Physics by Lamhi: Not Your Boring Physics

CLASS 12 · CHAPTER 5 · MAGNETISM

DETAILED NOTES

Magnetism and Matter

The complete chapter, section by section: the bar magnet as an equivalent solenoid, dipole moment and its axial field, torque and potential energy on a dipole (and the stable-vs-unstable trap that follows from it), Gauss's law for magnetism and why it rules out magnetic monopoles, Earth's magnetism, magnetisation, magnetic intensity and susceptibility (B = μ₀(H + M)), how diamagnetic, paramagnetic, and ferromagnetic materials differ, and the hysteresis loop that decides whether a ferromagnet belongs in an electromagnet or a permanent magnet, explained in plain language with original worked examples.

5.1 The Bar Magnet as an Equivalent Solenoid

A bar magnet has no literal “magnetic charge” sitting at its ends, unlike the electric charges at the heart of electrostatics. Instead, its field is produced by countless tiny atomic current loops inside the material, all lining up. Viewed from outside, a bar magnet produces a field pattern essentially identical to that of a finite current-carrying solenoid of the same length, this is the equivalent solenoid picture.

Just as a current loop of NN turns, current II, and area AA has a magnetic dipole moment m=NIAm=NIA, pointing along the loop’s axis (right-hand rule), the bar magnet is assigned a dipole moment mm of the same kind, directed from its south pole to its north pole:

Magnetic dipole moment (equivalent solenoid)
m=NIAm = NIA

(The dipole moment is written as a lowercase mm, as NCERT does, because the capital MM is needed in Section 5.6 for something different: magnetisation.)

This equivalence is more than a loose analogy: a bar magnet and a solenoid of matching mm experience the identical torque and potential-energy laws in an external field (Section 5.3), and produce the same field shape far from either one.

Worked example

Dipole moment of a solenoid

A solenoid has 300 turns, carries 2 A, and has a cross-sectional area of 4 cm². Find its magnetic dipole moment.

m=NIA=(300)(2)(4×10−4)=0.24 A⋅m2m = NIA = (300)(2)(4\times10^{-4}) = \textcolor{#e08a1e}{0.24\text{ A·m}^2}

5.2 Dipole Moment & Axial Field

A bar magnet can be modelled as two equal and opposite pole strengths ±qm\pm q_m, separated by a length 2l2l, with dipole moment m=qm(2l)m = q_m(2l), the direct magnetic analogue of an electric dipole’s p=q(2a)p=q(2a) (see Electric Charges and Fields). Along the magnet’s own axis, at a distance rr large compared to its length, the field strength is:

Axial field of a magnetic dipole (r ≫ l)
Baxis=μ04π2mr3B_{axis} = \dfrac{\mu_0}{4\pi}\dfrac{2m}{r^3}

Just like the electric dipole’s field, this falls off as 1/r31/r^3, faster than a single pole’s 1/r21/r^2 would, because the two opposite poles’ fields very nearly cancel at large distance, leaving only a small residual that shrinks faster with distance than either pole’s field alone. On the equatorial line (perpendicular bisector of the magnet), the field is half this, and points the opposite way:

Equatorial field of a magnetic dipole (r ≫ l)
Beq=μ04πmr3B_{eq} = \dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}

5.3 Torque & Potential Energy in a Uniform Field

Place a bar magnet of moment mm in a uniform external field BB, at angle θ\theta between them. Each pole feels a force of magnitude qmBq_mB, equal and opposite, separated by a perpendicular lever arm 2lsin⁡θ2l\sin\theta. The resulting couple is:

Torque on a dipole in a uniform field
τ=mBsin⁡θ\tau = m B \sin\theta

The potential energy, defined so that U=0U=0 at θ=90°\theta=90° (the reference angle at which the torque is maximum, not zero), comes from integrating the torque needed to rotate the dipole from 90° to θ\theta:

Potential energy of a dipole
U=−mBcos⁡θU = -mB\cos\theta

5.3.1 Case by case: is zero torque the same as stability?

No, and this is the single most important distinction in the chapter. Both θ=0°\theta=0° and θ=180°\theta=180° give sin⁡θ=0\sin\theta=0, hence τ=0\tau=0 at both. But they behave oppositely under a small nudge:

  • θ = 0° (m ∥ B): U=−mBU=-mB, the minimum possible value of UU. Nudge θ\theta slightly away and UU rises; a restoring torque appears and pushes the dipole back toward 0°. Stable equilibrium.
  • θ = 180° (m antiparallel to B): U=+mBU=+mB, the maximum possible value of UU. Nudge θ\theta slightly away and UU falls; the torque that appears grows rather than restoring, carrying the dipole all the way around to 0°, never back to 180°. Unstable equilibrium.

