy, most unstable
  • Work done in turning the dipole from θ1\theta_1 to θ2\theta_2 is W=mB(cos⁡θ1−cos⁡θ2)W = mB(\cos\theta_1 - \cos\theta_2). Most stable to most unstable costs 2mB2mB. Turning through 90∘90^\circ from the meridian costs mBmB, through 60∘60^\circ costs mB2\dfrac{mB}{2} — the ratio is 2.
  • Oscillations: the restoring torque τ=−mBsin⁡θ\tau = -mB\sin\theta with τ=Id2θdt2\tau = I\dfrac{d^2\theta}{dt^2} gives, for small θ\theta, d2θdt2=−(mBI)θ\dfrac{d^2\theta}{dt^2} = -\left(\dfrac{mB}{I}\right)\theta. This is angular S.H.M. with ω2=mBI\omega^2 = \dfrac{mB}{I} and T=2πImBT = 2\pi\sqrt{\dfrac{I}{mB}}.
  • Vibration magnetometer: works on the principle that a freely suspended magnet disturbed from equilibrium in a uniform field vibrates about the mean position. It is used to compare magnetic moments and magnetic fields, and to determine the horizontal component of the earth's magnetic field.
  • Poles of a current loop: view a loop with clockwise current from the front and the face nearest you is a south pole; view the same loop from behind, where the current appears anticlockwise, and that face is a north pole. Right-hand rule, and nothing about the current itself has changed.

11.3 Origin of magnetism in materials

  • Magnetism in bulk matter comes from the atoms — specifically from the electrons, which contribute an orbital magnetic moment (revolution about the nucleus) and a spin magnetic moment (intrinsic spin angular momentum). The vector sum of the two is the magnetic moment of the atom.
  • A revolving electron is equivalent to a tiny current loop: I=ev2πrI = \dfrac{ev}{2\pi r} and morb=IA=evr2m_{orb} = IA = \dfrac{evr}{2}.
  • With L=mevrL = m_e v r this becomes morb=eL2mem_{orb} = \dfrac{eL}{2m_e}, and vectorially m⃗orb=−(e2me)L⃗\vec{m}_{orb} = -\left(\dfrac{e}{2m_e}\right)\vec{L} — the two are antiparallel because the electron's charge is negative, and both are perpendicular to the orbital plane.
  • Gyromagnetic ratio morbL=e2me=8.8×1010 C kg−1\dfrac{m_{orb}}{L} = \dfrac{e}{2m_e} = 8.8\times10^{10}\ \text{C kg}^{-1}. The same result follows from quantum mechanics.
  • Using Bohr's postulate L=nh2πL = \dfrac{nh}{2\pi} gives morb=neh4πmem_{orb} = \dfrac{neh}{4\pi m_e}. For n=1n = 1 this is the Bohr magneton μB=eh4πme=9.274×10−24 A m2\mu_B = \dfrac{eh}{4\pi m_e} = 9.274\times10^{-24}\ \text{A m}^2.
  • Pauli's exclusion principle: no two electrons share the same set of quantum numbers nn, ll, mlm_l, msm_s. A pair of electrons in the same orbital have opposite spins and their moments cancel, so completely filled orbits contribute nothing.
  • An atom has a net magnetic moment only if it has unpaired electrons in its outermost orbit. Test in three steps: write the electronic configuration, draw the valence orbitals, look for an unpaired electron. Chlorine (1s22s22p63s23p51s^2 2s^2 2p^6 3s^2 3p^5) has one unpaired electron, so it is paramagnetic.
  • Effective magneton numbers (experimental, from Kittel): Fe3+\text{Fe}^{3+} (3d53d^5) 5.9, Fe2+\text{Fe}^{2+} (3d63d^6) 5.4, Co2+\text{Co}^{2+} (3d73d^7) 4.8, Ni2+\text{Ni}^{2+} (3d83d^8) 3.2 Bohr magnetons.

