Chapter 5: Magnetism and Matter

Magnetism is the macroscopic manifestation of orbital and spin angular momenta of electrons. After the discovery by Oersted (1820) that a current produces a magnetic field, the equivalence between a current loop and a bar magnet became the unifying idea of this chapter. In Class XII we move from the source (currents, in Ch.4) to matter — how materials respond to and store magnetic fields. The chapter develops three layered descriptions of the same physics: (i) a phenomenological bar magnet, (ii) the dipole-moment picture (analogue of electric dipole), and (iii) the microscopic susceptibility/permeability picture that classifies materials as diamagnetic, paramagnetic or ferromagnetic.

Concept Map

• Bar magnet $\to$ two equal & opposite poles, dipole moment $\vec m = q_m \cdot 2\vec l$
• Equivalent solenoid — a finite solenoid of $N$ turns carrying $I$ behaves identically; $\vert \vec m\vert = NIA$
• Field of a bar magnet — axial $B_{\text{ax}} = \dfrac{\mu_0}{4\pi}\dfrac{2m}{r^3}$, equatorial $B_{\text{eq}} = \dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}$
• Torque & energy in uniform field: $\vec\tau = \vec m\times\vec B$, $U=-\vec m\cdot\vec B$
• Gauss for magnetism: $\oint \vec B\cdot d\vec A = 0$ — no magnetic monopoles
• Earth's field: declination $D$, dip $I$, horizontal component $H_E = B_E\cos I$
• Material response: $\vec B = \mu_0(\vec H + \vec M)$, $\vec M = \chi\vec H$, $\mu_r = 1+\chi$
• Classification: diamagnetic ($\chi< 0$, small), paramagnetic ($\chi>0$, small, $\propto 1/T$), ferromagnetic ($\chi\gg 0$, hysteresis)
• Hysteresis loop — retentivity, coercivity, area = energy loss/cycle

5.1 Bar Magnet — Properties, Field Lines, Dipole Moment

Definition

A bar magnet is a permanently magnetised body in which atomic dipoles are aligned, producing a net dipole moment $\vec m$. The two ends are called the North (N) and South (S) magnetic poles. Like poles repel, unlike attract. Unlike electric charges, magnetic poles cannot be isolated — cutting a bar magnet always produces two smaller magnets, each with both N and S poles. This statement is the macroscopic content of the absence of magnetic monopoles.

Magnetic dipole moment (idealised pole model): if the two poles carry "pole strengths" $\pm q_m$ separated by a distance $2l$,

$\vec m = q_m\,(2\vec l)\quad\text{(directed from S to N)}.$

The unit of $\vert \vec m\vert$ is $\text{A}\,\text{m}^2$ (more fundamentally, the current-loop definition $m = IA$ shows the SI unit). Pole strength $q_m$ has units $\text{A}\,\text{m}$.

Properties of magnetic field lines

1. Field lines form closed loops: they emerge from N outside the magnet and re-enter at S, continuing inside from S to N. Contrast this with electric field lines from a dipole, which begin on $+q$ and end on $-q$.
2. They never intersect (the field at a point has a unique direction).
3. Tangent at any point gives the direction of $\vec B$ at that point.
4. Density (number per unit area perpendicular to the lines) is proportional to $\vert \vec B\vert$.
5. Outside the magnet, lines run N $\to$ S; inside, they run S $\to$ N — this closure is the geometric face of $\nabla\cdot\vec B = 0$.

Derivation: dipole moment of a current loop

For a planar loop of area $A$ carrying current $I$, the magnetic moment is

$\vec m = I\vec A,$

where $\vec A$ is normal to the loop in the right-hand-rule sense. This is the most fundamental definition, valid for atomic currents as well. For a coil with $N$ turns, $\vec m = NI\vec A$.

Worked Example

A circular coil of radius $r = 5.0\,\text{cm}$ has $N=200$ turns and carries $I = 0.40\,\text{A}$. Its magnetic moment is

$m = NIA = 200 \times 0.40 \times \pi(0.05)^2 = 200\times 0.40\times 7.854\times 10^{-3}\;\text{A m}^2 = 0.628\;\text{A m}^2.$

Pitfalls

• "Pole strength" is a convenient fiction; isolated monopoles have never been observed.
• $\vec m$ points from S $\to$ N inside the magnet — students often draw it the other way.
• Magnetic field lines are continuous including inside the magnet; if a sketch shows them stopping at the poles, it is wrong.

