Magnetism and Matter
- § Force and Torque on a Magnetic Dipole We carry the result \(\vec{\tau}=\vec{\mu}\times\vec{B}\) from Chapter 5 forward: its potential energy, and the new effect a non-uniform field produces — a net force that pulls a dipole bodily through space.
- § Bar Magnetism The bar magnet is not a new kind of object. It is a magnetic dipole wearing a tangible face — poles, closed field lines, an axial and an equatorial field already met as the far field of a current loop.
- § Gauss's Law for Magnetism One short, exact law, \(\oint\vec{B}\cdot d\vec{A}=0\), that encodes the deepest fact in this chapter: there are no magnetic monopoles.
- § Magnetisation and Magnetic Intensity What makes matter magnetic. We separate the field that free currents drive (\(\vec{H}\)) from the field that matter adds (\(\vec{M}\)), and join them in \(\vec{B}=\mu_0(\vec{H}+\vec{M})\).
- § Magnetic Materials Why aluminium is feebly drawn in, bismuth feebly pushed out, and iron seized violently — diamagnetism, paramagnetism, ferromagnetism, sorted by a single number, the susceptibility \(\chi\).
- § Hysteresis and Permanent Magnets Why magnetising and demagnetising iron around a closed loop costs energy, and how that loop divides the world into soft cores and hard permanent magnets.
- § Terrestrial Magnetism Earth as a giant, slightly crooked dipole, read off at any point through three numbers: declination, dip, and the horizontal field.
Perplexing Questions
- The pole you cannot hold alone. Break a bar magnet in half and you do not get a loose north pole in one hand and a loose south in the other. You get two smaller, complete magnets. Cut again: two smaller still, each with both poles. A positive charge can be isolated; a magnetic pole, it seems, never can. Why?
- The iron that multiplies a field a thousandfold. Slide a soft-iron core into a current-carrying coil and the field inside leaps by a factor of hundreds or thousands. The iron carries no battery and no current you supplied. Where does the enormous extra field come from?
- Three materials, three fates. Hang aluminium, bismuth, and iron near the pole of a strong magnet. The aluminium is drawn feebly in; the bismuth is pushed feebly out; the iron is seized violently. To the hand all three are “non-magnetic.” What single property sorts them into three families?
- The magnet that forgets when heated. Heat a permanent magnet past a certain temperature and its magnetism does not merely weaken — it disappears, and does not come back on cooling unless the magnet is re-magnetised. What is special about that temperature?
- The loop that loses energy going nowhere. Magnetise a ring of iron, then demagnetise it, then return it to exactly the state it started in. The round trip ends where it began — yet energy was spent, and came out as heat. How can a closed cycle, returning to its start, dissipate energy?
- The planet-sized magnet whose north is south. A compass needle's north-seeking end is pulled toward geographic north — which can only mean that geographic north is a magnetic south. And the needle does not lie flat; it dips. What is the Earth, magnetically, and why does the needle tilt?
Look for the [Resolution of PQ N] callout as you read.
Force and Torque on a Magnetic Dipole in a Magnetic Field
Release the magnet and let it swing: small oscillations give SHM with T = 2π√(I/mB) — the classic way to compare moments and fields.
Drag the needle in a uniform field — watch τ = mB sinθ and the U = −mB cosθ energy well track the angle, and find the stable and unstable equilibria.
Chapter 5 established (Section ) that a current loop in a magnetic field experiences a torque tending to align its dipole moment \(\vec{\mu}\) with \(\vec{B}\). That analysis assumed a uniform field, and there the net translational force on the loop was zero. A natural question follows: does a magnetic dipole — a current loop, or equally a bar magnet — ever feel a net force that moves it bodily through space? And what decides whether it rotates, translates, or both?
The whole answer lies in one distinction: whether the field is uniform or not.
Uniform Field: Torque Without Net Force
In a perfectly uniform field, the forces on opposite sides of a current loop are equal in magnitude and opposite in direction — the explicit result of Section . They form a pure couple: net force zero, but a nonzero torque whenever \(\vec{\mu}\) is not aligned with \(\vec{B}\). The torque relation derived there holds unchanged: \[ \vec{\tau} = \vec{\mu}\times\vec{B}, \qquad \tau = \mu B\sin\theta, \] with \(\theta\) the angle between \(\vec{\mu}\) and \(\vec{B}\). This torque rotates the dipole toward \(\theta=0\), where \(\vec{\mu}\parallel\vec{B}\).
