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CalibratePhysics

Chapter 5

Magnetism and Matter

CBSE · 3 subtopics

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

Constants, units and conversions

  • μ0=4π×10−7 T m A−1\mu_0 = 4\pi \times 10^{-7}\ \text{T m A}^{-1} — permeability of free space. CBSE expects μ04π=10−7 T m A−1\dfrac{\mu_0}{4\pi} = 10^{-7}\ \text{T m A}^{-1} to be quoted directly in dipole-field numericals.
  • Magnetic field (magnetic induction / flux density) BB is in tesla (T), with 1 T=104 G1\ \text{T} = 10^{4}\ \text{G}, i.e. 1 G=10−4 T1\ \text{G} = 10^{-4}\ \text{T}. Dimensions [M T−2A−1][\text{M T}^{-2}\text{A}^{-1}].
  • Magnetic moment mm is in A m2\text{A m}^{2}, equivalently J T−1\text{J T}^{-1}. Dimensions [L2A][\text{L}^{2}\text{A}].
  • Magnetic flux ϕB\phi_B is in weber (Wb), with 1 Wb=1 T m21\ \text{Wb} = 1\ \text{T m}^{2}. Dimensions [M L2T−2A−1][\text{M L}^{2}\text{T}^{-2}\text{A}^{-1}].
  • Magnetisation MM and magnetic intensity HH are both in A m−1\text{A m}^{-1}, dimensions [L−1A][\text{L}^{-1}\text{A}].
  • Susceptibility χ\chi and relative permeability μr\mu_r are pure numbers (no units); permeability μ\mu has the same units as μ0\mu_0, i.e. T m A−1\text{T m A}^{-1} (also written N A−2\text{N A}^{-2}).

The bar magnet — dipole moment

FormulaSymbols, units and conditions
m⃗=NIA⃗\vec{m} = N I \vec{A}
  • Magnetic moment of a flat coil of NN turns, current II, face area AA; direction normal to the plane of the coil, fixed by the right-hand rule (curl fingers along the current, thumb gives m⃗\vec{m}). For a circular coil of radius aa, m=NIπa2m = N I \pi a^{2}.
m=n (2l) i (πa2)m = n\,(2l)\,i\,(\pi a^{2})
  • Magnetic moment of a solenoid of length 2l2l, radius aa, nn turns per unit length, current ii. Since n(2l)=Nn(2l) = N is the total number of turns, this is just m=NiAm = NiA again — use it when the question gives turns per metre.

Far field of a bar magnet (short dipole, r≫lr \gg l)

FormulaSymbols, units and conditions
BA=μ04π⋅2mr3B_A = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2m}{r^{3}}
  • Axial (end-on) field at distance rr from the centre, along the magnet's axis. Direction: parallel to m⃗\vec{m} (from S to N inside, i.e. along the axis away from the N pole).
B⃗E=−μ04π⋅m⃗r3\vec{B}_E = -\dfrac{\mu_0}{4\pi}\cdot\dfrac{\vec{m}}{r^{3}}
  • Equatorial (broadside-on) field at distance rr on the perpendicular bisector. Magnitude μ0m4πr3\dfrac{\mu_0 m}{4\pi r^{3}}; the minus sign means it is antiparallel to m⃗\vec{m}.
BABE=2\dfrac{B_A}{B_E} = 2
  • At the same distance the axial field is twice the equatorial field, and points the opposite way. True only for a short dipole, r≫lr \gg l.

