Magnetism and Gauss's law

This section sits inside Unit V, "Magnetism and Matter," of CBSE Class 12 Physics, and it deliberately links three ideas the board treats at the same qualitative level: the closed-surface statement of Gauss's law for magnetism, the axial and equatorial field of a magnetic dipole (formulae quoted for reference, no derivation expected), and the torque on a dipole in a uniform field. CBSE explicitly keeps the vector-calculus form $\nabla\cdot\vec B = 0$ out of the syllabus — you are examined only on the closed-surface flux statement in words and simple flux-sum form. Two related sub-topics that older textbooks bundle in here are firmly out of scope for 2025-26: Earth's magnetism as a quantitative topic (declination, dip, horizontal component) has been removed entirely, and hysteresis-curve area/energy-loss calculations and Curie's law as a formal $\chi\propto 1/T$ relation are likewise excluded — keep magnetic-material behaviour at the qualitative "temperature affects magnetism" level only.

Start with the definition every flux calculation in this chapter reduces to. For a small patch of area $d\vec A$ sitting in a magnetic field $\vec B$, the magnetic flux through it is the component of $\vec B$ perpendicular to the patch, multiplied by the patch's area:

where: $\vec B$ is the magnetic field at that patch (in tesla, T), $d\vec A$ is a tiny area vector pointing along the outward normal to the surface, and $d\phi_B$ carries units of weber (Wb) $=\ \text{T}\,\text{m}^2$. The physical reading is simple: flux measures how much of the field "passes through" a surface, and the dot product means only the part of $\vec B$ that is perpendicular to the surface counts — a field running exactly parallel to a patch contributes nothing. Now close the surface — imagine wrapping it all the way around a volume, like a balloon or a cardboard box with no gaps. Adding up $\vec B\cdot d\vec A$ over every patch of that closed surface gives the total flux, and Gauss's law for magnetism makes a bold claim about that total:

where: the circle on the integral sign is only a reminder that the sum runs over a closed surface — no formal surface-integral machinery is required to use the statement at this level; you simply need to know that this total is always zero, whatever shape you draw and wherever you draw it, whether it encloses a magnet, part of a magnet, or nothing at all. Compare this with the electric case you already know: $\oint \vec E\cdot d\vec A = q_{enclosed}/\varepsilon_0$. There, the right-hand side is nonzero whenever net charge sits inside, because isolated positive and negative charges genuinely exist and can be pulled apart. Electric field lines are allowed to start on a positive charge and end on a negative one, so a surface that traps more starting points than ending points (or vice versa) shows a nonzero net flux. Magnetic field lines behave completely differently — they never start or stop anywhere; every one of them closes on itself, looping from the north pole, out through the surrounding space, back in through the south pole, and then continuing inside the magnet from south back to north to complete the loop. A field line that enters a closed surface at one point must, sooner or later, leave that same surface at another point, because it cannot simply vanish inside. Entries and exits always balance, and the sum $\oint\vec B\cdot d\vec A$ collapses to zero no matter how the surface is drawn.

This is exactly why a bar magnet behaves, seen from outside, like a solenoid carrying a steady current — the equivalent solenoid picture: a cylinder wound with current loops produces the same closed-loop field pattern as a bar magnet, with field lines emerging from one end, curving around outside, and re-entering the other end, then continuing inside the coil back to where they started. There is no genuine "source" of field lines anywhere on a solenoid or on a magnet — only regions where the lines happen to be more or less bunched together, which is what we perceive as a stronger or weaker pole.