5.2 Bar Magnet as Equivalent Solenoid

Definition

A solenoid of finite length carrying a current produces, on its axis, a field whose far-field behaviour is identical to that of a bar magnet of the same dipole moment. This is the cornerstone of Ampere's "molecular currents" picture of magnetism.

Derivation: axial field of a finite solenoid at a far point

Consider a solenoid of length $2L$, radius $a$, with $n$ turns per unit length carrying current $I$. Consider a point $P$ on the axis at a distance $r$ from the centre, with $r \gg L,\,a$. A thin element of width $dx$ at distance $x$ from the centre carries $dN = n\,dx$ turns and acts as a circular loop. Its axial field at $P$ is

$dB = \frac{\mu_0}{4\pi}\,\frac{2(n\,dx)\,I\,a^2}{[(r-x)^2 + a^2]^{3/2}}.$

For $r \gg L,a$ we approximate $(r-x)^2+a^2 \approx r^2$ in the denominator. The total field becomes

$B = \frac{\mu_0}{4\pi}\,\frac{2\,n\,I\,a^2}{r^3}\int_{-L}^{+L} dx = \frac{\mu_0}{4\pi}\,\frac{2\,n\,I\,a^2\,(2L)}{r^3}.$

The total number of turns is $N = n(2L)$ and the loop area is $A = \pi a^2$, so $NIA = m$:

$\boxed{\;B_{\text{axial}}=\frac{\mu_0}{4\pi}\,\frac{2m}{r^3}\;}$

This is identical to the axial field of a bar magnet of dipole moment $m$. Therefore a finite solenoid behaves like a bar magnet, with N at the end out of which current appears to flow anticlockwise (right-hand rule).

Worked Example

A solenoid 6.0 cm long, of 0.5 cm radius, has 400 turns and carries 4.0 A. Its moment is

$m = NIA = 400\times 4.0\times \pi(0.005)^2 = 0.126\;\text{A m}^2.$

Far from the solenoid, on the axis at 30 cm, the field equals

$B = 10^{-7}\,\frac{2\times 0.126}{(0.30)^3} = 9.3\times 10^{-7}\;\text{T}.$

Pitfalls

• The equivalence is far-field; near the solenoid the field is not dipolar.
• A short, fat solenoid does not look like a long thin magnet; geometry still matters for near-field problems.
• Don't confuse $n$ (turns per unit length) with $N$ (total turns); $N = n\cdot \ell$.

5.3 Magnetic Field due to a Bar Magnet on Axial and Equatorial Lines

Definition

For a magnetic dipole of moment $\vec m$ centred at the origin, the axial line is the line passing through both poles; the equatorial line is the perpendicular bisector of the segment joining the poles. We derive $B$ on both, far from the dipole, and compare with the electric dipole.

Derivation 1: Axial field

Place the dipole along the $x$-axis with the N pole at $+l$ and S pole at $-l$, dipole moment $m = q_m (2l)$ along $+\hat x$. Point $P$ lies on the axis at distance $r$ from the centre.

The fields at $P$ due to the two poles (using the "pole" analogue of Coulomb's law $B = \dfrac{\mu_0}{4\pi}\dfrac{q_m}{r^2}$):

$B_N = \frac{\mu_0}{4\pi}\,\frac{q_m}{(r-l)^2}\quad\text{(away from N, toward }P\text{ if }P\text{ beyond N)},$

$B_S = \frac{\mu_0}{4\pi}\,\frac{q_m}{(r+l)^2}\quad\text{(toward S, away from }P\text{)}.$

Net axial field (taking outward as positive):

$B_{\text{ax}} = \frac{\mu_0\,q_m}{4\pi}\left[\frac{1}{(r-l)^2}-\frac{1}{(r+l)^2}\right] = \frac{\mu_0\,q_m}{4\pi}\cdot\frac{(r+l)^2-(r-l)^2}{(r^2-l^2)^2} = \frac{\mu_0\,q_m}{4\pi}\cdot\frac{4rl}{(r^2-l^2)^2}.$