The dipole in a uniform magnetic field

FormulaSymbols, units and conditions
τ⃗=m⃗×B⃗,τ=mBsin⁡θ\vec{\tau} = \vec{m}\times\vec{B}, \qquad \tau = mB\sin\theta
  • Torque on a magnet of moment mm in a uniform field BB; θ\theta is the angle between m⃗\vec{m} and B⃗\vec{B}. Maximum mBmB at θ=90∘\theta = 90^\circ, zero at θ=0∘\theta = 0^\circ and 180∘180^\circ.
F⃗net=0\vec{F}_{\text{net}} = 0
  • Net force on a magnetic dipole in a uniform field is always zero — equal and opposite forces on the two poles. A net force appears only in a non-uniform field.
U=−m⃗⋅B⃗=−mBcos⁡θU = -\vec{m}\cdot\vec{B} = -mB\cos\theta
  • Potential energy, with the zero fixed at θ=90∘\theta = 90^\circ. Minimum U=−mBU = -mB at θ=0∘\theta = 0^\circ (stable equilibrium); maximum U=+mBU = +mB at θ=180∘\theta = 180^\circ (unstable equilibrium).
W=U2−U1=−mB(cos⁡θ2−cos⁡θ1)W = U_2 - U_1 = -mB(\cos\theta_2 - \cos\theta_1)
  • Work done by an external agent in turning the magnet from θ1\theta_1 to θ2\theta_2 (slowly, no kinetic energy). Standard values: 0∘→90∘0^\circ \to 90^\circ gives W=mBW = mB; 0∘→180∘0^\circ \to 180^\circ gives W=2mBW = 2mB.
F r=mBsin⁡θF\,r = mB\sin\theta
  • Equilibrium of a pivoted magnet: an external force FF applied perpendicular to the axis at distance rr from the pivot balances the magnetic restoring torque.
T=2πImBT = 2\pi\sqrt{\dfrac{I}{mB}}
  • Time period of small angular oscillations of a magnet (moment of inertia II) about the field direction; used with the arrangement of Fig. 5.3(b) to measure BB or mm. The rationalised NCERT keeps the arrangement but no longer derives this; carry it only as a supporting result.

Gauss's law for magnetism

FormulaSymbols, units and conditions
ϕB=∑allB⃗⋅ΔS⃗=0\phi_B = \sum_{\text{all}} \vec{B}\cdot\Delta\vec{S} = 0
  • The net magnetic flux through any closed surface is zero. ΔS⃗\Delta\vec{S} is an outward area element of the closed surface. Holds for every closed surface, always.
ΔϕB=B⃗⋅ΔS⃗=B ΔScos⁡θ\Delta\phi_B = \vec{B}\cdot\Delta\vec{S} = B\,\Delta S\cos\theta
  • Flux through one small area element; θ\theta is the angle between B⃗\vec{B} and the outward normal.
∑E⃗⋅ΔS⃗=qε0\sum \vec{E}\cdot\Delta\vec{S} = \dfrac{q}{\varepsilon_0}
  • Gauss's law of electrostatics, for the contrast CBSE asks for: electric charge is a source of E⃗\vec{E}, but there is no magnetic charge, so the magnetic right-hand side is zero.
∮B⃗⋅dS⃗=μ0 qm\oint \vec{B}\cdot d\vec{S} = \mu_0\, q_m
  • What Gauss's law of magnetism would become if magnetic monopoles existed, qmq_m being the enclosed magnetic charge. Quote this only when the question asks how the law would be modified.

Magnetisation and magnetic intensity

FormulaSymbols, units and conditions
M⃗=m⃗netV\vec{M} = \dfrac{\vec{m}_{\text{net}}}{V}
  • Magnetisation: net magnetic moment per unit volume of the sample, in A m−1\text{A m}^{-1}.
B0=μ0nIB_0 = \mu_0 n I
  • Field inside a long air-cored solenoid, nn turns per unit length, current II. With a core, the total field is B=B0+BmB = B_0 + B_m, where Bm=μ0MB_m = \mu_0 M is the contribution of the material.
H⃗=B⃗μ0−M⃗\vec{H} = \dfrac{\vec{B}}{\mu_0} - \vec{M}
  • Definition of magnetic intensity (magnetising field), in A m−1\text{A m}^{-1}. HH measures the external cause, MM the material's response.
B⃗=μ0(H⃗+M⃗)\vec{B} = \mu_0(\vec{H} + \vec{M})
  • Total field inside a magnetised material. Rearranged form of the definition of HH.
H=nIH = nI
  • Magnetic intensity inside a solenoid: it depends only on the winding and the current, not on the core material. Use this first in every core numerical.
M⃗=χH⃗\vec{M} = \chi \vec{H}
  • Definition of magnetic susceptibility χ\chi (dimensionless) for a linear material. χ>0\chi > 0 and small: paramagnetic; χ<0\chi < 0 and small: diamagnetic (M⃗\vec{M} opposite to H⃗\vec{H}).
B⃗=μ0(1+χ)H⃗=μ0μrH⃗=μH⃗\vec{B} = \mu_0(1+\chi)\vec{H} = \mu_0\mu_r\vec{H} = \mu\vec{H}
  • Chain relating the three constants. Read off: μr=1+χ\mu_r = 1+\chi and μ=μ0μr=μ0(1+χ)\mu = \mu_0\mu_r = \mu_0(1+\chi). Only one of χ\chi, μr\mu_r, μ\mu is independent.
M=(μr−1)HM = (\mu_r - 1)H
  • Working form for magnetisation when the core's relative permeability is given; follows from M=χHM = \chi H with χ=μr−1\chi = \mu_r - 1.
B=μ0n(I+IM)B = \mu_0 n (I + I_M)
  • Magnetising current IMI_M: the extra current the same windings would need, with no core, to produce the field the core actually produces. Solve for IMI_M after finding BB.