At \(\theta=0\) the torque is zero and the alignment is complete: stable equilibrium. At \(\theta=\pi\) the torque is again zero, but the equilibrium is unstable — any small perturbation grows. Between these, the torque is a restoring agency driving the dipole back toward \(\theta=0\).
The orientation carries a potential energy. Because the field does work on the rotating dipole, the dipole stores energy by orientation alone: \[ U = -\vec{\mu}\cdot\vec{B} = -\mu B\cos\theta. \] Stable equilibrium (\(\theta=0\)) minimises \(U=-\mu B\); unstable equilibrium (\(\theta=\pi\)) maximises \(U=+\mu B\). This energy expression is not a side note — it is the lever that opens the rest of the chapter, because it is what lets a gradient in the field produce a force.
Non-Uniform Field: Force as Well as Torque
In a non-uniform field, the field is stronger on one side of the dipole than the other. The forces on opposite faces no longer match, they no longer cancel, and a net translational force results. We can find that force exactly from the energy, without re-summing forces face by face.
[Resolution of PQ 3, partial]. The force \(F=\mu\,(dB/dz)\) already hints at why three materials meet three fates near a magnet: each acquires an induced moment in the field, and whether that moment points along or against \(\vec{B}\) decides whether the sample is pulled up the gradient or pushed down it. Sections and make the sorting precise.
- A magnetic dipole of moment \(\mu = 0.25\,\mathrm{A\cdot m^2}\) is placed in a uniform field \(B = 0.80\,\mathrm{T}\). Find the torque when (a) \(\theta=30^\circ\); (b) \(\theta=90^\circ\); (c) \(\theta=0^\circ\).\(\tau=\mu B\sin\theta\). (a) \(0.25\times0.80\times\sin30^\circ=0.10\,\mathrm{N\cdot m}\). (b) \(0.25\times0.80\times1=0.20\,\mathrm{N\cdot m}\) (maximum). (c) \(0\) (aligned, stable equilibrium).
- A current loop sits in a uniform field with \(\vec{\mu}\parallel\vec{B}\). (a) What is the torque? (b) What is the net force? (c) Stable or unstable equilibrium?(a) \(\tau=\mu B\sin0^\circ=0\). (b) Net force is zero in a uniform field, for any orientation. (c) \(\theta=0\) is stable: \(U=-\mu B\) is a minimum, and a small twist gives a restoring torque.
- Using energy, explain why a bar magnet near a strong non-uniform field is attracted toward the pole when its north pole points toward the field source.The moment \(\vec{m}\) points \(S\to N\), so with the north pole toward the source \(\vec{m}\) is roughly aligned with the external field. Then \(U=-\vec{m}\cdot\vec{B}\) grows more negative where \(B\) is larger (closer to the source). The system moves to minimise \(U\), i.e. toward the stronger-field region: attraction.
- A dipole is held perpendicular to a non-uniform field (\(\theta=90^\circ\)) and released from rest. Describe the torque, the net force, and the subsequent motion qualitatively.Torque \(\tau=\mu B\sin90^\circ=\mu B\) (maximum, turning \(\vec{\mu}\) toward \(\vec{B}\)). In a non-uniform field a translational force acts as well, even at \(90^\circ\). After release the dipole both rotates toward alignment and translates toward stronger field, the translation becoming more effective as \(\vec{\mu}\) aligns. The combined motion is genuinely two-fold.
Worked Examples
Find. \(\tau\) and \(U\).
Setup. \(\tau=\mu B\sin\theta\); \(U=-\mu B\cos\theta\).
Solve. \[ \tau=0.40\times0.25\times0.80=0.080\,\mathrm{N\cdot m},\qquad U=-0.40\times0.25\times0.60=-0.060\,\mathrm{J}. \] Answer. \(\tau=0.080\,\mathrm{N\cdot m}\) (turning \(\vec\mu\) toward \(\vec B\)); \(U=-0.060\,\mathrm{J}\).