Substitute $m = q_m(2l)$:

$B_{\text{ax}} = \frac{\mu_0}{4\pi}\,\frac{2mr}{(r^2-l^2)^2}.$

For $r \gg l$,

$\boxed{\;B_{\text{ax}}=\frac{\mu_0}{4\pi}\,\frac{2m}{r^3}\quad\text{(along }\vec m\text{)}\;}$

Derivation 2: Equatorial field

For point $P$ on the equator at distance $r$ from the centre, the distances from N and S are equal: $\sqrt{r^2+l^2}$. The magnitudes are equal,

$B_N = B_S = \frac{\mu_0}{4\pi}\,\frac{q_m}{r^2+l^2},$

but the vertical components (along the equator) cancel and the components along $-\hat m$ (parallel to the axis of the dipole, but pointing from N to S, i.e., opposite to $\vec m$) add. The angle each field makes with the dipole axis satisfies $\cos\theta = l/\sqrt{r^2+l^2}$.

$B_{\text{eq}} = 2\,B_N\cos\theta = 2\cdot \frac{\mu_0}{4\pi}\,\frac{q_m}{r^2+l^2}\cdot\frac{l}{\sqrt{r^2+l^2}} = \frac{\mu_0}{4\pi}\,\frac{2q_m l}{(r^2+l^2)^{3/2}} = \frac{\mu_0}{4\pi}\,\frac{m}{(r^2+l^2)^{3/2}}.$

For $r \gg l$,

$\boxed{\;B_{\text{eq}}=\frac{\mu_0}{4\pi}\,\frac{m}{r^3}\quad\text{(antiparallel to }\vec m\text{)}\;}$

Comparison with electric dipole

Worked Example

A bar magnet has $m = 0.50\,\text{A m}^2$. Field on axis at 10 cm:

$B = 10^{-7}\cdot \frac{2\times 0.50}{(0.10)^3} = 10^{-4}\;\text{T} = 1\;\text{G}.$

Field on the equator at the same distance is half: $0.5\,\text{G}$.

Pitfalls

• On the equator the field is opposite to $\vec m$, not zero.
• The ratio of axial to equatorial field at the same distance is exactly 2 — a quick sanity check.
• Formulas in the boxes are valid only for $r \gg l$; for short distances use the full expressions.

5.4 Torque on a Magnetic Dipole in a Uniform Field & Potential Energy

Definition

When a magnetic dipole $\vec m$ is placed in a uniform external field $\vec B$, it experiences a couple but no net translational force. The couple tends to align $\vec m$ with $\vec B$.

Derivation: torque

Take a bar magnet of moment $m = q_m(2l)$ making an angle $\theta$ with $\vec B$. Each pole experiences a force $q_m B$, but in opposite directions, forming a couple of arm $2l\sin\theta$:

$\tau = (q_m B)(2l\sin\theta) = (q_m\cdot 2l)\,B\sin\theta = m B\sin\theta.$

In vector form:

$\boxed{\;\vec\tau = \vec m \times \vec B\;}$

Equivalent for a current loop of moment $\vec m = I\vec A$ — already derived in Ch.4 from $\vec F = I\vec L\times\vec B$.

Derivation: potential energy

Work done by external agent in rotating the dipole from $\theta_0$ to $\theta$ against the field:

$W = \int_{\theta_0}^{\theta} \tau\,d\theta' = \int_{\theta_0}^{\theta} mB\sin\theta'\,d\theta' = -mB[\cos\theta - \cos\theta_0].$

Define $U(\theta_0 = \pi/2) = 0$ (the conventional zero), giving

$\boxed{\;U(\theta) = -\vec m\cdot\vec B = -mB\cos\theta\;}$

Equilibria:

• $\theta = 0$: $U_{\min} = -mB$ — stable equilibrium (aligned).
• $\theta = \pi$: $U_{\max} = +mB$ — unstable equilibrium (anti-aligned).

Small-oscillation period

For small angular displacements about the aligned position, $\sin\theta \approx \theta$ so

$I_{\text{moi}}\ddot\theta = -mB\theta\implies T = 2\pi\sqrt{\frac{I_{\text{moi}}}{mB}}.$

This is the basis of the vibration magnetometer used to measure the horizontal component of Earth's field.

