Does a reciprocity theorem hold for Ampère's law when exchanging the current loop and the integration path?

Question

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1. What the question is asking (plain language)

We have two closed curves in space:

(C_{1}) – a thin wire that actually carries a steady current (I).
It creates a magnetic field (\mathbf B_{1}(\mathbf r)).
(C_{2}) – an imaginary closed curve that we may draw anywhere.
If we integrate the field (\mathbf B_{1}) round this curve we obtain, by Ampère’s law, [ \oint_{C_{2}}\mathbf B_{1}!\cdot d\boldsymbol\ell_{2}= \mu_{0}\,I_{\text{enclosed by }C_{2}} . \tag{1} ]

Now we swap the roles:

The current is removed from the real wire (C_{1}) and is instead forced to flow along the previously‑imaginary curve (C_{2}) (still the same magnitude (I)).
This produces a new magnetic field (\mathbf B_{2}(\mathbf r)).
We now integrate (\mathbf B_{2}) round the original physical wire (C_{1}): [ \oint_{C_{1}}\mathbf B_{2}!\cdot d\boldsymbol\ell_{1}= \mu_{0}\,I_{\text{enclosed by }C_{1}} . \tag{2} ]

The question is: Are the two line integrals always equal? In symbols, [ \boxed{\;\displaystyle \oint_{C_{2}}\mathbf B_{1}!\cdot d\boldsymbol\ell_{2} = \oint_{C_{1}}\mathbf B_{2}!\cdot d\boldsymbol\ell_{1}\;} \tag{3} ]

In other words, does Ampère’s law possess a “reciprocity” property analogous to the well‑known mutual‑inductance symmetry (M_{12}=M_{21})?

2. Detailed derivation

2.1 Ampère’s law in integral form

For a steady (time‑independent) current distribution in free space, [ \oint_{C}\mathbf B\cdot d\boldsymbol\ell =\mu_{0}\,I_{\text{enc}}(C) . \tag{4} ]

The right‑hand side is the total current that threads any surface (S) whose boundary is the curve (C): [ I_{\text{enc}}(C)=\int_{S}\mathbf J\cdot d\mathbf a . ]

Because (\nabla!\cdot!\mathbf J =0) (steady current), the value of the surface integral is independent of the particular surface as long as the surface has the same edge (C).

Consequently the line integral (4) depends only on how many times the curve (C) winds around the physical current‑carrying wire(s).

2.2 Linking number

Define the linking number (L(C_{1},C_{2})) of two closed curves as the (signed) number of times one curve winds around the other. It can be written as a surface integral:

[ L(C_{1},C_{2})=\frac{1}{4\pi}\oint_{C_{1}}!!\oint_{C_{2}} \frac{(\mathbf r_{1}-\mathbf r_{2})\cdot \bigl(d\mathbf r_{1}\times d\mathbf r_{2}\bigr)} {|\mathbf r_{1}-\mathbf r_{2}|^{3}} . \tag{5} ]

(L) is an integer (or zero) and is unchanged by any smooth deformation of either curve that does not let one pass through the other.

2.3 The line integral produced by a filament current

Suppose a thin filament carries a current (I) along the curve (C_{1}).
Its magnetic field is given by the Biot–Savart law

[ \mathbf B_{1}(\mathbf r)=\frac{\mu_{0} I}{4\pi} \oint_{C_{1}}\frac{d\mathbf r_{1}\times(\mathbf r-\mathbf r_{1})} {|\mathbf r-\mathbf r_{1}|^{3}} . \tag{6} ]

Now evaluate the Ampèrian line integral around an arbitrary second closed curve (C_{2}):

[ \begin{aligned} \oint_{C_{2}}\mathbf B_{1}!\cdot d\boldsymbol\ell_{2} &= \frac{\mu_{0} I}{4\pi} \oint_{C_{2}}!!\oint_{C_{1}} \frac{d\mathbf r_{1}\times(\mathbf r_{2}-\mathbf r_{1})} {|\mathbf r_{2}-\mathbf r_{1}|^{3}} \cdot d\mathbf r_{2} \[4pt] &= \mu_{0} I\; \underbrace{\frac{1}{4\pi}\oint_{C_{2}}!!\oint_{C_{1}} \frac{(\mathbf r_{2}-\mathbf r_{1})\cdot \bigl(d\mathbf r_{2}\times d\mathbf r_{1}\bigr)} {|\mathbf r_{2}-\mathbf r_{1}|^{3}}}{\displaystyle =\,L(C{1},C_{2})} . \end{aligned} \tag{7} ]

Thus

[ \boxed{\displaystyle \oint_{C_{2}}\mathbf B_{1}!\cdot d\boldsymbol\ell_{2}= \mu_{0} I\,L(C_{1},C_{2}) } . \tag{8} ]

