Find magnetic flux linkage of a planar coil

Question

I have got a planar spiral shown in the graph. C1 is the spiral body, and C2, C3 and C4 are the wires to make it a loop, where C2 is under the coil body.

Assume I have known the geometric expressions of C1, C2, C3 and C4 in space.
For a single closed loop $l$, the magnetic flux linkage $psi$ is equal to the magnetic flux $Phi$ in terms of Magnetic vector potential $A$:
$$
psi=Phi=oint_lAcdot dl
$$
Question:
Now let's ignore the effect of average and inner lines of the pattern. So for the coil, can I tell that the magnetic flux linkage $psi$ is:
$$
psi=oint_{C_1}Acdot dC_1 + oint_{C_2}Acdot dC_2 + oint_{C_3}Acdot dC_3 + oint_{C_4}Acdot dC_4
$$
, where $A=sum^4_{n=1}oint_{C_n}frac{mu_0IdC_n}{4pi R}$?

ONLYA · Accepted Answer

Thanks for the help from @hyportnex.
It is noted that $R$ is the distance from the line to a point in space. Its expression is different for different curve piece so there should an index applied as $R_n$.
According to the property of line integral over a vector field, the line integral of the union of the curve is the sum of the line integral of each curve piece.
For magnetic potential vector $mathbf{A}$ to a point in space $(x, y, z)$:
$$
mathbf{A}=oint_Cfrac{mu I}{4pi R}dmathbf{C}=sum^4_{n=1}int_{C_n}frac{mu I}{4pi R_n}dmathbf{C}_n
$$
, where $R_n=|(x, y, z)-mathbf{C}_n|$ and $mathbf{C}_n$ can be parameterised with $t$.
The same applies to the magnetic flux linkage.
$$
psi=sum^4_{n=1}int_{C'_n}mathbf{A}cdot dmathbf{C'}_n
$$
, where $C'$ is the inner line if the width of the track takes account.

Find magnetic flux linkage of a planar coil

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