# Optimum Geometries for Wireless Power Transfer Between an Oblong Coil and a Rectangular Coil
## 3 Theorems
In order to compute the induced voltage on \(s\), [PERSON]'s equation of Faraday's Law can be used to express the induced voltage as the time derivative of the magnetic flux traversing coil \(s\):  
\[V_{ind}=\oint\limits_{s}\stackrel{{\rightarrow}}{{E}}\cdot \stackrel{{\rightarrow}}{{ds}}=-\frac{d}{dt}\int\limits_{A} \stackrel{{\rightarrow}}{{B}}\cdot\stackrel{{ \rightarrow}}{{dA}}, \tag{5}\]  
where \(\stackrel{{\rightarrow}}{{E}}\) is the electric field induced on the path corresponding to the thin wire coil \(s\), \(\stackrel{{\rightarrow}}{{ds}}\) is the differential element of the path of s, \(\stackrel{{\rightarrow}}{{A}}\) is the oriented rectangular surface enclosed by \(s\), and \(\stackrel{{\rightarrow}}{{dA}}\) is its oriented differential area element. Expression (5) can be rewritten using phasors, and considering the existing planar symmetry of \(\stackrel{{\rightarrow}}{{B}}\) due to the infinite length of \(p\), leading to  
\[\dot{V}_{ind}=-\frac{d}{dt}\int\limits_{-\frac{d}{2}}^{\frac{d}{2}}\dot{B}_{z }\;\ell\;dv \tag{6}\]  
Considering no relative movement of the coil \(s\) (quasi-stationary approximation) with respect to the wire \(p\), the magnetic flux on \(s\) is only a function of the time, so  
\[\dot{V}_{ind}=-\int\limits_{-\frac{d}{2}}^{\frac{d}{2}}\frac{d}{dt}\dot{B}_{z }\;\ell\;dv=-j\omega\epsilon\int\limits_{-\frac{d}{2}}^{\frac{d}{2}}\dot{B}_{z }\;dv \tag{7}\]  
For an infinitely long thin wire \(p\), the magnetic field \(\stackrel{{\rightarrow}}{{B}}\) on any point of space is perpendicular to \(p\), and thus contained in the plane parallel to the cross-section, being tangential to a circle of radius \(r\) centered at P passing on that point of space, with complex magnitude given by  
\[\dot{B}=\frac{\mu\dot{I}_{p}}{2\pi r} \tag{8}\]  
It can be observed that  
\[\dot{B}_{z}=\dot{B}\text{cos}\,\theta=\frac{\mu\;\dot{I}_{p}}{2\pi r}\;\frac{( v-u)}{r}=\frac{\mu\;\dot{I}_{p}}{2\pi}\;\frac{(v-u)}{h^{2}+(v-u)^{2}} \tag{9}\]  
Combining (7) and (9), and considering that \(\omega=2\pi f_{0}\),  
\[\dot{V}_{ind}=-j\;\mu\;f_{0}\;\dot{I}_{p}\;\epsilon\int\limits_{-\frac{d}{2}}^ {\frac{d}{2}}\frac{(v-u)}{h^{2}+(v-u)^{2}}dv \tag{10}\]  
The integral in (10) can be calculated by substitution, using \(r\) again:  
\[r =\sqrt{h^{2}+(v-u)^{2}} \tag{11}\] \[\Rightarrow r\frac{dr}{dv}=(v-u) \tag{12}\]Figure 2:0  
which can be integrated as  
Figure 2:1  
which is an equivalent expression to (4). \(\square\)  
In stationary IWPT (and, by approximation, in quasi-stationary DIWPT), the problem of calculating the induced voltage is equivalent to the calculation of the mutual inductance between the rectangular coil and the infinitely long parallel wire. A trivial case of Lemma 1, where \(h=0\), often referred in textbooks [33], is illustrated in Figure 2:2  
In this case, the magnetic flux traversing coil \(s\) (into the paper), \(\Phi_{s}\), is given by [33]  
Figure 2:3  
leading to the calculation of \(\dot{V}_{ind}\):  
Figure 2:4  
which is consistent with (2) when \(h=0\). The obtained expression for \(\dot{V}_{ind}\) in Lemma 1 (2) is implicitly associated with the reference polarity of the terminals of coil \(s\), where the right-hand rule is used for the closed curve integration of the induced electric field on \(s\). Considering the geometry in Figure 2, the real and imaginary parts of current \(\dot{I}_{p}\) are positive when coming out of paper, and \(\dot{V}_{ind}\) expresses the differential electric potential \(V_{1}-V_{2}\), between the left (\(S_{1}\)) and right (\(S_{2}\)) terminals of \(s\), seen from the front of the paper, when \(s\) is in open-circuit condition. If the polarity convention is reverse direction, that is, measuring \(V_{1}-V_{2}\), \(V_{ind}\) would alternatively be given by the phasor \(\dot{V}_{-ind}\):  
Figure 2:5  
Figure 2:6 Trivial case of Lemma 1 with \(h=0\).