Check. The largest possible values are \(\tau_{\max}=\mu B=0.10\, \mathrm{N\cdot m}\) and \(|U|_{\max}=\mu B=0.10\,\mathrm{J}\); both results sit below these bounds, as they must for \(\theta\neq90^\circ,0^\circ\).
Find. \(F_z\); ratio \(F_z/mg\).
Setup. \(F_z=\mu\,dB_z/dz\) (aligned dipole, pulled up the gradient); \(mg\) with \(g=9.8\,\mathrm{m\,s^{-2}}\).
Solve. \[ F_z=8.0\times10^{-3}\times15=0.12\,\mathrm{N},\qquad mg=2.0\times10^{-3}\times9.8=0.020\,\mathrm{N}. \] Answer. \(F_z=0.12\,\mathrm{N}\) toward the stronger field, about \(6\times\) the sample's weight, so it is lifted.
Check. Units: \(\mathrm{A\,m^2}\times\mathrm{T\,m^{-1}} =\mathrm{A\,m\,T}=\mathrm{N}\) (since \(\mathrm{T}=\mathrm{N\,A^{-1}\,m^{-1}}\)). A uniform field (\(dB_z/dz=0\)) would give \(F_z=0\), recovering the pure-couple case.
Bar Magnetism and the Magnetic Dipole Picture
Drag the field point around the dipole: the axial field is exactly twice the equatorial at equal distance, both falling as 1/r³, with tanα = ½ tanθ.
Cut the magnet and each half becomes a complete, weaker dipole (m → m/2). No lone pole ever appears — the content of ∇·B = 0.
Most students meet the bar magnet long before Ampère's law. The picture is concrete: one end grips nails, the other repels them; field lines spray from the north pole and re-enter at the south. The magnet seems primitive — an object that simply possesses magnetism the way a charge possesses charge.
That picture, useful early, hides the real story. The bar magnet is not primitive. It is a magnetic dipole. Its torque in an external field, the field it sets up around itself, the way its poles act at a distance — all of it follows from the dipole framework already built in Sections and . The bar magnet is not new physics; it is the earlier physics wearing a more tangible face. This section closes that gap.
Magnetic Poles and the Dipole Model
A bar magnet has a north pole (\(N\)) and a south pole (\(S\)). Like poles repel, unlike poles attract — experimental facts that need no further justification here. But the analogy with electric charge must be drawn carefully.
A positive charge can be isolated: strip the electron from a hydrogen atom and a free proton remains. No isolated magnetic pole has ever been found. Break a bar magnet in two and each piece has both a north and a south pole. Cut the pieces again: still two poles each. The poles are not separable objects; they are the two ends of a single dipole structure.
[Resolution of PQ 1]. This is why you cannot hold a lone pole. Cutting separates material, not poles. Each new cut face becomes a fresh pole because the field lines threading the magnet must still close — a consequence of \(\nabla\cdot\vec{B}=0\), formalised in Section .
The model consistent with all of this is the magnetic dipole. The bar magnet is characterised by its magnetic dipole moment \(\vec{m}\), directed from the south pole to the north pole along the axis. (Note the direction: \(\vec{m}\) points \(S\to N\) — the same sense as the field inside the magnet.) Its magnitude is \(m=q_m\,d\), where \(d\) is the pole separation. This \(\vec{m}\) is the same physical quantity as the current-loop moment \(\vec{\mu}=NI\vec{A}\) of Chapter 5: both are magnetic dipole moments, measured in \(\mathrm{A\cdot m^2}\), and both obey \(\vec{\tau}=\vec{m}\times\vec{B}\) and \(U=-\vec{m}\cdot\vec{B}\).
Field Lines, and the Axial and Equatorial Field
The field-line picture of a bar magnet closely resembles that of an electric dipole, with one decisive difference: the lines are closed. Outside, they run from north to south; inside, they continue from south to north. No line begins or ends anywhere. The field is strongest where the lines crowd — near the poles — and weakest at the equatorial midplane far away.