The rule: zero torque is necessary for an equilibrium but doesn’t say which kind it is. Check whether UU is at a minimum (stable) or a maximum (unstable) right around that point. This is exactly why a compass needle always settles pointing to magnetic north, never south, even though “pointing south” is momentarily also a zero-torque position.

Worked example

Torque and energy at two angles

A bar magnet of moment 0.6 A·m² sits in a uniform field of 0.2 T. Find the torque at θ = 60°, and the potential energy at both θ = 0° and θ = 180°.

τ=mBsin⁡60°=(0.6)(0.2)(0.8660)=0.1039 N⋅m\tau = mB\sin 60° = (0.6)(0.2)(0.8660) = \textcolor{#e08a1e}{0.1039\text{ N·m}}U(0°)=−mB=−(0.6)(0.2)=−0.12 J (stable)U(0°) = -mB = -(0.6)(0.2) = \textcolor{#e08a1e}{-0.12\text{ J (stable)}}U(180°)=+mB=+(0.6)(0.2)=+0.12 J (unstable)U(180°) = +mB = +(0.6)(0.2) = \textcolor{#e08a1e}{+0.12\text{ J (unstable)}}

5.4 Gauss’s Law for Magnetism

Section 5.2 treated a bar magnet as if it carried two “magnetic charges” at its ends. That picture gives the right field outside the magnet, but Section 5.1 already warned that no such charges really exist. Gauss’s law for magnetism is the precise form of that warning.

Take any closed surface (a balloon, a box, any shape at all) and chop it into small patches. For a patch of area ΔA\Delta A, with θ\theta the angle between the field B\mathbf{B} and the patch’s outward normal, the magnetic flux through it is

ΔΦB=B⋅ΔA=B ΔAcos⁡θ\Delta\Phi_B = \mathbf{B}\cdot\Delta\mathbf{A} = B\,\Delta A\cos\theta

positive where field lines leave the surface, negative where they enter, and measured in webers (1 Wb = 1 T·m²). Adding up every patch gives the net flux, and for a magnetic field that total always comes out to exactly zero, whatever the shape of the surface and whatever magnets or currents sit inside or outside it:

Gauss's law for magnetism
ΦB=∑all patchesB⋅ΔA=∮B⋅dA=0\Phi_B = \sum_{\text{all patches}}\mathbf{B}\cdot\Delta\mathbf{A} = \oint \mathbf{B}\cdot d\mathbf{A} = \textcolor{#e08a1e}{0}

5.4.1 Why the net flux must be zero

Magnetic field lines never start or stop anywhere. Every one of them is a closed loop: out of a magnet’s north end, round through the space outside, into the south end, and back through the inside of the magnet to where it began (for a solenoid, back through the inside of the coil). There are no magnetic monopoles, no isolated north or south “charges” for a line to begin on or end at.

Now follow one of those closed loops as it meets a closed surface. Every time it crosses outward, it has to cross back inward somewhere else to return to its starting point, so its crossings always come in out-in pairs. Flux counts an outward crossing as positive and an inward one as negative, so each pair cancels, every loop contributes exactly zero, and so does the total over all the loops. (A loop that never touches the surface contributes nothing in the first place.)

The most tempting counterexample is a surface drawn around just the north end of a bar magnet. Lines pour out of it, so surely the net flux is outward? It isn’t, because that surface also cuts through the magnet itself, and inside the magnet the field runs from the south end to the north end, carrying exactly as much flux into the surface as the outside lines carry out. This is also why cutting a magnet in half never isolates a pole: each piece turns out to have its own north and south end, just as cutting a solenoid in two gives two shorter solenoids. The pole-strength model of Section 5.2 is therefore a convenient fiction: it gets the field outside the magnet right, but it would wrongly predict an outward flux through a surface around one end, because it leaves out the field inside the magnet.