Magnetic properties of materials

FormulaSymbols, units and conditions
−1≤χ<0,0≤μr<1,μ<μ0-1 \le \chi < 0,\qquad 0 \le \mu_r < 1,\qquad \mu < \mu_0
  • Diamagnetic range. A superconductor is the limiting case, χ=−1\chi = -1, μr=0\mu_r = 0 (Meissner effect, perfect diamagnetism).
0<χ<ϵ,1<μr<1+ϵ,μ>μ00 < \chi < \epsilon,\qquad 1 < \mu_r < 1+\epsilon,\qquad \mu > \mu_0
  • Paramagnetic range, ϵ\epsilon a small positive number (typically χ∼+10−5\chi \sim +10^{-5}).
χ≫1,μr≫1,μ≫μ0\chi \gg 1,\qquad \mu_r \gg 1,\qquad \mu \gg \mu_0
  • Ferromagnetic range; μr\mu_r can exceed 10001000, and BB is a non-linear function of HH.
χ=CT⟹χ1χ2=T2T1\chi = \dfrac{C}{T} \qquad\Longrightarrow\qquad \dfrac{\chi_1}{\chi_2} = \dfrac{T_2}{T_1}
  • Curie's law for a paramagnetic material: susceptibility is inversely proportional to absolute temperature TT (in kelvin), CC being the Curie constant. The ratio form is what the one-mark numericals use.
χ∝1T−TC(T>TC)\chi \propto \dfrac{1}{T - T_C} \quad (T > T_C)
  • Curie-Weiss law: above the Curie temperature TCT_C a ferromagnet loses its domain structure and behaves as a paramagnet.
μ=BH\mu = \dfrac{B}{H}
  • Permeability as a ratio. For a ferromagnet, as HH is increased BB saturates, so μ\mu falls — that is why the permeability of a ferromagnetic material decreases with increasing magnetising field.

Earth's field — only what the NCERT exercises need

FormulaSymbols, units and conditions
μ04π⋅2md3=BH\dfrac{\mu_0}{4\pi}\cdot\dfrac{2m}{d^{3}} = B_H
  • Null-point condition on the axis of a short bar magnet laid along the magnetic north-south line: the magnet's axial field cancels the horizontal component BHB_H of the earth's field at distance dd.
μ04π⋅md′3=BH⟹d′=d21/3\dfrac{\mu_0}{4\pi}\cdot\dfrac{m}{d'^{3}} = B_H \qquad\Longrightarrow\qquad d' = \dfrac{d}{2^{1/3}}
  • Turn the same magnet through 180∘180^\circ and the null points move onto the equatorial line, at 2−1/32^{-1/3} times the old distance.
Short NotesThe chapter subtopic by subtopic, in the textbook's own order.3 subtopics
  1. 1The Bar Magnet7 board questions
  2. 2Magnetism and Gauss's Law1 board question
  3. 3Magnetic Properties of Materials17 board questions
The whole chapter in brief

What this chapter is about

  • Chapter 4 showed that moving charges make magnetic fields. Here magnetism is studied in its own right: the bar magnet, Gauss's law for magnetism, and how bulk matter responds to a field.
  • Five facts stated in the NCERT introduction, all directly examinable at one mark:
  • The earth behaves as a magnet, with its field pointing approximately from geographic south to geographic north.
  • A freely suspended bar magnet sets itself north-south; the tip pointing to geographic north is the north pole.
  • Like poles repel, unlike poles attract.
  • Poles cannot be isolated. Break a bar magnet and you get two complete magnets. Magnetic monopoles do not exist.
  • Magnets can be made from iron and its alloys.
  • The word "magnet" comes from Magnesia, an island in Greece where magnetic ore was found around 600 BC.