11.4 Magnetization and magnetic intensity

  • Magnetization M⃗=mnetV\vec{M} = \dfrac{m_{net}}{V}: net magnetic dipole moment per unit volume. Vector, unit A m−1\text{A m}^{-1}, dimensions [M0L−1T0I1][M^0L^{-1}T^0I^1].
  • In most bulk materials the atomic moments point in random directions and the vector sum is zero; in some, such as Fe3O4\text{Fe}_3\text{O}_4, it is not, and the material has a net moment.
  • Magnetic intensity (magnetising field) H=nIH = nI for a solenoid of nn turns per unit length. It is the ratio of the strength of the magnetising field to the permeability of free space, and — the point of introducing it — it does not depend on the material placed inside the coil.
  • Total field inside a specimen: B=B0+Bm=μ0nI+μ0M=μ0(H+M)B = B_0 + B_m = \mu_0 nI + \mu_0 M = \mu_0(H + M), so H=Bμ0−MH = \dfrac{B}{\mu_0} - M. HH and MM share the unit A m−1\text{A m}^{-1} and the same dimensions.
  • If HH is not too strong, M∝HM \propto H: M=χHM = \chi H, where χ\chi is the magnetic susceptibility — a dimensionless, unitless number measuring how much magnetization a given material acquires in a given magnetising field.
  • χ\chi is negative when the atomic dipole moments oppose the field, positive when they align with it. It is large for soft iron (χ>1000\chi > 1000).
  • Substituting: B=μ0(1+χ)H=μ0μrH=μHB = \mu_0(1+\chi)H = \mu_0\mu_r H = \mu H, with μr=1+χ=μμ0\mu_r = 1 + \chi = \dfrac{\mu}{\mu_0}.
  • Permeability μ\mu tells you how many lines of force pass through a substance kept in an external field; it is the magnetic analogue of permittivity. For a superconductor χ=−1\chi = -1, so μ=0\mu = 0 and no line of force passes through it.
  • Filling a solenoid or toroid core with a material raises the field by the fraction χ\chi: percentage increase =χ×100= \chi \times 100. Aluminium (χ=2.3×10−5\chi = 2.3\times10^{-5}) gives only 0.0023 per cent.
Substanceχ\chiSubstanceχ\chi
Silicon−4.2×10−6-4.2\times10^{-6}Aluminium2.3×10−52.3\times10^{-5}
Bismuth−1.66×10−5-1.66\times10^{-5}Calcium1.9×10−51.9\times10^{-5}
Copper−9.8×10−6-9.8\times10^{-6}Chromium2.7×10−42.7\times10^{-4}
Diamond−2.2×10−5-2.2\times10^{-5}Lithium2.1×10−52.1\times10^{-5}
Gold−3.6×10−5-3.6\times10^{-5}Magnesium1.2×10−51.2\times10^{-5}
Lead−1.7×10−5-1.7\times10^{-5}Niobium2.6×10−52.6\times10^{-5}
Mercury−2.9×10−5-2.9\times10^{-5}Oxygen (STP)2.1×10−62.1\times10^{-6}
Nitrogen−5.0×10−9-5.0\times10^{-9}Platinum2.9×10−42.9\times10^{-4}

11.5 Magnetic properties of materials

Type of materialRange of χ\chi
Diamagnetic−10−3-10^{-3} to −10−5-10^{-5}
Paramagnetic00 to 10−310^{-3}
Ferromagnetic10210^{2} to 10310^{3}

Diamagnetism

  • Atoms or molecules with completely filled electron orbits have no net magnetic moment. Placed in an external field, the field induces a current in the electron orbits (Lenz's law) whose own field opposes the applied field.
  • Of the paired orbits, the one whose loop current runs the same way as the induced current gains moment and the one running the opposite way loses moment, so a net moment appears opposite to the applied field.
  • Consequences: the field inside the material is reduced (lines of force are pushed out), the material is repelled by a magnet, and in a non-uniform field it moves from the stronger to the weaker part.
  • A diamagnetic liquid in a U-tube is pushed into the arm outside the field. A freely suspended diamagnetic rod sets itself perpendicular to the field, because that is the position of least magnetic potential energy for an induced dipole pointing against the field.
  • χ\chi is small and negative, μr<1\mu_r < 1, and it does not depend on temperature.
  • Examples: copper, gold, bismuth, lead, silicon, diamond, glass, water, wood, plastic, nitrogen. Diamagnetism is present in every material but is usually masked by the stronger paramagnetism or ferromagnetism.
  • Superconductors have χ=−1\chi = -1: the field lines are completely expelled. This perfect diamagnetism is the Meissner effect. An ordinary good conductor lets the lines penetrate its surface region.