5.4.2 Side by side with Gauss's law for electric fields

Compare this with Gauss’s law for electric fields (see Electric Charges and Fields):

∮E⋅dA=qencε0versus∮B⋅dA=0\oint \mathbf{E}\cdot d\mathbf{A} = \dfrac{q_{\text{enc}}}{\varepsilon_0} \qquad\text{versus}\qquad \oint \mathbf{B}\cdot d\mathbf{A} = 0

Electric field lines can start and stop: they begin on positive charges and end on negative ones. Wrap a surface around a positive charge and lines leave through it with nothing to bring them back, so the net flux is positive, in proportion to the charge enclosed. Magnetic field lines have no such start or end points, so the right-hand side has nothing to count and is always zero. The entire difference between the two laws is whether or not there is a “charge” for field lines to begin and end on.

One way to see that this is the only difference: if a magnetic monopole of strength qmq_m were ever found, the magnetic law would gain a right-hand side shaped exactly like the electric one, ∮B⋅dA=μ0qm\oint \mathbf{B}\cdot d\mathbf{A} = \mu_0 q_m. Physicists have searched for monopoles, and none has ever been found, so the right-hand side stays at zero.

Worked example

The flux you can't see: inside the magnet

A closed box surrounds the north end of a bar magnet, with the magnet poking in through the bottom face. Field lines leave through the top face carrying +2.4×10⁻⁵ Wb, and through each of the four side faces carrying +1.1×10⁻⁵ Wb. Find the flux through the bottom face, and estimate the field inside the magnet if its cross-section is 2.0 cm².

Gauss’s law says the fluxes through all six faces must add to zero:

Φbottom=−[2.4×10−5+4(1.1×10−5)]=−6.8×10−5 Wb\Phi_{\text{bottom}} = -\left[2.4\times10^{-5} + 4(1.1\times10^{-5})\right] = \textcolor{#e08a1e}{-6.8\times10^{-5}\text{ Wb}}

The minus sign means this flux goes into the box: it is the field inside the magnet, running from the south end up to the north end. If the bottom face fits closely around the magnet, nearly all of that flux passes through the magnet’s own 2.0 cm² cross-section, so the field in there is roughly:

Binside≈6.8×10−52.0×10−4=0.34 TB_{\text{inside}} \approx \dfrac{6.8\times10^{-5}}{2.0\times10^{-4}} = \textcolor{#e08a1e}{0.34\text{ T}}

The electric version of this box, wrapped around a positive charge, would have a positive net flux. The magnetic one cannot, which is the whole content of Gauss’s law for magnetism.

5.5 Earth’s Magnetism

The Earth itself behaves roughly like a giant bar magnet tilted from its rotation axis, and its field at any location is described by three elements:

5.5.1 Declination

The angle between geographic north and the direction a compass needle actually points (magnetic north), measured in the horizontal plane. It varies from place to place on the Earth’s surface.

5.5.2 Inclination (dip)

The angle the Earth’s total field makes with the horizontal plane at a location. Near the magnetic equator the field is nearly horizontal (dip ≈ 0°); near the magnetic poles it points almost straight down into the Earth (dip ≈ 90°).

5.5.3 Horizontal component

The part of the Earth’s total field BEB_E lying in the horizontal plane, the component a horizontal compass needle actually responds to:

HE=BEcos⁡(dip),ZE=BEsin⁡(dip)H_E = B_E\cos(\text{dip}), \qquad Z_E = B_E\sin(\text{dip})

where ZEZ_E is the vertical component. Together, HEH_E and ZEZ_E (plus declination, which fixes the horizontal direction) completely specify the Earth’s field at a point. (The subscript EE keeps this horizontal component from being confused with the magnetic intensity HH of Section 5.6, a different quantity that happens to share the letter.)

5.6 Magnetisation and Magnetic Intensity

Slide a rod of some material into a current-carrying solenoid and the field inside changes: very slightly down for copper, very slightly up for aluminium, and up by a factor of hundreds or thousands for iron. To describe this cleanly, the field inside is split into two parts: the part produced by the current we control in the windings, and the part the material adds on its own.

5.6.1 Magnetisation M

Every atom contains electrons circulating around its nucleus, and each circulating electron is a tiny current loop with a tiny magnetic dipole moment. In a bulk sample these moments add as vectors. With no applied field they cancel overall (either each atom’s moment is zero to begin with, or the atoms’ moments point every which way). An applied field upsets that balance, partly lining moments up or inducing small new ones, and leaves the sample with a net moment mnetm_{\text{net}}. The magnetisation of the sample is that net moment per unit volume:

Magnetisation
M=mnetV\mathbf{M} = \dfrac{\mathbf{m}_{\text{net}}}{V}

It is a vector, pointing along the net moment, and its unit is A·m² per m³, i.e. A/m.