Worked Example

A bar magnet of $m = 0.32\,\text{A m}^2$ placed at $60^\circ$ to a uniform field of $0.15\,\text{T}$ experiences a torque

$\tau = mB\sin\theta = 0.32\times 0.15\times \sin 60^\circ = 0.0416\;\text{N m}.$

Work to rotate it from $0^\circ$ to $90^\circ$:

$W = U(90^\circ) - U(0^\circ) = 0 - (-mB) = mB = 0.048\;\text{J}.$

Pitfalls

• Net force is zero only in a uniform field; in a non-uniform field there is a translational force $\vec F = \nabla(\vec m\cdot\vec B)$.
• Sign of $U$ depends on the reference; the convention $U(\pi/2)=0$ gives $U=-\vec m\cdot\vec B$.
• The moment of inertia in $T = 2\pi\sqrt{I/mB}$ is the moment of inertia about the suspension axis, not the geometrical centre alone.

5.5 Gauss's Law for Magnetism

Definition

For any closed surface $S$,

$\boxed{\;\oint_S \vec B\cdot d\vec A = 0\;}$

The net magnetic flux through any closed surface is zero. Equivalently, $\nabla\cdot\vec B = 0$ — magnetic field lines have no beginning and no end; they form closed loops.

Physical content

1. No magnetic monopoles: Coulomb's law for magnetism cannot be written as $\oint \vec B\cdot d\vec A = \mu_0 q_m^{\text{enc}}$ because there are no monopoles to enclose.
2. Field lines close on themselves: every line leaving a region must re-enter it — this is why we can split a bar magnet but never isolate a single pole.
3. Foundation for Maxwell's second equation.

Contrast with Gauss for electricity

$\oint \vec E\cdot d\vec A = \frac{q_{\text{enc}}}{\varepsilon_0}\quad\text{vs}\quad \oint \vec B\cdot d\vec A = 0.$

The asymmetry between the two laws is one of the most fundamental experimental facts about electromagnetism, and the search for magnetic monopoles (predicted by some grand-unified theories) remains an open problem.

Worked Example

Consider any Gaussian surface enclosing a bar magnet. The flux from the N pole "out" exactly cancels the flux from the S pole "in"; net flux through the closed surface is zero.

Pitfalls

• The law says total flux is zero, not that $\vec B = 0$ everywhere on the surface.
• Don't try to apply this for a closed surface that intersects current-carrying wires — the law still holds, but it's $\vec B$ that has the divergence-free property, not $\vec H$ in magnetised matter (where $\nabla\cdot\vec H = -\nabla\cdot\vec M$).

5.6 Earth's Magnetism — Elements

Definition

The Earth behaves, to leading order, as a giant magnetic dipole tilted at about $11.3^\circ$ to its rotation axis. The geomagnetic poles are where the dipole axis meets the surface; the magnetic poles (where the dip is $\pm 90^\circ$) are slightly different points and drift over time.

Convention: The Earth's magnetic south pole lies near the geographic north (in Canadian Arctic) and vice versa — that is why the N-seeking end of a compass points toward the geographic north.

The three elements

At any location on Earth's surface, the local field $\vec B_E$ is fully specified by three numbers:

1. Declination $D$ — the angle between the geographic meridian (true north) and the magnetic meridian (the vertical plane containing $\vec B_E$).
2. Inclination (dip) $I$ — the angle that $\vec B_E$ makes with the horizontal plane.
3. Horizontal component $H_E$ — the projection of $\vec B_E$ on the horizontal plane.

Derivation: components from the total field

Resolve $\vec B_E$ into horizontal and vertical components:

$H_E = B_E\cos I,\qquad V_E = B_E\sin I.$

Therefore

$\boxed{\;B_E = \sqrt{H_E^2 + V_E^2},\qquad \tan I = \frac{V_E}{H_E}\;}$

Variation with latitude

For a perfect dipole, at magnetic latitude $\lambda$,

$\tan I = 2\tan\lambda.$

So at the magnetic equator $I = 0$ (field horizontal); at the magnetic poles $I = \pm 90^\circ$ (field vertical).