We do not need to re-derive the field of this dipole. Section found the far field of a current loop on its axis, and a bar magnet is a magnetic dipole of exactly the same kind. Reading off that result with moment \(m\), at distance \(r\) from the centre (\(r\gg \ell\)): \[\begin{aligned} B_{\text{axial}} &= \frac{\mu_0}{4\pi}\,\frac{2m}{r^3} \qquad(\text{along } \vec{m}), \\[2pt] B_{\text{equatorial}} &= \frac{\mu_0}{4\pi}\,\frac{m}{r^3} \qquad(\text{opposite to } \vec{m}). \end{aligned}\] The axial field is exactly twice the equatorial field at the same distance — the same \(2:1\) ratio as for an electric dipole, and for the same geometric reason. Both fall as \(1/r^3\), the signature of dipole decay.
Directive Torque and Neutral Points
Place a bar magnet in a uniform external field \(\vec{B}\). It feels a torque \[ \vec{\tau} = \vec{m}\times\vec{B}, \qquad \tau = mB\sin\theta, \] identical to the current-loop torque with \(\vec{m}\) in place of \(\vec{\mu}=NI\vec{A}\). The torque turns the north pole toward \(\vec{B}\); stable equilibrium is at \(\theta=0\), unstable at \(\theta=\pi\). This is the mechanism of the compass: the needle is a small dipole, and Earth's field supplies the external \(\vec{B}\) that torques it into alignment. The needle does not point north by sympathy — it points along the local \(\vec{B}\) because that is where the torque vanishes.
Neutral points. When a bar magnet is placed in a uniform external field (such as Earth's horizontal component \(B_H\)), there are points where the magnet's own field exactly cancels the external field. At these neutral points the total field is zero, and a small test needle placed there shows no deflection. Their location depends on orientation: with the north pole along \(B_H\), the neutral points lie on the equatorial line; with the axis across \(B_H\), they lie on the axial line. They are a standard way to measure a magnet's moment by balancing \(B_H\) against Eq. or .
- A bar magnet is broken into two equal pieces by a cut perpendicular to its axis. Describe the poles of each piece, and state whether each piece's moment equals, doubles, or halves the original.Each piece has both an \(N\) and an \(S\) pole — no isolated pole appears. The moment halves: \(m'=q_m(\ell)=\tfrac12 q_m(2\ell)=m/2\), since the pole strength \(q_m\) is unchanged but the length is halved. Each piece is a weaker dipole.
- On the axial line of a short magnet at \(r=10\,\mathrm{cm}\) the field is \(B_0\). Find the field at an equatorial point at the same distance.\(B_{\text{axial}}=(\mu_0/4\pi)(2m/r^3)=B_0\); \(B_{\text{equatorial}}=(\mu_0/4\pi)(m/r^3)=B_0/2\). The equatorial field at equal distance is exactly half the axial field.
- A bar magnet with \(m=0.25\,\mathrm{A\cdot m^2}\) is held perpendicular to a uniform field \(B=0.40\,\mathrm{T}\). (a) Find the torque. (b) Is this stable, unstable, or neither?(a) \(\tau=mB\sin90^\circ=0.25\times0.40=0.10\,\mathrm{N\cdot m}\) (maximum torque). (b) Neither: \(\theta=90^\circ\) is the maximum-torque orientation, not a zero-torque equilibrium. Equilibria are at \(\theta=0\) (stable) and \(\theta=\pi\) (unstable).
- Inside a bar magnet the field lines run from south to north. Explain how this is consistent with field lines forming closed loops.Outside, lines run \(N\to S\); inside, they continue \(S\to N\). Each line is therefore a closed loop \(N\to\text{outside}\to S\to \text{inside}\to N\). The apparent “start” at the north face is just the external portion of a loop that never terminates.
Worked Examples
Find. \(r\) on the axis.
Setup. \(B_{\text{axial}}=\dfrac{\mu_0}{4\pi}\dfrac{2m}{r^3}\) with \(\dfrac{\mu_0}{4\pi}=10^{-7}\,\mathrm{T\,m\,A^{-1}}\); invert for \(r\).
Solve. \[ r^3=\frac{(\mu_0/4\pi)\,2m}{B_{\text{axial}}} =\frac{10^{-7}\times2\times0.50}{2.5\times10^{-5}} =4.0\times10^{-3}\,\mathrm{m^3}, \quad r=(4.0\times10^{-3})^{1/3}=0.159\,\mathrm{m}. \] Answer. \(r\approx0.16\,\mathrm{m}\) (\(15.9\,\mathrm{cm}\)).