5.6.2 Where the material’s extra field comes from

Picture a long rod, length LL and cross-sectional area AA, magnetised uniformly along its length with magnetisation MM. Its total dipole moment is MM times its volume, M(AL)M(AL).

Microscopically, that moment comes from countless atomic current loops, all circulating the same way round the rod’s axis. Wherever two neighbouring loops touch, their currents run in opposite directions and cancel. Only at the rod’s curved outer surface does a loop have no neighbour to cancel against, so the net effect of all those atomic currents is a thin sheet of current circulating round the outside of the rod: exactly what the windings of a solenoid look like. Call this the magnetising current, KK amperes for each metre of the rod’s length.

A solenoid of length LL carrying KK amperes per metre has KLKL ampere-turns wrapped round area AA, so by m=NIAm=NIA (Section 5.1) its moment is (KL)A(KL)A. That has to be the rod’s moment:

KLA=M(AL)  ⇒  K=MKLA = M(AL) \;\Rightarrow\; K = M

So the magnetisation is the surface current per unit length. The field inside a long solenoid is μ0nI\mu_0 nI, that is, μ0\mu_0 times the current per unit length (see Moving Charges and Magnetism), so the field the material contributes is:

Field contributed by the material
Bm=μ0K=μ0MB_m = \mu_0 K = \mu_0 M

5.6.3 Magnetic intensity H

Now fill a long solenoid (nn turns per metre, current II) with the magnetised material. The total field inside is the coil’s own field plus the material’s:

B=B0+Bm=μ0nI+μ0M=μ0(nI+M)B = B_0 + B_m = \mu_0 nI + \mu_0 M = \mu_0(nI + M)

The first term in the bracket, nInI, depends only on the current we send through the windings; it is the same whether the coil is empty or packed with iron. That quantity is the magnetic intensity HH (here, H=nIH = nI). Its general definition is

Magnetic intensity
H=Bμ0−M\mathbf{H} = \dfrac{\mathbf{B}}{\mu_0} - \mathbf{M}

which rearranges into the key relation of this section:

Total field inside a material
B=μ0(H+M)\textcolor{#e08a1e}{\mathbf{B}} = \mu_0(\mathbf{H} + \mathbf{M})

HH has the same unit as MM, A/m. A useful way to hold the two apart: HH is the cause, set by the free currents in the wires that we control, and MM is the material’s response to it. In empty space M=0M=0 and the relation shrinks to B=μ0HB=\mu_0H.

5.6.4 Magnetic susceptibility χ

How strongly does a material respond? For most materials MM is simply proportional to HH, and the constant of proportionality is the magnetic susceptibility:

Magnetic susceptibility
χ=MH\chi = \dfrac{M}{H}

a pure number, since MM and HH are both in A/m. Substituting M=χHM=\chi H into B=μ0(H+M)B=\mu_0(H+M):

B=μ0(H+χH)=μ0(1+χ)H=μ0μrH,μr=1+χB = \mu_0(H + \chi H) = \mu_0(1+\chi)H = \mu_0\mu_r H, \qquad \mu_r = 1+\chi

where μr\mu_r, the relative permeability, is the factor by which filling the coil with the material multiplies the field (the empty coil gives B0=μ0HB_0=\mu_0H). So the sign and size of χ\chi tell you directly what a material does to the field, and they sort materials into the three families of Section 5.7:

  • Diamagnetic, χ small and negative (around −10−5-10^{-5}): MM points against HH, so BB ends up a hair smaller than μ0H\mu_0H, by about one part in 10510^5. The extreme case, χ=−1\chi=-1, gives B=0B=0: the field is cancelled completely, which is what happens inside a superconductor.
  • Paramagnetic, χ small and positive (around 10−510^{-5} to 10−310^{-3}): MM points along HH, and BB is a hair larger than μ0H\mu_0H.
  • Ferromagnetic, χ large and positive (around 10210^2 to 10510^5): MM is hundreds to many thousands of times bigger than HH, so almost all of BB comes from the material, not the coil. For these materials MM is not even strictly proportional to HH: it depends on the sample’s history, as Section 5.8 shows, so a single value of χ\chi is only a rough guide.