Paramagnetism

  • Atoms or molecules with unpaired electrons carry a permanent magnetic dipole moment, but thermal agitation keeps these moments randomly oriented, so the material has no net moment in the absence of a field.
  • In a strong external field at low temperature most of the dipoles align along the field (the lower-energy orientation), the lines of force get closer together inside the material, and the material is weakly magnetised along the field.
  • In a non-uniform field a paramagnetic material moves from the weaker to the stronger part; a paramagnetic liquid in a U-tube rises in the arm nearer the magnet.
  • Remove the field and the dipoles randomise again — the magnetization is temporary.
  • χ\chi is small and positive and depends on temperature. Examples: aluminium, magnesium, lithium, molybdenum, tantalum, calcium, platinum, MnSO4\text{MnSO}_4, oxygen gas.
  • Curie law (Pierre Curie, 1895): M=CBTM = \dfrac{CB}{T}, hence χ=μ0CT\chi = \dfrac{\mu_0 C}{T} and χ∝1T\chi \propto \dfrac{1}{T}. Raising the field or lowering the temperature aligns more moments and increases the magnetization.

Ferromagnetism

  • Atoms carry moments as in a paramagnetic, but a strong exchange coupling (exchange interaction) between neighbouring moments — an order of magnitude stronger than ordinary dipole–dipole interaction — locks them parallel.
  • This creates domains: regions in which all atomic moments point the same way, along the domain axis. A domain is 10−610^{-6} to 10−4 m10^{-4}\ \text{m} across and holds 101010^{10} to 101710^{17} atoms. The boundary between two domains of different orientation is a domain wall.
  • Unmagnetised state: domain axes are random, so the net moment of the whole material is zero even though each domain has a large moment.
  • In a field, domains flip and rotate towards the field direction and favourably oriented domains grow; at a high enough field all the domains coalesce into one giant domain (saturation).
  • Removing the field does not restore the original domain boundaries, so a net moment is retained — the material is permanently magnetised.
  • Ferromagnetics are strongly attracted, move from the weaker to the stronger part of a non-uniform field, and have large positive χ\chi. Examples: iron, cobalt, nickel, gadolinium and some transition-metal alloys.
  • Beyond the three classes there are also ferrimagnetic, antiferromagnetic and spin-glass materials, which the syllabus only mentions.

Effect of temperature

  • Heating weakens the exchange coupling and distorts the domain structure. At the Curie temperature TCT_C the domain structure collapses completely and the material becomes paramagnetic.
  • Above TCT_C, χ=CT−TC\chi = \dfrac{C}{T - T_C}.
MaterialTCT_C (K)
Metallic cobalt1394
Metallic iron1043
Fe2O3\text{Fe}_2\text{O}_3893
Metallic nickel631
Metallic gadolinium317

The three distinctions the board sets

PropertyDiamagneticParamagneticFerromagnetic
Behaviour near a magnetWeakly repelledWeakly attractedStrongly attracted
In a non-uniform fieldMoves from stronger to weaker partMoves from weaker to stronger partMoves strongly from weaker to stronger part
Rod suspended in a uniform fieldRests slowly, length perpendicular to the fieldRests with length parallel to the fieldRests quickly, length parallel to the field
Atomic magnetic momentZeroNon-zero, randomly orientedNon-zero, aligned within domains
Direction of induced magnetizationOpposite to the fieldAlong the fieldAlong the field
χ\chiSmall, negativeSmall, positiveLarge, positive
μr\mu_rLess than 1Slightly more than 1Much greater than 1
Effect of temperatureNoneObeys Curie law, χ∝1T\chi \propto \dfrac{1}{T}Becomes paramagnetic above TCT_C
Permanent magnetisationNot possibleNot possible (temporary only)Possible