Check. Substituting back, \(10^{-7}\times2\times0.50/(0.159)^3 =2.5\times10^{-5}\,\mathrm{T}\), as required. Halving \(r\) would raise the field eightfold (\(1/r^3\)).
Find. Neutral-point distance \(d\).
Setup. With \(\vec m\) along \(B_H\), the magnet's equatorial field points opposite \(\vec m\) (hence opposite \(B_H\)); the neutral points lie on the equatorial line where \(B_{\text{equatorial}}=B_H\): \(\dfrac{\mu_0}{4\pi}\dfrac{m}{d^3}=B_H\).
Solve. \[ d^3=\frac{(\mu_0/4\pi)\,m}{B_H} =\frac{10^{-7}\times0.60}{3.0\times10^{-5}} =2.0\times10^{-3}\,\mathrm{m^3}, \quad d=(2.0\times10^{-3})^{1/3}=0.126\,\mathrm{m}. \] Answer. The neutral points lie \(\approx12.6\,\mathrm{cm}\) from the centre, on the equatorial (perpendicular-bisector) line.
Check. There the magnet's field exactly cancels \(B_H\), so a test needle shows no deflection. Had the magnet been laid across \(B_H\), the neutral points would instead fall on the axial line, using \(B_{\text{axial}}=B_H\).
Gauss's Law for Magnetism
The recurring refrain of the last section — closed loops, no isolated poles, \(\nabla\cdot\vec{B}=0\) — deserves to be stated once, cleanly, as a law in its own right. It is one of Maxwell's four equations, and the shortest.
Recall Gauss's law for the electric field (Section ): the flux of \(\vec{E}\) through any closed surface counts the enclosed charge, \(\oint\vec{E}\cdot d\vec{A}=Q_{\text{enc}}/\varepsilon_0\). The right-hand side is nonzero precisely because electric charge — a source at which field lines can begin and end — exists. The magnetic case is governed by the same logic running the opposite way.
- A closed surface is drawn enclosing only the north pole of a bar magnet (the south pole is outside). What is the net magnetic flux through the surface? Justify.Zero. By \(\oint\vec{B}\cdot d\vec{A}=0\) the net flux through any closed surface is zero, monopole or not. The lines emerging from the north face curve around and re-enter the same surface; outflux and influx cancel exactly. (Contrast the electric case, where enclosing a single charge gives nonzero flux.)
- State the differential form of Gauss's law for magnetism and put it in words.\(\nabla\cdot\vec{B}=0\): the magnetic field has zero divergence everywhere; it has no point sources or sinks. Field lines neither begin nor end — they close on themselves.
- Why does \(\oint\vec{E}\cdot d\vec{A}\) have a nonzero right-hand side while \(\oint\vec{B}\cdot d\vec{A}\) does not?Electric field lines begin and end on charges, so a surface can enclose a net source. Magnetic field lines form closed loops with no terminals, so no surface can enclose a net magnetic source: there are no magnetic monopoles.
Worked Examples
Find. Net outward flux through the remaining faces.
Setup. Gauss's law for magnetism: \(\oint_S\vec B\cdot d\vec A=0\) over any closed surface.
Solve. Total flux \(=0\), so (outward through the rest) \(-\,(8.0\times10^{-3})=0\), giving outward flux \(=8.0\times10^{-3}\,\mathrm{Wb}\).
Answer. \(8.0\times10^{-3}\,\mathrm{Wb}\) leaves through the remaining surface — exactly what entered.
Check. Every field line that enters a closed surface must leave it (lines are closed loops, no monopoles): influx and outflux always balance.
Find. Flux through the curved (dome) surface.
Setup. Because \(\nabla\cdot\vec B=0\), the flux through any two surfaces sharing the same rim is equal. So the dome flux equals the flat-disk flux \(\Phi=B\,\pi r^2\).
Solve. \[ \Phi=B\,\pi r^2=0.30\times\pi\times(0.10)^2=9.4\times10^{-3}\,\mathrm{Wb}. \] Answer. \(\Phi\approx9.4\times10^{-3}\,\mathrm{Wb}\) through the dome.