Worked example

Finding M and χ from a measured field

A long solenoid has 600 turns per metre and carries 1.5 A. With an iron-alloy rod filling it, the field inside is measured as 0.90 T. Find H, the magnetisation M of the rod, and its susceptibility χ. What would M be if the rod were copper (χ ≈ −1.0×10⁻⁵) instead?

The magnetic intensity depends only on the windings:

H=nI=(600)(1.5)=900 A/mH = nI = (600)(1.5) = \textcolor{#e08a1e}{900\text{ A/m}}

Rearranging B=μ0(H+M)B=\mu_0(H+M) for MM:

M=Bμ0−H=0.904π×10−7−900≈716,197−900≈7.15×105 A/mM = \dfrac{B}{\mu_0} - H = \dfrac{0.90}{4\pi\times10^{-7}} - 900 \approx 716{,}197 - 900 \approx \textcolor{#e08a1e}{7.15\times10^{5}\text{ A/m}}χ=MH=7.153×105900≈795\chi = \dfrac{M}{H} = \dfrac{7.153\times10^{5}}{900} \approx \textcolor{#e08a1e}{795}

Large and positive, so the rod is ferromagnetic. The coil alone would give only B0=μ0H≈1.13 mTB_0=\mu_0H\approx1.13\text{ mT}; the rod multiplies that by μr=1+χ≈796\mu_r=1+\chi\approx796. By Section 5.6.2, its aligned atomic currents act like a surface current of K=M≈7.15×105 A/mK=M\approx7.15\times10^5\text{ A/m}, the same as an extra M/n≈1190 AM/n\approx1190\text{ A} pushed through every turn of the winding on top of the real 1.5 A.

With a copper rod instead:

M=χH=(−1.0×10−5)(900)=−9.0×10−3 A/mM = \chi H = (-1.0\times10^{-5})(900) = \textcolor{#e08a1e}{-9.0\times10^{-3}\text{ A/m}}

Tiny, and negative: it points against HH, lowering the field by just one part in 10510^5.

5.7 Classification of Magnetic Materials

Every material responds to an applied magnetic field to some degree, and Section 5.6 turned that response into a single number, the magnetic susceptibility χ=M/H\chi = M/H. Its sign and size sort materials into three families:

5.7.1 Diamagnetic

Small, negative χ\chi (of order −10−5-10^{-5}). Weakly repelled by an external field; every material shows this effect at some level, but it’s usually masked by stronger effects when present. Examples: bismuth, copper, water.

5.7.2 Paramagnetic

Small, positive χ\chi (of order 10−310^{-3} to 10−510^{-5}). Weakly attracted into a field, because each atom carries its own small permanent dipole moment, which thermal motion keeps mostly randomised. Examples: aluminium, sodium, oxygen gas.

Susceptibility falls as temperature rises, since more thermal agitation disrupts the partial alignment a field can induce. This is Curie’s law:

Curie's law (paramagnetic materials)
χ∝1T\chi \propto \dfrac{1}{T}

5.7.3 Ferromagnetic

Large, positive χ\chi (of order 10210^2 to 10510^5). Strongly attracted into a field, and can retain magnetisation after the external field is removed (this is what makes permanent magnets possible; Section 5.8 explains how, and why some ferromagnets hold on far better than others). Examples: iron, cobalt, nickel. Above a material-specific Curie temperature, thermal agitation overcomes the internal alignment entirely, and the material becomes merely paramagnetic.

5.8 Hysteresis: Soft Iron vs Permanent Magnets

For diamagnetic and paramagnetic materials, MM simply follows HH: double HH and MM doubles, switch HH off and MM drops back to zero. Ferromagnets break this rule. Their magnetisation depends not just on the field applied now but on what was done to them before, and the record of that memory is the B–H loop, or hysteresis loop.

The memory comes from domains. Inside a ferromagnet, neighbouring atomic dipoles line up with each other on their own, over small regions called domains, each one fully magnetised in some direction. In an unmagnetised lump the domains point in random directions and cancel. An applied field magnetises the lump by letting the domains that already point roughly its way grow at the expense of their neighbours (the boundaries between domains, called domain walls, shift sideways), and by turning the rest to face it.

