University Physics II · Maxwell's Equations and Electromagnetic Waves · 12.2
Gauss's Law for Magnetism
Electric field lines start and stop on charge. Magnetic lines do neither — they close on themselves — so every closed surface has exactly as much magnetic flux leaving it as entering.
Build the model
Connect the measurement to the mechanism.
Gauss's law for electricity carries a source on the right: ∮E⃗·dA⃗ = qenc/ε₀. Its magnetic partner has nothing there. ∮B⃗·dA⃗ = 0 holds for every closed surface, in every field, at every instant, which is the field-theoretic way of saying that no isolated magnetic charge has ever been found. The geometry follows at once: magnetic field lines cannot begin or end, so they close on themselves, and whatever leaves a closed surface re-enters it.
Cut a bar magnet and each piece grows a fresh pair of poles, because the real sources are current loops and spins, and a loop's field is a dipole with zero divergence built in. Three working results come out of the same line: the normal component of B⃗ is continuous across any interface, the flux through an open surface depends only on its boundary curve, and B⃗ in a plane wave has no component along the direction of travel.
- Simple definition
- Gauss's law for magnetism states that the net magnetic flux through any closed surface is zero: ∮B⃗·dA⃗ = 0. There is no magnetic charge to act as a source, so field lines close on themselves rather than beginning or ending.
- Example
- Draw a closed surface around the north end of a bar magnet. Lines stream outward into the air, and exactly as many enter through the cut across the bar, where B⃗ runs from south to north inside the iron.
No source term: as much flux enters the surface as leaves it, always.
Any closed surface, outward normal; flux in Wb = T m²
B⃗ is solenoidal — no point in space acts as a source or a sink.
Holds point by point; units T m⁻¹
B⊥ never jumps across a surface, whatever materials sit on the two sides.
From a flat pillbox straddling the interface; both in T
Where a tube of field lines narrows, the field strengthens in the same ratio.
Same tube of lines, no leakage; Φ in Wb, A in m²
Why 'the flux through the loop' in Faraday's law needs no choice of surface.
Same closed curve, same orientation; Φ in Wb
Zero divergence is exactly the condition that lets B⃗ be written as a curl.
A⃗ in T m; possible because ∇·(∇ × A⃗) = 0 identically
The equation with no source term
Gauss's law for electricity reads ∮E⃗·dA⃗ = qenc/ε₀: draw any closed surface, sum E⃗·dA⃗ over it with the outward normal, and the total counts the charge inside. Write the same statement for the magnetic field and the right-hand side is zero: ∮B⃗·dA⃗ = 0, for every closed surface, in every field, at every instant. Nothing has been left out. The equation is empty on the right because no experiment has isolated the magnetic counterpart of charge. Term by term, B⃗·dA⃗ is positive where the field leaves the surface and negative where it enters, so the law says the two tallies always match. This is not a statement about symmetric geometry or about weak fields: it holds for a surface of any shape, drawn around anything — a magnet, a coil, half a transformer, a patch of empty space in a radio wave.
Lines that cannot end
A field line ends where flux is created. Shrink a closed surface onto a point and ∮B⃗·dA⃗ = 0 says nothing is created there, so magnetic field lines have no ends: each closes on itself or runs off to infinity. That is the entire topology of a dipole. Outside a bar magnet the lines arc from the north end round to the south end; inside the iron they carry on, from south to north, and join up. Test it. Take a bar of 2.0 cm² cross-section carrying B = 0.80 T along its axis, and draw a closed surface that swallows the north half, cutting straight across the middle. The cut face has an outward normal pointing back toward the south end while B⃗ there points the other way, so that face contributes −0.80 × 2.0 × 10⁻⁴ = −1.6 × 10⁻⁴ Wb. The law then fixes the rest with no calculation of the external field at all: the flux leaving through everything outside the iron is +1.6 × 10⁻⁴ Wb. Move the cut, or reshape the surface, and the two numbers change together.