11.6 Hysteresis

  • Take an unmagnetised ferromagnetic rod inside a solenoid. Set H=nIH = nI, get M=χHM = \chi H, and plot B=μ0(H+M)B = \mu_0(H+M) against HH through one full cycle of the current.
  • The graph is non-linear and not retraced — the value of BB depends on the previous magnetic history of the specimen. That lag of BB behind HH is hysteresis.
  • Path of the loop: O→aO \to a initial magnetisation up to saturation (all domains merged); a→ba \to b reduce HH to zero and BB stays at ObOb, the retentivity or remanence; b→cb \to c reverse the current until B=0B = 0 at H=OcH = Oc, the coercivity (domain axes now randomly oriented); c→dc \to d reverse saturation; d→ed \to e reduce HH to zero, leaving retained magnetisation in the reverse sense; e→f→ae \to f \to a closes the loop.
  • Retentivity: the magnetic induction remaining in the material when the magnetising field is reduced to zero. Coercivity: the reverse magnetising field needed to reduce that induction to zero.
  • The area enclosed by the loop is the energy lost as heat per unit volume per cycle. This is the analogue of the stress–strain hysteresis of Std XI, where the loop area gave the energy dissipated in deformation.
PropertySoft ferromagnetic (soft iron)Hard ferromagnetic (steel, alnico)
Hysteresis loopNarrow, small areaWide, large area
RetentivityLowHigh
CoercivityLowHigh
PermeabilityVery high, above 1000Lower
Energy loss per cycleSmallLarge
Magnetised and demagnetisedEasilyWith difficulty
UseElectromagnets, transformer coresPermanent magnets

11.7 Permanent magnet and electromagnet

  • Electromagnet: a soft iron rod (large permeability, above 1000, and little retained magnetisation) is placed as the core of a solenoid. Switching the current on raises the solenoid's field about a thousandfold; switching it off leaves the field effectively zero.
  • Uses of electromagnets: electric bells, loudspeakers, circuit breakers, research laboratories, and giant electromagnets on cranes to lift heavy iron loads. Superconducting magnets give fields of a few tesla and are used in NMR spectroscopy.
  • Permanent magnet: the same arrangement but with a hard ferromagnetic rod. The current magnetises it, and because a hard material retains magnetization to a large extent, the rod stays magnetised after the current is switched off.
  • What a permanent magnet needs: high retentivity (so it stays strongly magnetised) and high coercivity (so a stray field does not demagnetise it).
  • What a transformer core or an electromagnet needs: low retentivity and low coercivity, plus high permeability and a narrow loop for small energy loss.

11.8 Magnetic shielding

  • A soft ferromagnetic material placed in a uniform field draws a large number of lines of force into itself and leaves few outside.
  • Make a closed structure of it — for example a spherical iron shell — and almost all the lines crowd into the walls of the shell, so very few pass through the enclosed space. The space inside is nearly field-free. This is magnetic shielding.
  • Application: an instrument that must be protected from an external magnetic field is completely surrounded by a soft ferromagnetic case. Used in spacecraft, and in laboratories where an experiment or a high-field magnet must be isolated.
  • The field inside a hollow iron box is therefore less than that outside — less, not zero.
  • Other kinds of shielding on the same idea: electrical shielding (insulation round household wiring) and acoustic shielding (sound-insulated recording studios).

Traps that cost marks

  • Writing Um=mBcos⁡θU_m = mB\cos\theta without the minus sign. The board's model answer is Um=−mBcos⁡θU_m = -mB\cos\theta, giving −mB-mB at θ=0∘\theta = 0^\circ (stable) and +mB+mB at θ=180∘\theta = 180^\circ (unstable).
  • Forgetting to convert gauss to tesla (1 gauss=10−4 T1\ \text{gauss} = 10^{-4}\ \text{T}) or g cm2\text{g cm}^2 to kg m2\text{kg m}^2 (multiply by 10−710^{-7}) before using T=2πImBT = 2\pi\sqrt{\dfrac{I}{mB}}.
  • Using tt in ∘C^\circ\text{C} in Curie's law. χT\chi T is constant only with TT in kelvin, so 27 ∘C27\ ^\circ\text{C} must become 300 K.
  • Confusing BB, HH and MM. H=nIH = nI depends only on the coil; MM is the response of the material; B=μ0(H+M)B = \mu_0(H+M) is the total field. HH and MM are in A m−1\text{A m}^{-1}, BB in tesla.
  • Quoting a unit for χ\chi or μr\mu_r. Both are pure numbers. μ\mu does have a unit, T m A−1\text{T m A}^{-1} or H m−1\text{H m}^{-1}.
  • Writing μr=χ\mu_r = \chi instead of μr=1+χ\mu_r = 1 + \chi. For a strongly magnetic material the 1 hardly matters, but for a paramagnetic it is the whole answer.
  • Saying a diamagnetic rod sets itself parallel to the field. It sets itself perpendicular; only para- and ferromagnetic rods align along the field.
  • Saying the field inside a hollow iron box is zero. The board's answer is that it is less than that outside.
  • Mixing up retentivity and coercivity. Retentivity is a value of BB (at H=0H=0); coercivity is a value of HH (at B=0B=0).
  • Choosing high retentivity for a transformer core. The core must be soft iron: low coercivity and low retentivity.
  • Dropping the minus sign in m⃗orb=−(e2me)L⃗\vec{m}_{orb} = -\left(\dfrac{e}{2m_e}\right)\vec{L}, or forgetting to say that m⃗orb\vec{m}_{orb} and L⃗\vec{L} are antiparallel and perpendicular to the orbital plane.
  • Reporting the magnetic moment of a coil as IAIA when it has NN turns; it is NIANIA.
  • Halving a magnet and expecting the period to be unchanged. Both II and mm change, and T′=T2T' = \dfrac{T}{2}.
Solved Numerical ExamplesWorked numericals to attempt first, each with a hint before the full solution.Coming soon