5.8.1 Tracing the loop

Put an unmagnetised ferromagnetic rod inside a long solenoid. The current sets H=nIH=nI (reversing the current reverses HH), and a probe measures BB inside the rod. Now take HH through a full cycle:

  1. Up from zero. As HH increases, BB climbs steeply at first, while domain walls slide easily, then levels off. Eventually every domain is aligned and MM can grow no further. This is saturation: from here on, BB creeps up only by the small μ0ΔH\mu_0\Delta H added by the coil itself.
  2. Back down to H = 0. BB does not come back down the same curve. Domain walls that moved on the way up get snagged on impurities and defects in the crystal, so many domains stay aligned. With the current switched off completely, BB is still well above zero. This leftover field is the retentivity (or remanence), BrB_r: the rod has become a magnet.
  3. Reverse the current. It takes a field in the opposite direction to break the alignment up again. BB only falls to zero once the reverse field reaches H=−HcH=-H_c. The size of that reverse field, HcH_c, is the coercivity.
  4. Keep going, then come back. Pushing HH further negative saturates the rod the other way round. Bringing HH back up through zero to positive values traces the mirror image (B=−BrB=-B_r at H=0H=0, then B=0B=0 at H=+HcH=+H_c) and closes the loop at the original saturation point. The first climb from the unmagnetised state is never repeated: from now on the rod goes round the loop on every cycle.

“Hysteresis” comes from the Greek for “lagging behind”, and that is exactly what BB does: on every half-cycle, HH passes through zero first and BB only gets there later. It also means a ferromagnet has no single susceptibility. At the point H=0H=0, B=BrB=B_r, the magnetisation is large while HH is zero, so M/HM/H has no sensible value at all.

5.8.2 Retentivity, coercivity, and the energy lost each cycle

Two numbers read straight off the loop describe a ferromagnet for practical purposes. The retentivity BrB_r, where the loop crosses the BB-axis, says how strong a field the material keeps once the magnetising field is gone. The coercivity HcH_c, where the loop crosses the HH-axis, says how hard that leftover magnetism is to destroy: how big a reverse field it takes to wipe it out.

The loop’s area matters too. Every trip round the loop drags the domain walls back and forth past the snags that hold them, rather like dragging furniture across a rough floor, and the work done against that resistance ends up as heat in the material. The energy lost per unit volume in one cycle equals the area enclosed by the loop. (The units agree: HH in A/m times BB in T, where 1 T = 1 N/(A·m) from F=BILF=BIL, gives N/m², which is the same as J/m³, an energy per unit volume.) A fat loop wastes a lot of energy every cycle; a thin one wastes very little.

5.8.3 Why soft iron for electromagnets, and steel or alnico for permanent magnets

Ferromagnets come in two broad kinds, told apart by the shape of their loops. Soft ferromagnets such as soft iron have a tall, narrow loop: a steep climb (a small HH produces a large BB, i.e. a high permeability), a very small coercivity, and a small loop area. Hard ferromagnets such as steel and alnico (an alloy of iron with aluminium, nickel, cobalt, and copper) have a wide, fat loop: a coercivity tens to hundreds of times larger, and a large area. (The current NCERT textbook names these two kinds without drawing the loop; the loop is what explains the difference.) Each job calls for one shape or the other:

  • Electromagnets → soft iron. An electromagnet (a scrapyard lifting magnet, a relay, an electric bell) must be strong when switched on and must let go when switched off. The steep climb gives a strong field from a modest current. The tiny coercivity means any leftover magnetism is wiped out by the slightest reverse field, even the weak reverse field that the core’s own poles set up inside it, so when the current stops, the core loses almost all its magnetism. A scrapyard magnet with a steel core would keep hold of some of its load even after the current was cut.
  • Transformer cores → soft iron. A transformer core (like the core of a motor or generator) is magnetised one way, then the other, 50 times every second on Indian mains supply, so it goes all the way round its loop 50 times a second. Each lap turns energy equal to the loop area (per unit volume) into heat. A thin loop keeps that waste small and stops the core from overheating; with a fat loop, the core would heat up and waste far more energy every second. The high permeability also means a small current produces a large flux.
  • Permanent magnets → steel or alnico. A permanent magnet (a compass needle, a loudspeaker magnet) is magnetised once and must then stay magnetised for years. A high retentivity makes the field it keeps strong; a high coercivity makes that field survive stray fields, knocks, and temperature changes, since only a large reverse field can undo it. The large loop area costs nothing here, because the magnet is never taken round the loop again once it has been made.