Check. Cap the dome with the disk to make a closed surface: the two fluxes are equal and opposite, summing to zero, consistent with \(\oint\vec B\cdot d\vec A=0\).
Solved examples
Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.
Find: \(\tau\), \(U\).
Setup: A magnetic dipole obeys \(\tau = mB\sin\theta\) and \(U = -mB\cos\theta\) (with \(U=0\) at \(\theta=90^\circ\), the natural zero of \(-\cos\theta\)).
Solve: \[ \tau = mB\sin30^\circ = (0.60)(0.25)(0.5) = \boxed{0.075\,\mathrm{N\cdot m}}, \] \[ U = -mB\cos30^\circ = -(0.60)(0.25)(0.866) = \boxed{-0.13\,\mathrm{J}}. \] Answer: \(\tau = 0.075\,\mathrm{N\cdot m}\), \(U = -0.13\,\mathrm{J}\). The negative energy places the magnet below the reference level, on the side of the stable equilibrium \(\theta=0\); the torque turns its north pole further toward \(\vec{B}\). Check. The deepest possible well is \(|U|_{\max}=mB=0.15\,\mathrm{J}\) and the largest torque is \(\tau_{\max}=mB=0.15\,\mathrm{N\cdot m}\); the found \(|U|=0.13\,\mathrm{J}\) and \(\tau=0.075\,\mathrm{N\cdot m}\) sit safely below both.
Find: \(B_{\text{axial}}\), \(B_{\text{equatorial}}\), directions.
Setup: For a short magnet (\(r\gg\ell\)), \(B_{\text{axial}} = (\mu_0/4\pi)(2m/r^3)\) along \(\vec{m}\), and \(B_{\text{equatorial}} = (\mu_0/4\pi)(m/r^3)\) opposite to \(\vec{m}\) (Eqs. –).
Solve: \[ B_{\text{axial}} = 10^{-7}\times\frac{2(0.40)}{(0.20)^3} = 10^{-7}\times100 = \boxed{1.0\times10^{-5}\,\mathrm{T}}, \] \[ B_{\text{equatorial}} = 10^{-7}\times\frac{0.40}{(0.20)^3} = 10^{-7}\times50 = \boxed{5.0\times10^{-6}\,\mathrm{T}}. \] Answer: Axial \(1.0\times10^{-5}\,\mathrm{T}\) along \(\vec{m}\); equatorial \(5.0\times10^{-6}\,\mathrm{T}\) antiparallel to \(\vec{m}\). The axial field is exactly twice the equatorial at equal distance — the dipole \(2:1\) ratio. Check. \(B_{\text{axial}}/B_{\text{equatorial}}=2\) exactly, independent of \(m\) and \(r\) — the dipole signature; both fall as \(1/r^3\).
Find: \(H\), \(B\).
Setup: In a solenoid the free current alone sets \(H = nI\); the core then gives \(B = \mu_0\mu_r H\).
Solve: \[ H = nI = (800)(2.5) = \boxed{2000\,\mathrm{A\,m^{-1}}}, \] \[ B = \mu_0\mu_r H = (4\pi\times10^{-7})(500)(2000) \approx \boxed{1.26\,\mathrm{T}}. \] Answer: \(H = 2000\,\mathrm{A\,m^{-1}}\), \(B \approx 1.26\,\mathrm{T}\). Without the core, \(B_0 = \mu_0 H \approx 2.5\,\mathrm{mT}\): the core multiplies the field by \(\mu_r = 500\). Check. Independently, \(M=(\mu_r-1)H=499\times2000\approx9.98\times10^{5}\,\mathrm{A\,m^{-1}}\) gives \(B=\mu_0(H+M)\approx1.26\,\mathrm{T}\), matching \(\mu_0\mu_r H\).
Find: \(M\), material class.
Setup: For a linear material \(M = \chi H\); the sign of \(\chi\) fixes the class (small \(\chi\gt 0 \Rightarrow\) paramagnetic).