Nothing to cut out
Cut the bar in half at that surface and you do not end up with a north pole in one hand and a south pole in the other. Each piece is a complete magnet with two poles of its own, and the pattern survives all the way down, because the poles were never objects. Every magnetic field we can make comes from a current loop or from an intrinsic magnetic moment — the Bohr magneton that sets its scale is 9.27 × 10⁻²⁴ J T⁻¹ — and both are dipoles. A current loop returns as much field as it sends out, so the divergence of its field is zero exactly, not approximately. In a permanent magnet the aligned atomic moments add to a magnetization whose uncompensated surface currents run around the bar; 'north pole' names the face those lines leave from, not a lump of magnetic charge sitting there. Break the magnet and the surface currents simply reroute around the new faces.
Continuity across a boundary, and flux tubes
Flatten a closed surface into a pillbox straddling the boundary between two materials, faces of area ΔA on either side. Let the thickness go to zero so the rim contributes nothing, and the law leaves (B₂⊥ − B₁⊥)ΔA = 0: the component of B⃗ normal to an interface is continuous, whatever sits on the two sides. Note what does jump. For a linear material H⃗ = B⃗/μ, so crossing from air into iron with μr = 5000 at normal incidence leaves B⊥ untouched and drops H⊥ by a factor of 5000. Apply the same argument to the walls of a tube of field lines and you get B₁A₁ = B₂A₂: pinch the bundle into a smaller cross-section and the field there is stronger in exact proportion. That is the working rule for magnetic circuits. An iron core of cross-section 4.0 cm² at B = 1.2 T carries Φ = 1.2 × 4.0 × 10⁻⁴ = 4.8 × 10⁻⁴ Wb, and across an air gap where fringing spreads that same flux over 4.6 cm² the field is (4.8 × 10⁻⁴)/(4.6 × 10⁻⁴) = 1.04 T, about 13 % lower.
What a zero divergence lets you do
Three consequences you will use constantly. First, the magnetic flux through an open surface depends only on its boundary curve: two surfaces spanning the same loop — a flat disc and a deep bag — together bound a closed volume whose net flux is zero, so their fluxes are equal. That is what makes 'the flux through the circuit' in Faraday's law well defined; otherwise you would have to say which soap film you meant. Second, ∇·B⃗ = 0 everywhere is exactly the condition for B⃗ to be the curl of something, since ∇·(∇ × A⃗) = 0 for any A⃗. The vector potential exists for that reason alone, and the machinery of magnetostatics and radiation is built on it. Third, take a plane wave B⃗ = B⃗₀ cos(k⃗·r⃗ − ωt). Then ∇·B⃗ = −k⃗·B⃗₀ sin(k⃗·r⃗ − ωt), and requiring that to vanish at every position and time forces k⃗·B⃗₀ = 0. The magnetic field of a plane electromagnetic wave has no component along the direction of travel; transversality is not an extra assumption, it is this equation.
The search for the missing charge
The zero is an experimental result, so it is worth knowing how hard it has been pushed. Searches have looked for trapped poles in rock, sediment and lunar samples, for monopoles made in accelerator collisions — MoEDAL sits at the LHC for this purpose — and for a single pole drifting through a superconducting ring, where it would leave a permanent step in the trapped flux. Blas Cabrera recorded one such event in 1982; it never repeated, and no detector has seen another. Cosmic-ray searches now hold any monopole flux below roughly 10⁻¹⁵ per square centimetre per second per steradian. Were one found, the law would read ∮B⃗·dA⃗ = qₘ, enc with magnetic charge measured in webers, ∇·B⃗ = ρₘ would replace the zero, and Faraday's law would gain a magnetic-current term. One reason to keep looking: Dirac showed in 1931 that a single pole of strength g would force every electric charge to be a multiple of one unit, through the condition eg = nh. A pole of h/e = 4.14 × 10⁻¹⁵ Wb would account for the charge quantisation we already observe.
Change one variable at a time
Make the relationship visible.
Drag the side-leak share from 0 to 0.9: the two outward arrows and the two hollow bars trade length, but the hollow pair always spans exactly the shaded bar. That equality is ∮B·dA = 0.