Worked Example

At a place, $H_E = 0.25\,\text{G}$ and dip $I = 60^\circ$. The total field:

$B_E = \frac{H_E}{\cos I} = \frac{0.25}{0.5} = 0.50\;\text{G},$

and the vertical component:

$V_E = B_E\sin I = 0.50\times 0.866 = 0.433\;\text{G}.$

Tangent law (for two perpendicular fields)

If a small magnet free to rotate in the horizontal plane is in equilibrium under $H_E$ (geographic N–S) and an external horizontal field $B$ (perpendicular to $H_E$, E–W), then making an angle $\theta$ with the meridian,

$B = H_E\tan\theta.$

This is used in tangent galvanometers.

Pitfalls

• Magnetic and geographic poles are not the same.
• Declination East or West must always be specified.
• The horizontal component is what affects compass needles; the vertical component is "wasted" for horizontal compasses.
• The dip angle is measured from the horizontal, not the vertical.

5.7 Magnetic Intensity, Magnetisation, Susceptibility, Permeability

Definition

When matter is placed in a magnetic field, it gets magnetised. To describe this we use four related fields/quantities:

• Magnetic intensity $\vec H$: the auxiliary field produced by free currents alone. Units A/m.
• Magnetisation $\vec M$: net magnetic dipole moment per unit volume. Units A/m.
• Magnetic susceptibility $\chi$ (dimensionless): describes how easily a material magnetises:

$\vec M = \chi\,\vec H.$

• Relative permeability $\mu_r$ and permeability $\mu$:

$\mu_r = 1 + \chi,\qquad \mu = \mu_0\mu_r.$

The fundamental relation

The total magnetic field inside a magnetised material is the sum of the field due to free currents and that due to magnetisation currents:

$\boxed{\;\vec B = \mu_0(\vec H + \vec M)\;}$

Combined with $\vec M = \chi\vec H$:

$\vec B = \mu_0(1+\chi)\vec H = \mu_0\mu_r\vec H = \mu\vec H.$

Solenoid filled with magnetic material

For an ideal solenoid with $n$ turns/m carrying current $I_f$ (free current):

• The vacuum field would have been $B_0 = \mu_0 n I_f$.
• The auxiliary field $H = nI_f$ inside.
• With a core of relative permeability $\mu_r$, the field becomes $B = \mu_0\mu_r n I_f$.
• The magnetisation $M = (\mu_r - 1)nI_f = \chi nI_f$.

Worked Example

A material has $\chi = 599$ (a soft iron). For an external $H = 100\,\text{A/m}$:

$\mu_r = 1 + 599 = 600;\quad M = \chi H = 5.99\times 10^4\;\text{A/m};\quad B = \mu_0\mu_r H = 4\pi\times 10^{-7}\times 600\times 100 = 0.0754\;\text{T}.$

Pitfalls

• $\vec H$ and $\vec M$ have the same units (A/m); $\vec B$ has units of Tesla — be careful with mixed-unit problems.
• $\chi$ here is the dimensionless volume susceptibility; some books use mass susceptibility $\chi/\rho$ — verify units.
• For non-linear materials (ferromagnets), $\chi$ depends on $H$ — it is not a constant.

5.8 Classification of Magnetic Materials

The three classes

Materials are classified by their value and sign of susceptibility:

Curie's law (paramagnetism)

For paramagnets, thermal agitation disorders the moments; an external $H$ partially aligns them, giving

$\chi = \frac{C}{T},\qquad M = \frac{C}{T}H,$

where $C$ is the Curie constant of the material. At low $T$, $\chi$ is large; at high $T$, $\chi\to 0$.

Curie–Weiss law (ferromagnetism above $T_C$)

Above the Curie temperature $T_C$, a ferromagnet becomes paramagnetic with

$\chi = \frac{C}{T - T_C}.$

For iron, $T_C \approx 1043\,\text{K}$; for cobalt $\approx 1394\,\text{K}$; for nickel $\approx 631\,\text{K}$.

Why diamagnetism is universal

Every material has some diamagnetism (induced opposing moments by an external field — a microscopic Lenz's law). It is masked in paramagnets and ferromagnets by the much stronger alignment of permanent moments.