These are being added chapter by chapter. They are solved numericals from the prescribed book and the reference guides — not previous board questions, which live under Practice — so that you can work through the method before meeting it in a real paper.

Important DerivationsEvery derive / obtain / prove question in this chapter. 1 have come up in two or more papers.2
  1. 1
    Magnetic potential energy of a dipole; minimum and maximum values
    State the formula for magnetic potential energy of a dipole and hence obtain the minimum and maximum magnetic potential energy.

    Asked in: July 2022 (Q.7, 2 marks)

    Show the derivation
    Step 1
    Magnetic potential energy of a dipole:
    Um=−mBcos⁡θU_m=-mB\cos\theta
    Step 2
    Special cases:
    a. Minimum magnetic potential energy:
    Um=−mBU_m=-mB
    b. Maximum magnetic potential energy:
    Um=mBU_m=mB
  2. 2
    Period of a bar magnet vibrating in a magnetic field (angular S.H.M.) Repeats
    Obtain an expression for the period of a bar magnet vibrating in a uniform magnetic field and performing angular S.H.M.

    Asked in: March 2009; February 2026 (Q.17, 3 marks)

    Show the derivation
    Angular S.H.M. is defined as the oscillatory motion of a body in which the torque for angular acceleration is directly proportional to the angular displacement and its direction is opposite to that of angular displacement.
    Expression for frequency:
    Step 1
    If a bar magnet is freely suspended in the plane of a uniform magnetic field, it remains in equilibrium with its axis parallel to the direction of the field.
    Step 2
    If it is given a small angular displacement θ about an axis passing through its centre, perpendicular to itself and to the field and released, it performs angular oscillations.
    Step 3
    Let μ be the magnetic dipole moment and B the magnetic field.
    Step 4
    In the deflected position, a restoring torque acts on the magnet that tends to bring it back to its equilibrium position.
    Step 5
    The magnitude of this torque is τ=μBsin⁡θ\tau=\mu B\sin\theta
    If θ is small, sin⁡θ≈θ\sin\theta\approx\theta
    ∴τ=μBθ\therefore \tau=\mu B\theta
    Step 6
    For clockwise angular displacement θ, the restoring torque is in the anticlockwise direction.
    ∴τ=Iα=−μBθ\therefore \tau=I\alpha=-\mu B\theta
    Where, I is the moment of inertia of the bar magnet and α is its angular acceleration.
    ∴α=−(μBI)θ....(1)\therefore \alpha=-\left(\dfrac{\mu B}{I}\right)\theta\qquad....(1)
    Step 7
    Since μ, B and I are constants, equation (1) shows that angular acceleration is directly proportional to the angular displacement and directed opposite to the angular displacement. Hence the magnet performs angular S.H.M.
    Step 8
    The period of vibrations of the magnet is given by,
    T=2πangular acceleration per unit angular displacement=2πα/θT=\dfrac{2\pi}{\sqrt{\text{angular acceleration per unit angular displacement}}}=\dfrac{2\pi}{\sqrt{\alpha/\theta}}
    Thus, by considering magnitude of angular acceleration (α),
    T=2πIμBT=2\pi\sqrt{\dfrac{I}{\mu B}}
    Step 9
    Frequency =1T=12πμBI=\dfrac{1}{T}=\dfrac{1}{2\pi}\sqrt{\dfrac{\mu B}{I}}

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