IN THROUGH THE CUT160 μWb
OUT THROUGH THE SIDES88 μWb
OUT PAST THE POLE72 μWb
NET |∮B·dA|0.00 μWb
Live interpretationIN THROUGH THE CUT: 160 μWb. OUT THROUGH THE SIDES: 88 μWb. OUT PAST THE POLE: 72 μWb. NET |∮B·dA|: 0.00 μWb
Catch the common trap
Explain before calculating.
A closed surface encloses the north half of a bar magnet, cutting across the bar at its midpoint. Inside the iron the field is 0.80 T over the 2.0 cm² cross-section, directed toward the north end. What is the net outward magnetic flux through the part of that closed surface lying outside the iron?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA closed surface is drawn around the north half of a bar magnet, slicing straight across the iron. At the cut the field inside the iron is uniform at 0.65 T over the 3.0 cm² cross-section and points along the bar towards the enclosed north end. Find the net outward flux through the whole of the rest of that surface.
- Convert the area: A = 3.0 cm² = 3.0 × 10⁻⁴ m².
- On the cut face the outward normal points out of the enclosed piece, back towards the south end, while B⃗ points the other way, so that face contributes Φ(cut) = −BA = −0.65 T × 3.0 × 10⁻⁴ m² = −1.95 × 10⁻⁴ Wb.
- Gauss's law for magnetism sets the total over the closed surface to zero: Φ(cut) + Φ(rest) = 0.
- Φ(rest) = +1.95 × 10⁻⁴ Wb, outward — fixed without knowing anything about the shape of the field in the air.
AnswerΦ(rest) = +1.95 × 10⁻⁴ Wb outward (195 μWb)
MediumA transformer core of cross-section 6.0 cm² carries a uniform B = 1.35 T. A saw cut leaves a short air gap in which fringing spreads the same flux over 7.5 cm². Find the flux in the core, the mean field in the gap, and the percentage by which the field falls.
- Φ = BA = 1.35 T × 6.0 × 10⁻⁴ m² = 8.1 × 10⁻⁴ Wb.
- Core and gap are two cross-sections of one flux tube. Wrap a closed surface round the tube between them: ∮B⃗·dA⃗ = 0 leaves nothing unaccounted for, so B₁A₁ = B₂A₂ = Φ.
- B(gap) = Φ/A(gap) = 8.1 × 10⁻⁴ Wb ÷ 7.5 × 10⁻⁴ m² = 1.08 T.
- Drop = (1.35 − 1.08)/1.35 = 0.27/1.35 = 0.20, i.e. 20 % — exactly the area ratio 6.0/7.5 read the other way up.
AnswerΦ = 8.1 × 10⁻⁴ Wb; B(gap) = 1.08 T, 20 % below the core field
HardTwo fields are proposed: B⃗₁ = b(x î + y ĵ − 2z k̂) with b = 0.15 T m⁻¹, and B⃗₂ = c(y î + x ĵ + z k̂) with c = 0.20 T m⁻¹. Decide which can be a magnetic field, and for the one that cannot, find the net outward flux it would give through a cube of side 0.10 m centred on the origin.
- The point form of the law is ∇·B⃗ = ∂Bx/∂x + ∂By/∂y + ∂Bz/∂z = 0, and it must hold at every point.
- For B⃗₁: ∂(bx)/∂x + ∂(by)/∂y + ∂(−2bz)/∂z = b + b − 2b = 0 — allowed.
- For B⃗₂: ∂(cy)/∂x + ∂(cx)/∂y + ∂(cz)/∂z = 0 + 0 + c = 0.20 T m⁻¹ ≠ 0 — forbidden.
- The divergence theorem turns that constant into a flux: ∮B⃗₂·dA⃗ = ∫(∇·B⃗₂) dV = cV = 0.20 T m⁻¹ × (0.10 m)³ = 0.20 × 1.0 × 10⁻³ = 2.0 × 10⁻⁴ Wb, and T m⁻¹ × m³ = T m² = Wb.
- A net 2.0 × 10⁻⁴ Wb leaving a closed surface would be 2.0 × 10⁻⁴ Wb of magnetic charge sitting inside it — the thing no search has ever found.
AnswerOnly B⃗₁ is possible; B⃗₂ would drive 2.0 × 10⁻⁴ Wb out of the cube, which ∮B⃗·dA⃗ = 0 forbids