5.2 The bar magnet and its field lines

  • Iron filings sprinkled on glass over a short bar magnet arrange themselves in a pattern that mimics the field lines and shows the magnet is a magnetic dipole with a north and a south pole. A current-carrying solenoid gives a very similar pattern.
  • Four properties of magnetic field lines — the standard three-mark answer:
  • They are continuous closed curves. Outside the magnet they run from N to S; inside the magnet they run from S to N. (Unlike electric field lines of a dipole, which begin on the positive charge and end on the negative charge or escape to infinity.)
  • The tangent at any point on a field line gives the direction of the net field B⃗\vec{B} at that point.
  • The number of lines crossing unit area measures the strength of B⃗\vec{B}: lines crowd together where the field is strong (near the poles) and spread apart where it is weak (near the middle).
  • Two field lines never intersect. If they did, the field would have two directions at the point of intersection, which is impossible.
  • Field lines are not "lines of force". The magnetic force on a moving charge is qv⃗×B⃗q\vec{v}\times\vec{B}, which is perpendicular to B⃗\vec{B}, so a field line does not show the direction of the force.
  • To map the field experimentally, place a small compass needle at various points and note its orientation.

5.2.2 Bar magnet as an equivalent solenoid

  • Ampere's hypothesis: all magnetic phenomena can be explained by circulating currents. A bar magnet behaves as a large number of circulating currents, like a solenoid.
  • Cutting a bar magnet in half is like cutting a solenoid — two smaller, weaker magnets, each with both poles; the field lines stay continuous.
  • Calculating the axial field of a finite solenoid at a far point gives

B=μ04π⋅2mr3,m=n(2l) i (πa2)=NiA B = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2m}{r^{3}}, \qquad m = n(2l)\,i\,(\pi a^{2}) = NiA which is exactly the far axial field of a bar magnet. Hence the magnetic moment of a bar magnet equals the magnetic moment of the equivalent solenoid producing the same field.

5.2.3 The dipole in a uniform magnetic field

  • Torque τ⃗=m⃗×B⃗\vec{\tau} = \vec{m}\times\vec{B}, magnitude τ=mBsin⁡θ\tau = mB\sin\theta; it is a restoring torque, turning m⃗\vec{m} towards B⃗\vec{B}.
  • Net force is zero in a uniform field. So a compass needle turns but does not drift.
  • Potential energy U=−m⃗⋅B⃗=−mBcos⁡θU = -\vec{m}\cdot\vec{B} = -mB\cos\theta, with the zero chosen at θ=90∘\theta = 90^\circ.
  • θ=0∘\theta = 0^\circ: U=−mBU = -mB, minimum — stable equilibrium, τ=0\tau = 0.
  • θ=90∘\theta = 90^\circ: U=0U = 0, τ=mB\tau = mB maximum.
  • θ=180∘\theta = 180^\circ: U=+mBU = +mB, maximum — unstable equilibrium, τ=0\tau = 0.
  • Work done in rotating from θ1\theta_1 to θ2\theta_2: W=−mB(cos⁡θ2−cos⁡θ1)W = -mB(\cos\theta_2 - \cos\theta_1).
  • An iron nail near a bar magnet feels a force as well as a torque, because the magnet's field there is non-uniform and the nail acquires an induced moment; the attraction arises because the induced pole nearer the magnet is the unlike one.

5.2.4 The electrostatic analogue

  • The far field of a magnetic dipole is obtained from the electric dipole result by E⃗→B⃗\vec{E}\to\vec{B}, p⃗→m⃗\vec{p}\to\vec{m}, 14πε0→μ04π\dfrac{1}{4\pi\varepsilon_0}\to\dfrac{\mu_0}{4\pi}.
QuantityElectric dipoleMagnetic dipole
Constant1ε0\dfrac{1}{\varepsilon_0}μ0\mu_0
Dipole momentp⃗\vec{p}m⃗\vec{m}
Equatorial field, short dipole−p⃗4πε0r3-\dfrac{\vec{p}}{4\pi\varepsilon_0 r^{3}}−μ0m⃗4πr3-\dfrac{\mu_0\vec{m}}{4\pi r^{3}}
Axial field, short dipole2p⃗4πε0r3\dfrac{2\vec{p}}{4\pi\varepsilon_0 r^{3}}μ04π2m⃗r3\dfrac{\mu_0}{4\pi}\dfrac{2\vec{m}}{r^{3}}
Torque in an external fieldp⃗×E⃗\vec{p}\times\vec{E}m⃗×B⃗\vec{m}\times\vec{B}
Energy in an external field−p⃗⋅E⃗-\vec{p}\cdot\vec{E}−m⃗⋅B⃗-\vec{m}\cdot\vec{B}