Worked Example

A paramagnetic salt has $\chi = 6.0\times 10^{-4}$ at $T = 300\,\text{K}$. At $T = 100\,\text{K}$ (Curie's law):

$\chi(100) = \chi(300)\cdot\frac{300}{100} = 6.0\times 10^{-4}\times 3 = 1.8\times 10^{-3}.$

Pitfalls

• Don't confuse "magnetic permeability $\mu$" with "magnetic moment $m$".
• "Diamagnetic" $\neq$ "non-magnetic": $\chi$ is small but negative.
• Curie law breaks down at very low $T$ (saturation) and inside the ferromagnetic phase.

5.9 Hysteresis — B–H Curve, Coercivity, Retentivity

Definition

When a ferromagnet is taken through a complete cycle of magnetising field $H$, the magnetisation $M$ (or equivalently $B$) lags behind $H$. The $B$–$H$ curve traces a closed loop known as a hysteresis loop.

Key points on the loop

1. Initial magnetisation curve (OA): starting from unmagnetised state, $B$ rises non-linearly with $H$ to saturation at $A$.
2. Saturation $B_s$: all domains aligned with $H$; further increase in $H$ gives no further increase in $M$.
3. Retentivity $B_r$ (residual induction): the value of $B$ when $H$ is brought back to zero — the magnet is still magnetised.
4. Coercivity $H_c$: the reverse field needed to bring $B$ to zero.
5. The loop is symmetric on the other side.

Energy loss per cycle

The work done per unit volume by the source against the hysteresis is the area of the loop in the $B$–$H$ plane:

$W_{\text{cycle}} = \oint H\,dB\quad(\text{J/m}^3\text{ per cycle}).$

This energy is dissipated as heat inside the material — a major loss mechanism in AC transformers.

Soft vs hard ferromagnets

Permanent magnets require high $H_c$ (don't demagnetise easily) and high $B_r$ (retain strong field). Transformer cores require low $H_c$ (small loop area, low loss) and high permeability.

Worked Example

A transformer core has a hysteresis loop area $300\,\text{J/m}^3$ per cycle. If it operates at 50 Hz and has a volume of $0.02\,\text{m}^3$, the hysteresis power loss is

$P_h = \text{(area)}\times f\times V = 300\times 50\times 0.02 = 300\;\text{W}.$

Pitfalls

• Retentivity ($B_r$) and coercivity ($H_c$) refer to different axes — one is a $B$-value, one is an $H$-value.
• Saturation is not the same as retentivity — saturation is at the peak $H$; retentivity is at $H=0$.
• The "magnetic" and "iron" losses in a transformer include hysteresis loss + eddy-current loss; only the first is given by the loop area.

Solved Problems

Problem 1 — Equivalent solenoid

A short bar magnet has a magnetic moment $0.48\,\text{A m}^2$. Find $B$ at a point 10 cm from the centre on (a) the axis (b) the equator.

$B_{\text{ax}} = 10^{-7}\cdot\frac{2(0.48)}{(0.10)^3} = 9.6\times 10^{-4}\;\text{T} = 9.6\;\text{G},$

$B_{\text{eq}} = 10^{-7}\cdot\frac{0.48}{(0.10)^3} = 4.8\times 10^{-4}\;\text{T} = 4.8\;\text{G}.$

Problem 2 — Torque & energy

A magnet of moment $0.05\,\text{A m}^2$ is placed at $30^\circ$ to a field of $0.16\,\text{T}$. Compute torque and PE.

$\tau = mB\sin\theta = 0.05\times 0.16\times 0.5 = 4.0\times 10^{-3}\;\text{N m},$

$U = -mB\cos\theta = -0.05\times 0.16\times \cos 30^\circ = -6.93\times 10^{-3}\;\text{J}.$

Problem 3 — Dip and components

At a station, $V_E = 0.6\,\text{G}$, $H_E = 0.346\,\text{G}$. Find dip $I$ and total field $B_E$.

$\tan I = \frac{V_E}{H_E} = \frac{0.6}{0.346} = 1.732\implies I = 60^\circ,$

$B_E = \sqrt{H_E^2 + V_E^2} = \sqrt{0.346^2 + 0.6^2} = 0.693\;\text{G}.$

Problem 4 — Magnetisation

An iron rod has length 0.5 m, cross-section $1.0\times 10^{-4}\,\text{m}^2$, and develops a magnetic moment $5.0\,\text{A m}^2$ in a field. Its magnetisation is

$M = \frac{m}{V} = \frac{5.0}{0.5\times 10^{-4}} = 1.0\times 10^{5}\;\text{A/m}.$

If applied $H = 250\,\text{A/m}$, $\chi = M/H = 400$, $\mu_r = 401$.

