Magnetism and Matter

The chapter moves from a single object to a whole zoo of materials. It opens with the bar magnet itself: its field-line pattern, its equivalence to a current-carrying solenoid of the same magnetic moment, and the torque and potential energy it feels in a uniform external field. It then draws a precise electrostatic analogy, converting every electric-dipole formula into its magnetic counterpart by a simple substitution, before stating Gauss's law for magnetism — field lines never begin or end because isolated poles do not exist. The second half turns from a single dipole to bulk matter: it defines magnetisation M and magnetic intensity H, links them through susceptibility chi and relative permeability mu_r, and uses these to classify every material as diamagnetic, paramagnetic or ferromagnetic, explaining superconductors, iron, aluminium and bismuth with the same small set of ideas.

The bar magnet as a magnetic dipole

Sprinkle iron filings on glass over a bar magnet and they trace the same characteristic pattern a current-carrying solenoid produces: lines emerging from one pole, curving around, and entering the other. This pattern lets you plot magnetic field lines, whose properties distinguish magnetism from electrostatics. The tangent at any point gives the direction of $B$ there; line density gives its magnitude, so lines bunch closer where $B$ is larger; lines never intersect, since a crossing would make the field direction ambiguous; and, most importantly, magnetic field lines always form continuous closed loops. This last property is the direct signature that isolated magnetic poles, or monopoles, do not exist: unlike electric field lines, which begin on a positive charge and end on a negative one (or escape to infinity), magnetic field lines have no starting or ending point. Break a bar magnet transverse to its length or along its length and, either way, you get two smaller magnets, each already possessing its own north and south pole — a single pole can never be isolated by cutting.

The bar magnet as an equivalent solenoid

Ampere's hypothesis, met in the previous chapter, is that all magnetic phenomena reduce to circulating currents. A bar magnet's resemblance to a current-carrying solenoid — in field-line pattern, and in how cutting either one in half gives two smaller, weaker versions of the same object — suggests that a bar magnet can be modelled as a stack of tiny circulating currents. Making this precise (the first derivation below) gives the axial field of a finite solenoid, far from its centre, as

where: $m$ is the magnetic moment of the solenoid (or, equivalently, of the bar magnet it mimics), $r$ is the axial distance from the centre with $r \gg l$, the solenoid's half-length, and $\mu_0 = 4\pi \times 10^{-7}\ \text{T m A}^{-1}$ is the permeability of free space. Because this is also the measured far axial field of an actual bar magnet, a bar magnet's moment is defined to equal the moment of the equivalent solenoid, $m = NIA$, that reproduces the same far field — turns times current times cross-sectional area.

Torque and potential energy in a uniform field

Place a small compass needle of known magnetic moment $m$ in a uniform external field $B$ and let it swing freely (Fig. 5.3(b) of NCERT is exactly this setup). The field exerts a torque that tries to rotate the needle until $m$ lines up with $B$,

where: $\theta$ is the angle between the dipole moment $\vec m$ and the field $\vec B$, and $\tau$ is a restoring torque, largest when $m$ is perpendicular to $B$ and zero when the two are parallel or antiparallel. Integrating this torque over angle (carried out in full in the second derivation below) gives the magnetic potential energy of the dipole,

where: the zero of potential energy is fixed, by convention, at $\theta = 90^\circ$ (needle perpendicular to the field). Equation (3) shows $U_m$ is minimum, $-mB$, at $\theta = 0^\circ$ — the most stable orientation, with $m$ parallel to $B$ — and maximum, $+mB$, at $\theta = 180^\circ$, the most unstable orientation, from which the slightest disturbance flips the magnet toward alignment.

The electrostatic analogy

Comparing equations (1) to (3) with the electric-dipole results from earlier chapters reveals a one-to-one correspondence: the far field of a bar magnet of moment $m$ follows from the electric dipole of moment $p$ by simply replacing $E \to B$, $p \to m$ and $1/4\pi\varepsilon_0 \to \mu_0/4\pi$. In particular, the equatorial field (perpendicular bisector, $r \gg l$) and axial field of a short bar magnet are

where: the axial field is exactly twice the equatorial field in magnitude at the same distance $r$, and the equatorial field points opposite to $m$ while the axial field points along it — precisely mirroring the electric dipole. The full analogy is summarised below.