5.3 Gauss's law for magnetism

  • Statement: the net magnetic flux through any closed surface is zero, ϕB=∑B⃗⋅ΔS⃗=0\phi_B = \sum \vec{B}\cdot\Delta\vec{S} = 0.
  • Reason: magnetic field lines are continuous closed loops, so as many lines enter a closed surface as leave it.
  • Significance: isolated magnetic poles (monopoles) do not exist. There are no sources or sinks of B⃗\vec{B}; the simplest magnetic element is a dipole or a current loop, and all magnetic phenomena are explained by arrangements of dipoles and current loops.
  • Contrast with electrostatics: ∑E⃗⋅ΔS⃗=qε0\sum\vec{E}\cdot\Delta\vec{S} = \dfrac{q}{\varepsilon_0} — a non-zero right-hand side, because electric charge exists and can be isolated.
  • If monopoles existed the law would read ∮B⃗⋅dS⃗=μ0qm\oint\vec{B}\cdot d\vec{S} = \mu_0 q_m, with qmq_m the enclosed magnetic charge.
  • Around a bar magnet's N pole alone, the net flux is still zero — field lines pass through the magnet's interior and close the loops.

5.3 Reading field-line diagrams (the standard "what is wrong" question)

  • Lines emanating radially from a point: wrong for magnetism — the net flux through a surface around that point would not be zero. That is the pattern of a long charged wire's electric field.
  • Lines crossing: always wrong, for both B⃗\vec{B} and E⃗\vec{E} — the direction would be ambiguous.
  • A closed loop of static magnetic field lines around empty space: wrong. A closed magnetostatic field line must enclose a region through which current passes.
  • Lines confined entirely inside a toroid, forming closed loops: right — each loop encloses current.
  • Solenoid lines running perfectly straight and staying confined at the ends: wrong; they must curve out and close, or Ampere's law is violated.
  • Lines inside and outside a bar magnet forming closed loops: right. Not all lines emanate from the N pole; the net flux around either pole is zero.
  • Field lines between two pole pieces being perfectly straight right to the edges: wrong. Some fringing is inevitable.
  • Lines all emerging from one plate and ending on another: an electrostatic pattern, impossible for B⃗\vec{B}.
  • A bar magnet exerts no torque on itself due to its own field; nor does a current element feel a force from its own field (though it can feel one from other elements of the same wire).
  • A system with zero net charge can still have a magnetic moment — the atomic current loops need not cancel.

5.4 Magnetisation and magnetic intensity

  • Magnetisation M⃗=m⃗netV\vec{M} = \dfrac{\vec{m}_{\text{net}}}{V}: net magnetic moment per unit volume, in A m−1\text{A m}^{-1}.
  • Put a core inside a solenoid whose empty-core field is B0=μ0nIB_0 = \mu_0 nI. The field becomes B=B0+BmB = B_0 + B_m, where Bm=μ0MB_m = \mu_0 M is the core's contribution.
  • Magnetic intensity H⃗=B⃗μ0−M⃗\vec{H} = \dfrac{\vec{B}}{\mu_0} - \vec{M}, so that B⃗=μ0(H⃗+M⃗)\vec{B} = \mu_0(\vec{H}+\vec{M}). In a solenoid H=nIH = nI: HH carries the external cause (the current), MM the material's response.
  • For a linear material M⃗=χH⃗\vec{M} = \chi\vec{H}, giving B=μ0(1+χ)H=μ0μrH=μHB = \mu_0(1+\chi)H = \mu_0\mu_r H = \mu H.
SymbolNameRelationUnits
MMmagnetisationM=mnetVM = \dfrac{m_{\text{net}}}{V}A m−1\text{A m}^{-1}
HHmagnetic intensityH=Bμ0−MH = \dfrac{B}{\mu_0} - MA m−1\text{A m}^{-1}
χ\chimagnetic susceptibilityM=χHM = \chi Hdimensionless
μr\mu_rrelative permeabilityμr=1+χ\mu_r = 1+\chidimensionless
μ\mupermeabilityμ=μ0μr\mu = \mu_0\mu_rT m A−1\text{T m A}^{-1}
  • μr\mu_r is the magnetic analogue of the dielectric constant in electrostatics.
  • Only one of χ\chi, μr\mu_r, μ\mu is independent — given one, the other two follow.