Away from the pole faces, a bar magnet's field looks, to good approximation, like that of a magnetic dipole of moment $m$ (units A·m$^2$), the product of one pole's effective pole strength and the pole separation. CBSE asks you to quote, not derive, the field strength at two special locations a distance $r$ from the centre of a short dipole: on the axis (the line through both poles, extended) and in the equatorial plane (the perpendicular bisector of the magnet):

where: $\mu_0 = 4\pi\times10^{-7}\ \text{T}\,\text{m/A}$ is the permeability of free space (so $\mu_0/4\pi = 10^{-7}\ \text{T}\,\text{m/A}$, the constant you will actually plug numbers into), $m$ is the magnetic dipole moment, and $r$ is the distance from the magnet's centre, taken large compared with the magnet's own length. The physical reading is that both fields fall off as $1/r^3$ — magnetic dipole fields die away faster than a single-charge Coulomb field, which falls as $1/r^2$, because a dipole's near-equal-and-opposite poles cancel more of the field at a distance — and the axial value is always exactly twice the equatorial value at the same $r$, a ratio worth memorising outright. If instead of measuring the magnet's own field you place the whole magnet inside someone else's uniform external field $\vec B$, the field exerts a turning effect rather than a push:

where: $\vec\tau$ is the torque (N·m), and the cross product means the torque is largest when $\vec m$ and $\vec B$ are perpendicular ($\tau = mB\sin\theta$) and zero when they are already aligned or exactly opposed. The physical reading is intuitive — a compass needle ($\vec m$) feels no twist when it already points along the local field, and the maximum twist when it is crosswise to it, which is exactly why a compass settles pointing along the field rather than across it. CBSE treats this torque relation qualitatively too: you use the formula, you do not derive it from the force on each pole.

Board level - axial and equatorial field of a bar magnet

A short bar magnet has magnetic dipole moment $m = 0.60\ \text{A}\cdot\text{m}^2$. Find the magnetic field (a) on its axis and (b) in its equatorial plane, both at a distance $r = 10\ \text{cm}$ from the magnet's centre. Take $\mu_0/4\pi = 10^{-7}\ \text{T}\,\text{m/A}$.

Step 1. Convert and identify values. $r = 10\ \text{cm} = 0.10\ \text{m}$, so $r^3 = (0.10)^3 = 1.0\times10^{-3}\ \text{m}^3$. $m = 0.60\ \text{A}\cdot\text{m}^2$ is already in SI units, so no further conversion is needed.

Step 2. Axial field using the quoted formula: $B_{axial} = \dfrac{\mu_0}{4\pi}\dfrac{2m}{r^3} = 10^{-7}\times\dfrac{2\times0.60}{1.0\times10^{-3}} = 10^{-7}\times1200 = 1.2\times10^{-4}\ \text{T}$.

Step 3. Equatorial field at the same $r$: $B_{equatorial} = \dfrac{\mu_0}{4\pi}\dfrac{m}{r^3} = 10^{-7}\times\dfrac{0.60}{1.0\times10^{-3}} = 10^{-7}\times600 = 6.0\times10^{-5}\ \text{T}$. Answer: $B_{axial} = 1.2\times10^{-4}\ \text{T}$ (0.12 mT) and $B_{equatorial} = 6.0\times10^{-5}\ \text{T}$ (0.06 mT), confirming the axial value is exactly twice the equatorial one, as the derivation above predicts.

Competitive level - torque and potential energy of the same magnet in Earth's local field

The bar magnet from above ($m = 0.60\ \text{A}\cdot\text{m}^2$) is now placed in a uniform external field $B = 0.30\ \text{T}$ (a strong lab electromagnet, chosen for clean numbers), with its axis making an angle of $30°$ with the field. Find (a) the torque on the magnet and (b) the work needed to rotate it from this position to full alignment ($\theta = 0°$) with the field.

Step 1. Torque at $\theta = 30°$: $\tau = mB\sin\theta = 0.60\times0.30\times\sin30° = 0.18\times0.5 = 0.090\ \text{N}\cdot\text{m}$. This is the instantaneous twist the field exerts on the magnet at that orientation, tending to rotate $\vec m$ toward $\vec B$.

Step 2. Recall (quoted, not derived at this level) the potential energy of a dipole in a field, $U(\theta) = -mB\cos\theta$, lowest when aligned ($\theta = 0°$) and highest when anti-aligned ($\theta = 180°$). Evaluate it at the starting angle: $U(30°) = -0.60\times0.30\times\cos30° = -0.18\times0.866 = -0.156\ \text{J}$.