Problem 5 — Curie's law

A paramagnet has $\chi = 1.2\times 10^{-3}$ at $300\,\text{K}$. Find $\chi$ at $150\,\text{K}$.

$\chi(150) = \chi(300)\times\frac{300}{150} = 2.4\times 10^{-3}.$

Problem 6 — Solenoid with iron core

A solenoid with $n = 2000$ turns/m carries 1 A. Its core is iron with $\mu_r = 800$. Find $H$, $B$, $M$.

$H = nI = 2000\;\text{A/m},$

$B = \mu_0\mu_r H = 4\pi\times 10^{-7}\times 800\times 2000 = 2.01\;\text{T},$

$M = (\mu_r-1)H = 799\times 2000 = 1.6\times 10^{6}\;\text{A/m}.$

Problem 7 — Period of oscillation

A bar magnet of $m = 0.36\,\text{A m}^2$ and moment of inertia $I_{\text{moi}} = 7.5\times 10^{-6}\,\text{kg m}^2$ is suspended freely in $H_E = 0.36\,\text{G} = 3.6\times 10^{-5}\,\text{T}$. Its small-oscillation period:

$T = 2\pi\sqrt{\frac{I_{\text{moi}}}{mH_E}} = 2\pi\sqrt{\frac{7.5\times 10^{-6}}{0.36\times 3.6\times 10^{-5}}} = 2\pi\sqrt{578.7}\;\text{s} = 47.8\;\text{s}.$

JEE/NEET Edge Cases

• Sign of work done by external agent: when rotating against the torque, $W_{\text{ext}} > 0$; when rotated by the torque, $W_{\text{ext}} < 0$. Always check the sign by computing $\Delta U$.
• Cutting a magnet — if a bar magnet of moment $m$ is cut transversely into $n$ equal pieces, each piece has $m/n$ and the same pole strength; if cut longitudinally into $n$ pieces, each has $m/n$ but with pole strength $q_m/n$.
• Two magnets joined — moments add as vectors. If two equal magnets of moment $m$ are joined with poles in line, total moment is $2m$; in opposite, total is $0$; perpendicular, total is $m\sqrt 2$.
• Vibration magnetometer with two magnets — if periods $T_1$ and $T_2$ are taken with two configurations (sum and difference of moments),

$\frac{m_1}{m_2} = \frac{T_2^2 + T_1^2}{T_2^2 - T_1^2}.$

• Tangent law in a deflection magnetometer: $H_E\tan\theta = B_{\text{external}}$.
• Earth's field at magnetic equator: dip is zero, field is purely horizontal.
• Susceptibility temperature variation is a frequent NCERT exemplar trap — paramagnets follow $1/T$, diamagnets are nearly $T$-independent, ferromagnets follow $1/(T-T_C)$.

Quick Recap

• A bar magnet $\equiv$ a current solenoid; $\vec m = NI\vec A$.
• $B_{\text{axial}} = (\mu_0/4\pi)(2m/r^3)$, $B_{\text{equatorial}} = (\mu_0/4\pi)(m/r^3)$ — ratio 2:1.
• $\vec\tau = \vec m\times\vec B$, $U = -\vec m\cdot\vec B$.
• $\oint\vec B\cdot d\vec A = 0$ — no monopoles.
• Earth's field: $D$, $I$, $H_E$; $H_E = B_E\cos I$, $V_E = B_E\sin I$, $\tan I = 2\tan\lambda$.
• $\vec B = \mu_0(\vec H + \vec M)$, $\vec M = \chi\vec H$, $\mu_r = 1+\chi$.
• Three classes: dia ($\chi< 0$), para ($\chi>0$, $\propto 1/T$), ferro ($\chi\gg 0$, hysteresis).
• Hysteresis area = energy dissipated per cycle per unit volume; soft (low $H_c$) = transformer; hard (high $H_c$) = permanent magnet.

Formula Sheet