5.5 Classifying materials

PropertyDiamagneticParamagneticFerromagnetic
Susceptibility χ\chi−1≤χ<0-1 \le \chi < 0 (small, negative)0<χ<ϵ0 < \chi < \epsilon (small, positive)χ≫1\chi \gg 1 (large, positive)
Relative permeability μr\mu_r0≤μr<10 \le \mu_r < 11<μr<1+ϵ1 < \mu_r < 1+\epsilonμr≫1\mu_r \gg 1, over 10001000
Permeability μ\muμ<μ0\mu < \mu_0μ>μ0\mu > \mu_0μ≫μ0\mu \gg \mu_0
Atomic moment with no fieldzeropermanent, randomly orientedpermanent, aligned within domains
Field lines in the specimenpushed out, sparser insidepulled in, denser insidestrongly concentrated inside
In a non-uniform fieldmoves strong to weak (repelled)moves weak to strong (weakly attracted)moves weak to strong (strongly attracted)
Effect of temperatureessentially independent of TTχ=C/T\chi = C/T (Curie's law)paramagnetic above TCT_C
Examplesbismuth, copper, lead, silicon, nitrogen at STP, water, sodium chloridealuminium, sodium, calcium, oxygen at STP, copper chlorideiron, cobalt, nickel, gadolinium, alnico

5.5.1 Diamagnetism

  • Diamagnetic substances tend to move from the stronger to the weaker part of an external field: a magnet repels them.
  • Their atoms have zero resultant magnetic moment in the absence of a field. Orbiting electrons are equivalent current loops whose orbital moments cancel.
  • When a field is applied, electrons whose orbital moment is along the field slow down and those opposed speed up (an induced-current effect, in accordance with Lenz's law). The substance develops a net moment opposite to the applied field, hence the repulsion.
  • Inside the specimen the field is reduced, typically by about one part in 10510^{5}; the field lines are expelled.
  • Diamagnetism is universal — present in every substance — but so weak that paramagnetism or ferromagnetism masks it where those are present.
  • Superconductors are perfect diamagnets: a metal cooled to a very low temperature showing both perfect conductivity and perfect diamagnetism, with the field lines completely expelled, χ=−1\chi = -1, μr=0\mu_r = 0. This is the Meissner effect. Superconducting magnets are used, for example, in magnetically levitated trains.

5.5.2 Paramagnetism

  • Paramagnetic substances get weakly magnetised along an external field and tend to move from weak field to strong field: they are weakly attracted.
  • Each atom, ion or molecule has a permanent magnetic dipole moment, but ceaseless random thermal motion keeps them disordered, so there is no net magnetisation without a field.
  • In a strong enough field, and at low enough temperature, the atomic dipoles align with B⃗0\vec{B}_0. Raising the field or lowering the temperature increases the magnetisation until saturation, when every dipole is aligned.
  • Inside the specimen the field is enhanced, again typically by about one part in 10510^{5}; the field lines crowd in.
  • χ\chi and μr\mu_r depend on the sample temperature as well as on the material: Curie's law, χ=CT\chi = \dfrac{C}{T}.

5.5.3 Ferromagnetism

  • Ferromagnetic substances get strongly magnetised and are strongly attracted from weak field towards strong field.
  • Their atoms carry permanent dipole moments as in a paramagnet, but interact so as to align spontaneously over a macroscopic volume called a domain. Typical domain size is about 1 mm1\ \text{mm}, containing roughly 101110^{11} atoms.
  • With no applied field the domain magnetisations point randomly and there is no bulk magnetisation. Applying B⃗0\vec{B}_0 turns the domains towards the field and the favourably oriented domains grow, until the sample becomes a single giant domain. Domain motion is observable under a microscope using a suspension of powdered ferromagnetic material.
  • Field lines are highly concentrated inside a ferromagnet; μr\mu_r exceeds 10001000.
  • Hard ferromagnets retain their magnetisation when the field is removed — alnico (iron, aluminium, nickel, cobalt, copper) and naturally occurring lodestone. These make permanent magnets and compass needles.
  • Soft ferromagnets lose their magnetisation as soon as the field is removed — soft iron is the standard example.
  • Above the Curie temperature TCT_C the domain structure disintegrates and the ferromagnet becomes a paramagnet, following χ∝1T−TC\chi \propto \dfrac{1}{T-T_C}. The loss of magnetisation with temperature is gradual.
  • Because BB saturates while HH keeps rising, μ=BH\mu = \dfrac{B}{H} for a ferromagnet decreases as the magnetising field is increased.
  • Beyond the three main classes there also exist ferrimagnetic, anti-ferromagnetic and spin-glass materials.