Solve: \[ M = \chi H = (2.0\times10^{-5})(1.0\times10^{4}) = \boxed{0.20\,\mathrm{A\,m^{-1}}}. \] Answer: \(M = 0.20\,\mathrm{A\,m^{-1}}\), aligned with \(\vec{H}\). A small positive susceptibility identifies a paramagnetic material. Check. \(M/H=\chi=2\times10^{-5}\): the magnetisation is a hundred-thousandth of the drive, the feeble response of a paramagnet (a ferromagnet would give \(M\sim10^{3}H\)).
Find: \(\chi_2\).
Setup: Curie's law \(\chi = C/T\), so \(\chi T\) is constant: \(\chi_2 = \chi_1\,(T_1/T_2)\).
Solve: \[ \chi_2 = (3.0\times10^{-3})\times\frac{300}{450} = (3.0\times10^{-3})\times\tfrac23 = \boxed{2.0\times10^{-3}}. \] Answer: \(\chi_2 = 2.0\times10^{-3}\). Heating weakens paramagnetism: more thermal disorder, less alignment. Check. \(\chi T\) is invariant: \(3.0\times10^{-3}\times300=0.90=2.0\times10^{-3}\times450\), consistent with \(\chi=C/T\).
Problem bank
Five questions from this chapter’s 50-question bank, free — attempt each one before you reveal the answer. The rest of the bank, and the timed test that draws on all of it, are in the full book.
- Torque on a Dipole
A magnetic dipole of moment \(\mu = 0.30\,\mathrm{A\cdot m^2}\) lies perpendicular to a uniform field \(B = 0.50\,\mathrm{T}\). Find the torque.\(\tau = \mu B\sin90^\circ = (0.30)(0.50)(1) = 0.15\,\mathrm{N\cdot m}\) (maximum torque). - Orientation Energy
For a dipole of moment \(\mu = 0.40\,\mathrm{A\cdot m^2}\) in a field \(B = 0.20\,\mathrm{T}\), find the potential energy when it is (a) aligned and (b) anti-aligned with the field.\(U = -\mu B\cos\theta\). (a) Aligned (\(\theta=0\)): \(U = -\mu B = -0.08\,\mathrm{J}\). (b) Anti-aligned (\(\theta=180^\circ\)): \(U = +\mu B = +0.08\,\mathrm{J}\). The aligned state is the energy minimum. - Magnetic Moment from Pole Strength
A bar magnet has pole strength \(q_m = 12\,\mathrm{A\cdot m}\) and magnetic length \(2\ell = 0.10\,\mathrm{m}\). Find its magnetic moment.\(m = q_m(2\ell) = (12)(0.10) = 1.2\,\mathrm{A\cdot m^2}\). - Iron-Core Solenoid
A solenoid with \(n = 1000\) turns per metre carries \(I = 1.5\,\mathrm{A}\) and is filled with a core of \(\mu_r = 400\). Find (a) \(H\) and (b) \(B\) inside.(a) \(H = nI = 1500\,\mathrm{A\,m^{-1}}\). (b) \(B = \mu_0\mu_r H = (4\pi\times10^{-7})(400)(1500) \approx 0.754\,\mathrm{T}\). The core multiplies the empty-solenoid field by \(400\). - Reading a Hysteresis Loop
Sketch a \(B\)–\(H\) loop for a ferromagnet. Mark the retentivity and the coercivity, and explain why the enclosed area represents an energy loss.The loop rises to saturation, returns at \(H=0\) to a positive \(B = B_r\) (retentivity), and crosses \(B=0\) at \(H = -H_c\) (coercivity). The work done per unit volume per cycle is \(\oint H\,dB\), the loop area. Since the material returns to its starting state, this work is not stored but dissipated as heat in shifting and rotating domains — an energy loss per cycle.
Chapter test
A paper drawn at random from this chapter's bank. Choose the exam you are training for — the marking scheme, pace and difficulty mix follow the real pattern. Work on paper; when you finish (or the clock runs out), the answers are revealed and you mark yourself honestly.
The chapter continues.
You’ve read the opening, the first three theory sections, the opening run of worked examples and five bank questions — all free, with no account. The rest of the chapter is behind the pass.
- Magnetisation and Magnetic Intensity
- Magnetic Materials: Dia-, Para-, and Ferromagnetism
- Hysteresis and Permanent Magnets
- Terrestrial Magnetism
- Common Pitfalls and Exam Strategy
- Summary and Resolution of Perplexing Questions
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