The one place the analogy breaks down is the source itself: an electric dipole is built from two separated, genuinely isolated charges $+q$ and $-q$, while a magnetic dipole has no isolated pole to separate — it is, at bottom, always a current loop. That difference is exactly what the next section makes precise.

Gauss's law for magnetism

For an electric dipole, a closed surface enclosing only the positive charge shows a net outward flux. For a bar magnet or solenoid, no closed surface ever shows a net flux, however it is drawn — as many field lines leave as enter, since the lines are continuous closed loops with no source or sink. Formally, summing the flux $\Delta\phi_B = \vec B \cdot \Delta \vec S$ through every element of a closed surface $S$ gives

where: the sum runs over all elements of any arbitrary closed surface, and the result is always exactly zero. Compare this with Gauss's law of electrostatics, $\sum \vec E \cdot \Delta \vec S = q/\varepsilon_0$, where $q$ is the enclosed charge; the magnetic version has no such source term because there is no magnetic charge to enclose. Gauss's law for magnetism states: the net magnetic flux through any closed surface is zero. If magnetic monopoles of pole strength $q_m$ were ever discovered, the right-hand side would become $\mu_0 q_m$ instead of zero — but no such particle has ever been observed.

Magnetisation, magnetic intensity and susceptibility

A bulk material is built from atoms, and a circulating electron inside an atom carries a small magnetic moment of its own. In most materials these atomic moments point every which way and cancel on average, but an applied field can coax them into partial or full alignment, giving the material a net moment. The magnetisation $M$ of a sample is defined as its net magnetic moment per unit volume,

where: $M$ is a vector with dimensions $\text{L}^{-1}\text{A}$, measured in $\text{A m}^{-1}$. Now consider a long solenoid of $n$ turns per unit length carrying current $I$; empty, its interior field is $B_0 = \mu_0 n I$. Fill the core with a magnetic material and the interior field becomes $B = B_0 + B_m$, where the extra contribution turns out to be proportional to the material's magnetisation, $B_m = \mu_0 M$. It is convenient to isolate the part of the field traceable purely to the free (externally supplied) current, called the magnetic intensity $H$,

where: $H$ has the same dimensions and unit as $M$ ($\text{A m}^{-1}$), and inside a solenoid $H = nI$ regardless of what fills the core — it depends only on the winding and the free current, never on the material. Equation (7) cleanly splits the total field into an externally controlled part ($H$) and a material-dependent part ($M$). For most materials in modest fields, $M$ responds linearly to $H$,

where: $\chi$, the magnetic susceptibility, is a dimensionless number measuring how readily a material magnetises in response to an applied field; it is small and positive for paramagnetic materials, small and negative for diamagnetic materials (where $M$ and $H$ point in opposite directions), and large and positive for ferromagnetic materials. Combining equations (7) and (8),

where: $\mu_r = 1+\chi$ is the dimensionless relative magnetic permeability, the direct magnetic analogue of the dielectric constant, and $\mu = \mu_0\mu_r$ is the material's magnetic permeability, carrying the same dimensions and unit as $\mu_0$. Only one of $\chi$, $\mu_r$ and $\mu$ is independent; given any one, the other two follow immediately.

Classifying magnetic materials: diamagnetic, paramagnetic, ferromagnetic

The sign and size of $\chi$ sort every material into one of three families.

Diamagnetic substances have zero net atomic magnetic moment to begin with. An applied field slightly speeds up orbiting electrons circulating opposite to it and slows those circulating along it (a Lenz's-law effect, met properly next chapter), so the material develops a small net moment opposing the field and is repelled — pushed from strong field toward weak. Bismuth, copper, lead, silicon, nitrogen at STP, water and sodium chloride are diamagnetic, though the effect (typically one part in $10^5$) is masked whenever paramagnetism or ferromagnetism is also present; diamagnetism is universal but easily the weakest of the three. Its most striking form is the superconductor, cooled to very low temperature, combining perfect conductivity with perfect diamagnetism: $\chi = -1$, $\mu_r = 0$, field totally excluded — the Meissner effect, which underlies magnetic levitation.

Paramagnetic substances have atoms (or ions or molecules) with a genuine permanent dipole moment. Thermal motion randomises these in zero field, so bulk magnetisation averages to zero; a strong enough field, especially at low temperature, partially aligns them, giving weak attraction toward stronger field. Aluminium, sodium, calcium, oxygen at STP and copper chloride are paramagnetic; both $\chi$ and $\mu_r$ depend on temperature, with magnetisation rising toward saturation as field increases or temperature falls.