Diagrams described in words

  • Bar magnet field lines: closed loops leaving the N pole, curving round outside to enter the S pole, and continuing inside the magnet from S to N. Lines crowded near both poles, sparse near the middle. The same picture, drawn for a finite current-carrying solenoid, is indistinguishable at large distances.
  • Diamagnetic bar in a uniform field: draw the uniform horizontal lines bending away around the bar, so that inside the bar the lines are fewer and more widely spaced than outside — the specimen expels the field.
  • Paramagnetic bar in the same field: the lines bend inwards towards the bar and crowd through it, so the density inside is greater than outside.
  • Gaussian surface round one pole of a magnet: draw a closed surface enclosing just the N pole; every line leaving it through the outside is matched by a line entering it through the magnet's body, so the net flux is zero.

Numbers worth remembering

  • μ04π=10−7 T m A−1\dfrac{\mu_0}{4\pi} = 10^{-7}\ \text{T m A}^{-1}; 1 G=10−4 T1\ \text{G} = 10^{-4}\ \text{T}.
  • Diamagnetic: χ≈−10−5\chi \approx -10^{-5}. Paramagnetic: χ≈+10−5\chi \approx +10^{-5}. Superconductor: χ=−1\chi = -1, μr=0\mu_r = 0.
  • Copper: μr=0.999994\mu_r = 0.999994, diamagnetic.
  • Ferromagnet: μr>1000\mu_r > 1000.
  • Axial field is twice the equatorial field at the same distance, and oppositely directed.

Traps that cost marks

  • Do not write that field lines start at N and end at S. They are closed loops: N to S outside, S to N inside. This exact wording has been the answer to a one-mark MCQ.
  • A uniform field gives torque but never a net force. Only a non-uniform field (a real bar magnet near a nail) gives a force as well.
  • With m⃗\vec{m} parallel to B⃗\vec{B}: torque is zero and force is zero. Do not write MBMB for either.
  • In W=−mB(cos⁡θ2−cos⁡θ1)W = -mB(\cos\theta_2 - \cos\theta_1), θ1\theta_1 is the initial and θ2\theta_2 the final angle. Getting them the wrong way round flips the sign of the work.
  • Stable equilibrium is θ=0∘\theta = 0^\circ with U=−mBU = -mB; unstable is θ=180∘\theta = 180^\circ with U=+mBU = +mB. Both have zero torque, so "torque is zero" alone does not identify stability — quote the energy.
  • Gauss's law for magnetism gives zero for every closed surface, not just symmetric ones, and not just ones outside the magnet.
  • χ\chi negative means diamagnetic, however large its magnitude — a value of −0.5-0.5 is still diamagnetic, since μr=1+χ=0.5<1\mu_r = 1+\chi = 0.5 < 1.
  • Curie's law is χ∝1/T\chi \propto 1/T with TT in kelvin, and applies to paramagnets. For a ferromagnet above TCT_C use χ∝1/(T−TC)\chi \propto 1/(T-T_C).
  • Above the Curie temperature a ferromagnet becomes paramagnetic, not diamagnetic.
  • HH inside a solenoid is nInI and does not depend on the core. Only BB and MM change when the core changes.
  • Increasing HH on a ferromagnet decreases its permeability, because BB is approaching saturation.
  • Do not talk about isolated pole strengths as though monopoles exist; the rationalised NCERT treats the bar magnet as a dipole equivalent to a solenoid.
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.2
  1. 1
    Derivation 1
    (a) Derive the expression for the force acting between two long parallel current-carrying conductors. Hence, define 1 A current. (b) A bar magnet of dipole moment 3 A m23\,\text{A m}^2 rests with its centre on a frictionless pivot. A force FF is applied at right angles to the axis of the magnet, 10 cm10\,\text{cm} from the pivot. It is observed that an external magnetic field of 0.25 T0.25\,\text{T} is required to hold the magnet in equilibrium at an angle of 30°30° with the field. Calculate the value of FF. How will the equilibrium be affected if FF is withdrawn?