Step 3. Evaluate at full alignment: $U(0°) = -0.60\times0.30\times\cos0° = -0.18\times1 = -0.180\ \text{J}$. The work needed to rotate the magnet from $30°$ to $0°$ equals the drop in potential energy, but since the field does positive work pulling the magnet into alignment (energy decreases on its own), no external work is actually required — the change is $\Delta U = U(0°) - U(30°) = -0.180-(-0.156) = -0.024\ \text{J}$. Answer: $\tau = 0.090\ \text{N}\cdot\text{m}$ at $30°$, and the magnet's energy drops by $0.024\ \text{J}$ as the field itself swings it into alignment — an external agent would need to supply $+0.024\ \text{J}$ only to rotate it the other way, back out to $30°$. Timing note: under ninety seconds once $mB$ is computed once and reused. Compute $mB = 0.18\ \text{J}$ a single time, then $\tau = mB\sin\theta$ and $U=-mB\cos\theta$ are one multiplication each — do not recompute $mB$ from scratch for each part. The one trap worth flagging under exam pressure: $\theta$ is measured from the field direction to $\vec m$, and a negative $\Delta U$ means the field is doing the work for you, not that you must supply that energy — read the sign before writing a conclusion.

• Writing or attempting to memorise the divergence form $\nabla\cdot\vec B = 0$ — this vector-calculus statement is outside the CBSE syllabus; the examinable form is the closed-surface flux statement in words or in $\oint\vec B\cdot d\vec A = 0$.
• Believing zero net flux means $\vec B = 0$ everywhere on the surface — it does not; $\vec B$ can be large at individual points on the surface, it is only the sum, with sign, over the whole closed surface that is always zero.
• Assuming the zero-flux result is special to spherical surfaces drawn around a bar magnet — it holds for a closed surface of any shape, anywhere, magnet inside or not.
• Confusing this law with Ampere's circuital law: Gauss's law for magnetism relates $\vec B$ to enclosed current through a closed loop, while Gauss's law for magnetism relates flux through a closed surface to (always zero) enclosed magnetic charge — different geometry, different statement, easy to muddle under time pressure.
• Forgetting that the axial field is always exactly twice the equatorial field at the same $r$ from the same dipole — a common numeric slip is to swap the factor of 2, or to apply it at two different distances.
• Trying to derive the axial/equatorial field formulae or the torque formula from first principles in an answer — CBSE wants these quoted and used, not derived; a derivation attempt wastes exam time and is not what is being asked.
• Dragging in Earth's magnetism (declination, dip) or a quantitative hysteresis/Curie-law argument to "explain" this topic — both are explicitly removed from the 2025-26 syllabus and do not belong in an answer here.

• Define magnetic flux $d\phi_B = \vec B\cdot d\vec A$ and state Gauss's law for magnetism, $\oint\vec B\cdot d\vec A = 0$, in words and in this simple flux-sum form.
• Explain, using the closed-loop nature of magnetic field lines, why the law implies isolated magnetic monopoles cannot exist.
• Quote (without deriving) the axial and equatorial dipole-field formulae and use them in a calculation, including the fixed $2:1$ axial-to-equatorial ratio at matched distance.
• Quote and apply the torque relation $\vec\tau=\vec m\times\vec B$ for a dipole in a uniform external field, and connect it to the bar-magnet-as-equivalent-solenoid picture.
• Contrast this law cleanly with Gauss's law for electricity and with Ampere's circuital law, stating precisely how each differs.
• Recognise which neighbouring topics (Earth's magnetism, quantitative hysteresis, Curie's law) are outside the current CBSE 2025-26 syllabus and should not be pulled into an answer here.

Ask the tutor: Want to see how the equivalent-solenoid picture of a bar magnet connects to the current loops you studied in Moving Charges and Magnetism — ask the tutor to walk through that link.