Ferromagnetic substances also have a permanent atomic dipole moment, but neighbouring moments interact strongly enough to spontaneously align over macroscopic domains, roughly a millimetre across, each holding about $10^{11}$ atoms. With no field the domains point randomly and net magnetisation is near zero; a field grows the favourably aligned domains until the sample becomes one giant domain, strongly concentrating field lines and producing powerful attraction. Iron, cobalt, nickel and gadolinium are ferromagnetic, with $\mu_r$ sometimes exceeding $1000$. Alignment that persists once the field is removed (alnico, lodestone) marks a hard ferromagnet, good for permanent magnets; alignment that vanishes (soft iron) marks a soft ferromagnet. Above a characteristic transition temperature (the Curie temperature) thermal agitation overwhelms the domain-aligning interaction and a ferromagnet turns into an ordinary paramagnet.

Board level - magnetic moment of a bar magnet from a measured torque

A short bar magnet, placed with its axis at $30^\circ$ to a uniform external magnetic field of $0.25\ \text{T}$, experiences a torque of magnitude $4.5\times10^{-2}\ \text{J}$. Find the magnitude of the magnetic moment of the magnet.

Step 1. Start from the torque formula, equation (2): $\tau = mB\sin\theta$, with $\tau = 4.5\times10^{-2}\ \text{J}$, $B = 0.25\ \text{T}$ and $\theta = 30^\circ$.

Step 2. Rearrange for $m$: $m = \dfrac{\tau}{B\sin\theta} = \dfrac{4.5\times10^{-2}}{0.25 \times \sin 30^\circ} = \dfrac{4.5\times10^{-2}}{0.25 \times 0.5}$.

Step 3. Simplify the denominator and divide: $0.25 \times 0.5 = 0.125$, so $m = 4.5\times10^{-2}/0.125 = 0.36\ \text{J T}^{-1}$. Answer: the magnetic moment is $m = 0.36\ \text{J T}^{-1}$ (equivalently, $\text{A m}^2$).

Competitive level (JEE / NEET) - a current-carrying solenoid as a bar magnet: moment and torque

A closely wound solenoid of $800$ turns and cross-sectional area $2.5\times10^{-4}\ \text{m}^2$ carries a current of $3.0\ \text{A}$. Find its associated magnetic moment. If this solenoid is then free to turn about a vertical axis and a uniform horizontal magnetic field of $0.25\ \text{T}$ is applied, find the torque on it when its axis makes an angle of $30^\circ$ with the field.

Step 1. The solenoid acts as a bar magnet along its axis, with $m = NIA$ (the identification from the first derivation above). Substituting, $m = 800 \times 3.0 \times 2.5\times10^{-4} = 0.60\ \text{A m}^2$.

Step 2. With the solenoid now free to rotate in the applied field, treat it as a bar magnet of moment $m = 0.60\ \text{A m}^2$ and apply equation (2), $\tau = mB\sin\theta$, with $B = 0.25\ \text{T}$ and $\theta = 30^\circ$.

Step 3. Substitute and simplify: $\tau = 0.60 \times 0.25 \times \sin 30^\circ = 0.60 \times 0.25 \times 0.5 = 0.075\ \text{N m}$. Answer: magnetic moment $m = 0.60\ \text{A m}^2$; torque at $30^\circ$ is $\tau = 7.5\times10^{-2}\ \text{N m}$. Timing note: under two minutes. The trap is remembering the solenoid's own internal field plays no role once treated as a dipole — just $m=NIA$ then the ordinary torque formula. Since the field is uniform, there is no net force, only a torque.

Competitive level (JEE / NEET) - H, M, B and the magnetising current for a solenoid with a magnetic core

A solenoid has a core of material with relative permeability $\mu_r = 400$. The windings, insulated from the core, carry a current of $2\ \text{A}$, with $1000$ turns per metre. Find (a) the magnetic intensity $H$, (b) the magnetisation $M$, (c) the magnetic field $B$, and (d) the magnetising current $I_m$.

Step 1. $H$ depends only on the free current and turns density, not on the core: $H = nI = 1000 \times 2.0 = 2\times10^{3}\ \text{A/m}$.