Exactly the same algebra, with the roles of the curves interchanged, gives

[ \boxed{\displaystyle \oint_{C_{1}}\mathbf B_{2}!\cdot d\boldsymbol\ell_{1}= \mu_{0} I\,L(C_{2},C_{1}) } . \tag{9} ]

But the linking number is symmetric:

[ L(C_{1},C_{2}) = L(C_{2},C_{1}) . \tag{10} ]

Combining (8)–(10) we obtain the desired reciprocity:

[ \boxed{\displaystyle \oint_{C_{2}}\mathbf B_{1}!\cdot d\boldsymbol\ell_{2} = \oint_{C_{1}}\mathbf B_{2}!\cdot d\boldsymbol\ell_{1} = \mu_{0} I \,L(C_{1},C_{2}) } . \tag{11} ]

2.4 Interpretation in terms of Ampère’s law

Equation (11) tells us that the line integral equals (\mu_{0}I) multiplied by the integer linking number. From the viewpoint of Ampère’s law, the quantity (I_{\text{enclosed}}) in (4) is precisely (\displaystyle I\;L(C_{1},C_{2})). Hence the “enclosed‑current’’ term is the same whichever curve we choose as the integration path, provided the same physical current filament is present.

2.5 When can the equality fail?

Situation	Reason it would break the equality
No linkage ((L=0))	Both integrals are zero, so equality still holds (trivial case).
Time‑varying currents	Ampère’s law must be supplemented by the displacement current term (\displaystyle \mu_{0}\varepsilon_{0}\,\partial\mathbf E/\partial t). The simple form (4) no longer applies; the reciprocity still holds for the magnetostatic part but an extra term appears.
Non‑linear, anisotropic, or inhomogeneous magnetic media	Ampère’s law in the form (\nabla\times\mathbf B = \mu_{0}\mathbf J) is replaced by (\nabla\times\mathbf H = \mathbf J) with (\mathbf B = \mu(\mathbf r)\mathbf H). The line integral of (\mathbf B) now depends on the path because (\mu) varies, and the simple link‑number result is lost.
Current distributed over a surface that is cut by one of the loops	If the “current filament’’ is not a thin wire but a sheet that intersects the integration surface, the notion of a single linking number is ambiguous; the integral equals (\mu_{0}) times the net current that actually threads the surface, which may differ for the two choices.
Topologically non‑trivial space (e.g., a space with a hole that the curves can wind around)	The derivation above uses only the linking number, which is a topological invariant defined in any three‑dimensional manifold. As long as the manifold is oriented and simply‑connected away from the currents, the result still holds. Exotic manifolds that prevent a surface bounded by a given curve from being defined would violate the premises of Ampère’s law.

In ordinary laboratory conditions—steady currents flowing in thin wires in ordinary (free‑space or linear, isotropic) media—the equality always holds.

3. Final answer

Yes.
For any two closed curves (C_{1}) and (C_{2}) in free space, each carrying the same steady current (I) (one at a time), the Ampèrian line integrals satisfy

[ \boxed{\displaystyle \oint_{C_{2}}\mathbf B_{1}!\cdot d\boldsymbol\ell_{2} = \oint_{C_{1}}\mathbf B_{2}!\cdot d\boldsymbol\ell_{1} = \mu_{0}\,I\;L(C_{1},C_{2}) } . ]

The common value is (\mu_{0}I) multiplied by the (signed) linking number of the two loops. Thus the “reciprocity’’ is a direct consequence of Ampère’s law together with the topological invariant linking number; it is equivalent to the well‑known symmetry of mutual inductance (M_{12}=M_{21}).

4. Common mistakes

Mistake	Why it is wrong	How to avoid it
Confusing the line integral of (\mathbf B) with the flux through a surface.	Ampère’s law involves a circulation (\oint\mathbf B!\cdot d\boldsymbol\ell), not (\int\mathbf B!\cdot d\mathbf a). The flux appears in the definition of mutual inductance, not directly here.	Remember that Stokes’ theorem converts the circulation into the surface integral of (\nabla\times\mathbf B), which equals (\mu_{0}\mathbf J).
Assuming the equality holds for arbitrary media.	In a material with spatially varying permeability, (\mathbf B) is not proportional to (\mathbf H) uniformly, so the integral can depend on the path.	Restrict the proof to free space (or a linear, homogeneous, isotropic medium) where (\nabla\times\mathbf B = \mu_{0}\mathbf J) holds.
Neglecting the sign (orientation) of the linking number.	The direction in which each loop is traversed determines the sign of the integral; swapping the orientation flips the sign.	Keep track of the right‑hand rule: the normal to the surface bounded by (C) follows the direction of (d\boldsymbol\ell) by the right‑hand rule.
Using the result for time‑varying currents without the displacement current term.	Ampère’s law in its static form omits (\mu_{0}\varepsilon_{0}\partial\mathbf E/\partial t); when fields vary, the extra term contributes.	State explicitly that the currents are steady (magnetostatic).
Thinking that a non‑linked pair of loops must give a non‑zero integral.	If the

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