    Asked in: CBSE Class 12 Physics — Long Answer · AI 2020 · Section E, 5 marks

    Show the derivation

    (a) Consider two long, straight, parallel conductors AA and BB, carrying currents IaI_a and IbI_b respectively, separated by a distance dd. Magnetic field produced by conductor AA at any point on conductor BB: BA=μ0Ia2πdB_A = \dfrac{\mu_0 I_a}{2\pi d} ...(i) (directed perpendicular to BB) This field exerts a force on conductor BB (carrying current IbI_b). The force per unit length on BB is Fl=IbBA=Ib×μ0Ia2πd\dfrac{F}{l} = I_b B_A = I_b\times\dfrac{\mu_0 I_a}{2\pi d} Fl=μ0IaIb2πd\dfrac{F}{l} = \dfrac{\mu_0 I_a I_b}{2\pi d} By Newton's third law, BB exerts an equal and opposite force on AA. The force is attractive if the currents are in the same direction, and repulsive if they are opposite. Definition of 1 ampere: One ampere is that constant current which, if maintained in each of two straight parallel conductors of infinite length and negligible cross-section, placed 1 m apart in vacuum, would produce between them a force of 2×10−7 N2\times10^{-7}\,\text{N} per metre of length. (b) Given: m=3 A m2m = 3\,\text{A m}^2, r=10 cm=0.10 mr = 10\,\text{cm} = 0.10\,\text{m}, B=0.25 TB = 0.25\,\text{T}, θ=30°\theta = 30° In equilibrium, the deflecting torque due to FF equals the restoring torque due to BB: F×r=mBsin⁡θF\times r = mB\sin\theta F×0.10=3×0.25×sin⁡30°=3×0.25×0.5F\times0.10 = 3\times0.25\times\sin30° = 3\times0.25\times0.5 F=0.3750.10=3.75 NF = \dfrac{0.375}{0.10} = 3.75\,\text{N} If FF is withdrawn, the magnet is no longer in equilibrium. The restoring torque due to B⃗\vec{B} makes it oscillate about the field direction for some time, and — due to damping — it finally comes to rest aligned along the direction of the external magnetic field.

  2. 2
    Derivation 2
    (a) Show that a current-carrying solenoid behaves like a small bar magnet. Obtain the expression for the magnetic field at an external point lying on its axis. (b) A steady current of 2 A2\,\text{A} flows through a circular coil having 55 turns of radius 7 cm7\,\text{cm}. The coil lies in the XX-YY plane with its centre at the origin. Find the magnitude and direction of the magnetic dipole moment of the coil.

    Asked in: CBSE Class 12 Physics — Long Answer · AI 2020, Ev · Section E, 5 marks

    Show the derivation

    (a) A solenoid may be regarded as a combination of a large number of identical circular current loops, each of which behaves like a magnetic dipole. Hence, a current-carrying solenoid behaves like a small bar magnet. Consider a solenoid of length 2l2l, radius aa, having nn turns per unit length and carrying current ii. Consider a circular element of thickness dxdx at a distance xx from the centre OO of the solenoid, at an axial point PP (distance rr from OO). Number of turns in this element =n dx= n\,dx. The magnetic field at PP due to this circular element: dB=μ0 n dx i a22[(r−x)2+a2]3/2dB = \dfrac{\mu_0\,n\,dx\,i\,a^2}{2\left[(r-x)^2+a^2\right]^{3/2}} Integrating over the length of the solenoid, B=μ0nia22∫−lldx[(r−x)2+a2]3/2B = \dfrac{\mu_0 n i a^2}{2}\displaystyle\int_{-l}^{l}\dfrac{dx}{\left[(r-x)^2+a^2\right]^{3/2}} For a point PP far away from the solenoid, r≫ar \gg a and r≫lr \gg l, so [(r−x)2+a2]3/2≈r3\left[(r-x)^2+a^2\right]^{3/2} \approx r^3, and B=μ0nia22r3∫−lldx=μ0ni(2l)a22r3B = \dfrac{\mu_0 n i a^2}{2r^3}\displaystyle\int_{-l}^{l}dx = \dfrac{\mu_0 n i (2l) a^2}{2r^3} B=μ04π⋅2mr3B = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2m}{r^3}, where m=n(2l) i (πa2)m = n(2l)\,i\,(\pi a^2) is the magnetic moment of the solenoid — identical in form to the axial field of a bar magnet, confirming the solenoid behaves as one. (b) Given: N=5N = 5, i=2 Ai = 2\,\text{A}, a=7 cm=7×10−2 ma = 7\,\text{cm} = 7\times10^{-2}\,\text{m} Magnetic dipole moment, M=Niπa2=5×2×227×(7×10−2)2M = Ni\pi a^2 = 5\times2\times\dfrac{22}{7}\times(7\times10^{-2})^2 M=5×2×227×49×10−4=154×10−3=0.154 A m2M = 5\times2\times\dfrac{22}{7}\times49\times10^{-4} = 154\times10^{-3} = 0.154\,\text{A m}^2 M⃗\vec{M} is directed perpendicular to the plane of the coil (the XX-YY plane), i.e., along the zz-axis — the direction given by the right-hand rule (curl the fingers along the current direction; the thumb points along M⃗\vec{M}).

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