Step 2. The field $B$ follows from equation (9): $B = \mu_r \mu_0 H = 400 \times (4\pi\times10^{-7}) \times 2\times10^{3} \approx 1.0\ \text{T}$.

Step 3. The magnetisation follows from equation (7) rearranged: $M = (B-\mu_0 H)/\mu_0 = (\mu_r-1)H = 399 \times 2\times10^{3} \approx 8\times10^{5}\ \text{A/m}$ — note that $M$ is nearly four hundred times larger than $H$, reflecting how strongly this core magnetises.

Step 4. The magnetising current, from the third derivation's result $I+I_m = B/(\mu_0 n)$: with $B \approx 1.0\ \text{T}$ and $n = 1000$, $I+I_m = 1.0/(4\pi\times10^{-7}\times1000) \approx 795.8\ \text{A}$, so $I_m \approx 794\ \text{A}$. Answer: $H = 2\times10^{3}\ \text{A/m}$, $M \approx 8\times10^{5}\ \text{A/m}$, $B \approx 1.0\ \text{T}$, and $I_m \approx 794\ \text{A}$. Timing note: about two minutes. The trap is mixing routes: computing $M$ from unrounded $\chi=399$ but $I_m$ from a rounded $B$, which is why the clean $I_m=(\mu_r-1)I=798\ \text{A}$ differs slightly from the rounded $794\ \text{A}$ — either is acceptable, just be consistent.

Thinking magnetic field lines are "lines of force" on a moving charge. They are not — the magnetic force on a moving charge is $q\vec v \times \vec B$, always perpendicular to $B$, so a field line never shows the direction a charge will actually be pushed.

Assuming every magnetic configuration must have a north pole and a south pole. True only if the source has a nonzero net magnetic moment. A toroid, or an infinite straight current-carrying wire, produces a perfectly good field with closed field lines but no identifiable poles at all.

Believing an iron nail near a bar magnet should feel no net force, since a dipole in a uniform field feels torque but no force. The field near a real magnet is highly non-uniform, so the nail's induced moment sees different field strengths at its two ends and feels a genuine net (always attractive) force in addition to a torque.

Confusing Gauss's law for magnetism with "the field is zero everywhere on a closed surface." It says the net flux is zero, not that $B$ vanishes; field lines cross the surface freely, as long as exactly as many leave as enter.

Forgetting the sign of $M$ relative to $H$ in diamagnetic materials, and treating any $\mu_r > 1$ as a "strong" magnet. Negative $\chi$ makes $M$ point opposite to $H$, which is why diamagnetic materials are repelled; and paramagnetic materials also have $\mu_r$ slightly above $1$ but are only weakly attracted — only $\chi \gg 1$ signals real (ferromagnetic) strength.

• State the four properties of magnetic field lines and explain why they always form closed loops, unlike electric field lines.
• Derive the axial field of a bar magnet, $B = \mu_0\,2m/4\pi r^3$, by treating it as an equivalent current-carrying solenoid.
• Write down and apply the torque, $\vec\tau = \vec m \times \vec B$, and potential energy, $U_m = -\vec m \cdot \vec B$, of a magnetic dipole in a uniform field, and identify the stable and unstable orientations.
• Use the electrostatic analogy to convert any electric dipole formula into its magnetic counterpart, including axial and equatorial field expressions.
• State Gauss's law for magnetism and explain why it follows directly from the non-existence of magnetic monopoles.
• Define magnetisation $M$, magnetic intensity $H$, and relate them to the total field via $B = \mu_0(H+M)$.
• Use $M = \chi H$ and $B = \mu_0\mu_r H = \mu H$ to move between susceptibility, relative permeability and permeability.
• Classify a given material as diamagnetic, paramagnetic or ferromagnetic from the sign and size of $\chi$, and describe the microscopic mechanism behind each.
• Explain the Meissner effect in a superconductor as the limiting case $\chi = -1$, $\mu_r = 0$.
• Describe magnetic domains and explain why ferromagnetism disappears above a characteristic temperature.

Ask the tutor: if you would like to walk through why the axial field of a solenoid collapses to the dipole form step by step, or want a hand sorting a list of materials into diamagnetic, paramagnetic and ferromagnetic, ask the tutor to work through it with you.