One trap worth knowing: soft iron’s retentivity, read off its loop, can be comparable to steel’s or even slightly higher. What rules soft iron out for permanent magnets is its tiny coercivity, not its retentivity: whatever magnetism it keeps is far too easy to destroy. When choosing a permanent-magnet material, coercivity is the property to check first.

Worked example

How hard is each material to demagnetise?

Typical round-number coercivities are about 80 A/m for soft iron and about 5.0×10⁴ A/m for alnico. A rod of each, magnetised to its retentivity, is placed in a long solenoid with 2000 turns per metre. What reverse current brings BB back to zero in each?

BB reaches zero when the reverse HH equals the coercivity, and inside a solenoid H=nIH=nI, so I=Hc/nI = H_c/n:

Isoft iron=802000=0.040 A,Ialnico=5.0×1042000=25 AI_{\text{soft iron}} = \dfrac{80}{2000} = \textcolor{#e08a1e}{0.040\text{ A}}, \qquad I_{\text{alnico}} = \dfrac{5.0\times10^{4}}{2000} = \textcolor{#e08a1e}{25\text{ A}}

A 40 mA trickle wipes out the soft iron; the alnico needs a current 625 times larger. For scale, the Earth’s own field, about 5×10−5 T5\times10^{-5}\text{ T}, corresponds in air to H=B/μ0≈40 A/mH=B/\mu_0\approx40\text{ A/m}, already half the coercivity of soft iron. A soft-iron “permanent magnet” would be at the mercy of every stray field it met, while alnico would need a reverse field more than a thousand times the Earth’s to erase it. That is the width of the hysteresis loop, in numbers.

— From the NCERT exercises

Two of the chapter’s own exercise questions, reworded slightly, with original worked solutions. Both lean on the idea at the heart of Section 5.4: a solenoid behaves like a bar magnet because its field lines close on themselves, and a magnet in a uniform field feels no net push because its poles only ever come in equal and opposite pairs.

NCERT Exercise 5.3

A closely wound solenoid has 800 turns and a cross-sectional area of 2.5 × 10⁻⁴ m², and carries a current of 3.0 A. In what sense does it act like a bar magnet, and what is its magnetic moment?

Solution

Outside the coil, its field lines follow the same pattern as a bar magnet’s: out of one end (which acts as a north pole), round through the space outside, into the other end (a south pole), and back along the inside of the coil, each line closing on itself exactly as Section 5.4 requires. Like a bar magnet, it swings to line up with an external field, feels the same torque τ=mBsin⁡θ\tau=mB\sin\theta, and cut in half gives two shorter “magnets”, never a lone pole. Its moment, from Section 5.1:

m=NIA=(800)(3.0)(2.5×10−4)=0.60 A⋅m2m = NIA = (800)(3.0)(2.5\times10^{-4}) = \textcolor{#e08a1e}{0.60\text{ A·m}^2}

directed along the solenoid’s axis: curl the fingers of your right hand along the current and your thumb points along mm, toward the end that behaves as the north pole.

NCERT Exercise 5.6

A closely wound solenoid of 2000 turns and cross-sectional area 1.6 × 10⁻⁴ m² carries 4.0 A and is hung from its centre so that it can turn in a horizontal plane. (a) What is its magnetic moment? (b) A uniform horizontal field of 7.5 × 10⁻² T is set up at 30° to the solenoid's axis. What are the net force and the torque on it?

Solution

(a) From m=NIAm=NIA:

m=(2000)(4.0)(1.6×10−4)=1.28 A⋅m2m = (2000)(4.0)(1.6\times10^{-4}) = \textcolor{#e08a1e}{1.28\text{ A·m}^2}

along the axis, in the direction given by the right-hand rule.

(b) Force: zero. In the pole picture, the two ends feel forces qmBq_mB of equal size in opposite directions, and they cancel. The deeper reason is Section 5.4’s: there is no isolated pole for a uniform field to push on, only equal and opposite pairs, so a uniform field can twist a magnet but never drag it. (A non-uniform field, like the one near the end of another magnet, can exert a net force.)

Torque: from Section 5.3,

τ=mBsin⁡θ=(1.28)(7.5×10−2)sin⁡30°=0.048 N⋅m\tau = mB\sin\theta = (1.28)(7.5\times10^{-2})\sin30° = \textcolor{#e08a1e}{0.048\text{ N·m}}

acting to turn the axis (the direction of mm) into line with BB, toward the stable θ=0°\theta=0° orientation of Section 5.3.1.

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