# Find magnetic flux linkage of a planar coil

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 $$A$$:

$$\psi=\Phi=\oint_lA\cdot dl$$

### 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:

$$\psi=\oint_{C_1}A\cdot dC_1 + \oint_{C_2}A\cdot dC_2 + \oint_{C_3}A\cdot dC_3 + \oint_{C_4}A\cdot dC_4$$ , where $$A=\sum^4_{n=1}\oint_{C_n}\frac{\mu_0IdC_n}{4\pi R}$$?

• hint: both $A$ and $C_k; k=1..4$ are vectors Commented Oct 30, 2020 at 10:07
• Yep. So you mean that I need to be aware of length addition rather than vector addition for the path integrals? In other words, I need to make it in the scalar form first and then add them together? Commented Oct 30, 2020 at 16:23
• no. if $A$ and $dC$ are "vectors" then the angle $\alpha_k$ between them is relevant: $\mathbf{A} \cdot d\mathbf{C_k} = AC_k cos(\alpha)$; $\alpha_k$ is the angle between $A$ and the tangent of the curve $C_k$. Anyhow, your formula for the vector potential $A=\sum_{n=1}^{4}....$ is wrong. Commented Oct 30, 2020 at 16:41
• @hyportnex I do not think $\mathbf{A}$ is wrong. If we have a look at the Magnetic Field $\mathbf{B}$ of each section, they can be added. $\mathbf{B}$ is the curl of $\mathbf{A}$. According to $\nabla \times (\mathbf{a}+\mathbf{b})=\nabla\times\mathbf{a}+\nabla\times\mathbf{b}$, $\mathbf{A}_{total}=\mathbf{A}_1+\mathbf{A}_2+\cdots$. In terms of vector line integral, piecewise smooth curve can be seen as the sum of each piece of curve said here. Commented Nov 22, 2020 at 20:26
• Your "$R$" in the denominator should depend on the index $n$ and the equation should be a vector one. Commented Nov 22, 2020 at 20:34

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 $$R_n$$.
For magnetic potential vector $$\mathbf{A}$$ to a point in space $$(x, y, z)$$: $$\mathbf{A}=\oint_C\frac{\mu I}{4\pi R}d\mathbf{C}=\sum^4_{n=1}\int_{C_n}\frac{\mu I}{4\pi R_n}d\mathbf{C}_n$$ , 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. $$\psi=\sum^4_{n=1}\int_{C'_n}\mathbf{A}\cdot d\mathbf{C'}_n$$ , where $$C'$$ is the inner line if the width of the track takes account.