advanced30 min studyLesson 244 in Physics

Magnetic boundary conditions and surface currents

Derive magnetic jumps with an explicit interface normal and distinguish free surface current from magnetization current.

Before you begin

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The question for today

What you’ll understand

Derive normal B continuity and the tangential H jump.
Use a declared normal to determine the sign of a surface current.
Account separately for free and bound current at a magnetic interface.

An interface can bend magnetic field lines and create a tangential field jump without creating magnetic monopoles. To find the jump, first choose the normal and distinguish the magnetic flux density from the auxiliary magnetic field. The normal component of B follows one Maxwell equation, while the tangential jump of H follows another. Treating them as interchangeable vectors gives incorrect refraction and frequently reverses the inferred surface-current direction.

A pillbox and a narrow loop measure different components

Let the unit normal point from medium 1 into medium 2. Integrating zero magnetic divergence over a pillbox makes its side flux vanish as the thickness shrinks, leaving equal normal B on its two faces. This condition does not require equal permeabilities and does not force the tangential fields to agree. It assumes ordinary finite bulk fields at the interface; a singular sheet requires its own properly interpreted source conditions.

An Ampère loop straddling the same interface probes the tangential H jump. In magnetostatics its enclosed free current survives as a surface current density measured in amperes per metre. The normal cross the difference H2 minus H1 equals that free sheet current. Derive the sign from a right-handed loop or evaluate the cross product directly. Reversing the normal also exchanges the medium labels, leaving the physical statement unchanged.

\hat{\mathbf n}\cdot(\mathbf B_2-\mathbf B_1)=0,\qquad \hat{\mathbf n}\times(\mathbf H_2-\mathbf H_1)=\mathbf K_f

The normal points from side 1 to side 2; B is in T, H in A/m, and free sheet current Kf in A/m. These are magnetostatic interface conditions, or their quasistatic limit without a singular displacement-current sheet.↔ Scroll wide equations horizontally.

Material response changes what continuity means

In SI the definition $\mathbf B=\mu_0(\mathbf H+\mathbf M)$ separates magnetization from the free-current equation. For a linear isotropic material, B equals permeability times H. If there is no free sheet current, tangential H is continuous, but tangential B generally changes when permeability changes. Normal B remains continuous while normal H can change. The frequently quoted statement that the magnetic field is continuous therefore needs both a specified vector and a specified component.

Magnetization currents account for the difference between applying Ampère's law to B and to H. For a single magnetic body with outward normal, the bound sheet current is magnetization crossed with that outward normal. At a two-material interface, contributions from both sides combine. This is a real source representation of the averaged magnetization; free and bound describe how sources are organized in the model, rather than whether their fields obey different force laws.

\mathbf K_b=(\mathbf M_1-\mathbf M_2)\times\hat{\mathbf n},\qquad \hat{\mathbf n}\times\frac{\mathbf B_2-\mathbf B_1}{\mu_0}=\mathbf K_f+\mathbf K_b

M and both sheet currents have units A/m. The formula uses the same 1-to-2 normal. It follows from summing each body's M crossed with its own outward normal; using only one side's magnetization can miss part of the current.↔ Scroll wide equations horizontally.

Boundary equations constrain a global solution

An ideal infinite current sheet in vacuum has oppositely directed tangential fields on its two sides when reflection symmetry rules out an added background field. The jump equation alone determines their difference, not their mean. An externally applied uniform field can be added to both sides without changing the sheet current. This is a useful warning that local boundary conditions do not by themselves determine an entire apparatus field.

For an interface with no free surface current, define the field angle from the normal. Continuity gives the same normal B and equal tangential B divided by permeability, so the tangent of the angle scales with permeability. This bending is magnetostatic flux redistribution, not optical propagation or a rule about wave impedance. The material law and distant source geometry must still be compatible with the assumed fields on both sides.

Work it through

A signed sheet-current calculation

At z=0 take medium 1 below and medium 2 above, with $\hat{\mathbf n}=\hat{\mathbf z}$ , $\mu_1=\mu_0$ , and $\mu_2=2\mu_0$ . Given $\mathbf H_1=5\hat{\mathbf y}+2\hat{\mathbf z}$ A/m and $\mathbf K_f=3\hat{\mathbf x}$ A/m, find H2 and the bound sheet current. Assume no x component.

Normal B continuity gives $\mu_0(2)=2\mu_0 H_{2z}$ , so $H_{2z}=1$ A/m. Equal normal B does not mean equal normal H when the permeabilities differ.
Since $\hat{\mathbf z}\times\hat{\mathbf y}=-\hat{\mathbf x}$ , the free-current condition is $-(H_{2y}-5)=3$ . Hence $H_{2y}=2$ A/m. The sign comes from the declared normal, not from assuming that every component must grow with permeability.
Here M1 is zero and $\mathbf M_2=\mathbf H_2$ because the second relative permeability is two. Thus $\mathbf K_b=(-2\hat{\mathbf y}-\hat{\mathbf z})\times\hat{\mathbf z}=-2\hat{\mathbf x}$ A/m. The total sheet current is +1 A/m along x, agreeing with the jump of tangential B divided by μ₀.

The result

The upper field is $\mathbf H_2=2\hat{\mathbf y}+\hat{\mathbf z}$ A/m. Bound current is −2 A/m along x, partially cancelling the +3 A/m free sheet current in the B jump.

The living notebook

Explore the model

Interactive lab

Interactive lab · vectors

Resolve normal and tangential components

Take the x projection as a normal component. Reset and set both controls to Angle 60° and Magnitude 2; predict normal 1 and tangential approximately 1.73. For a fresh comparison, Reset and set both controls to Angle 0° and Magnitude 1; predict the same normal component and zero tangential component. The exercise isolates which projection a pillbox constrains; it does not impose a material boundary law.

Vectorx / y projectionsTrace marker

x component: 1.393
y component: 0.975
Magnitude: 1.7

Angle35°Magnitude1.7

t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forEqual normal components can coexist with unequal tangential components. The vector lab contains no interface, permeability control, or free/bound-current readout.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The pillbox constrains normal B, while the Ampère loop constrains tangential H when Kf is zero. The other two components depend on permeability and magnetization. Bound surface currents can therefore coexist with continuous tangential H.

Your turn

Try it yourself

A planar interface with no free surface current separates relative permeabilities 1 and 4. In medium 1 the magnetic flux density has normal component 1.0 mT and tangential component 2.0 mT. Find those components in medium 2 and the change of field angle from the normal.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Keep normal B fixed, but match tangential H by dividing each tangential B by its own permeability.

Compare with the solution

Normal B remains 1.0 mT. Tangential continuity gives $B_{2t}/(4\mu_0)=(2.0\,\mathrm{mT})/\mu_0$ , so $B_{2t}=8.0$ mT. The first angle is $\arctan2\approx63.4^\circ$ and the second is $\arctan8\approx82.9^\circ$ . The field becomes more tangential in the higher-permeability medium for this locally compatible configuration. Neither angle follows from matching the full B vector across the interface.

Keep the idea

Take it with you

Declare a 1-to-2 normal before assigning any magnetic jump sign.
Normal B is continuous; tangential H jumps by the free sheet current.
The B jump contains both free current and the bound current from the magnetization difference.

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