## 2 System Design and Modeling
First, consider only the single-turn excitation coil in region 0, ignoring the effects of the receiving coil, Hall sensor, and metallic foreign objects. The boundary conditions at \(x=x_{0}\) for regions I and II can be obtained as:  
\[\widetilde{A}_{\mathrm{H}}^{\mathrm{I}}(\lambda_{i},x_{0})=\widetilde{A}_{ \mathrm{H}}^{\mathrm{II}}(\lambda_{i},x_{0}) \tag{6}\]  
\[\frac{\partial\widetilde{A}_{\mathrm{H}}^{\mathrm{I}}}{\partial x}\bigg{|}_{x =x_{0}}=\left.\frac{\partial\widetilde{A}_{\mathrm{H}}^{\mathrm{II}}}{ \partial x}\right|_{x=x_{0}}-\mu_{0}\widetilde{I}r_{0}\mathrm{J}_{1}(\lambda_ {i}r_{0}) \tag{7}\]  
Solving the equation, the excitation field coefficient is obtained as:  
\[C_{S}=C_{\mathrm{II}}=\frac{\mu_{0}\widetilde{I}}{2\lambda_{i}}r_{0}\mathrm{J }_{1}(\lambda_{i}r_{0})e^{-\lambda_{i}x_{0}} \tag{8}\]  
Next, merging regions I and II into region 0, and adding the effect of the metallic medium, the boundary conditions at the interface \(x=x_{k}\) (\(k=1\), 2) between regions \(k-1\) and \(k\) can be obtained as:  
\[\widetilde{A}_{\mathrm{H}\mathrm{k-1}}(\lambda_{i},x_{k})=\widetilde{A}_{ \mathrm{H}\mathrm{k}}(\lambda_{i},x_{k}) \tag{9}\]  
\[\frac{\partial A_{\varphi}}{\partial r}+\frac{\partial^{2}A_{\varphi}}{\partial x ^{2}}-\frac{A_{\varphi}}{r^{2}}-\mu\sigma\frac{\partial A_{\varphi}}{\partial t }+\mu I\delta(r-r_{0})\delta(x-x_{0})=0 \tag{1}\]0Solving the equation, the reflection field coefficient is obtained as:  
\[\frac{\partial A_{\varphi}}{\partial r}+\frac{\partial^{2}A_{\varphi}}{\partial x ^{2}}-\frac{A_{\varphi}}{r^{2}}-\mu\sigma\frac{\partial A_{\varphi}}{\partial t }+\mu I\delta(r-r_{0})\delta(x-x_{0})=0 \tag{1}\]1  
where \(\widetilde{R}=\frac{e^{2\gamma_{1}t_{2}}(\gamma_{1}\mu_{2}+\gamma_{2}\mu_{1})( \lambda_{i}\mu_{1}-\gamma_{1})+(\gamma_{1}\mu_{2}-\gamma_{2}\mu_{1})(\lambda_{ i}\mu_{1}+\gamma_{1})}{e^{2\gamma_{1}t_{2}}(\gamma_{1}\mu_{2}+\gamma_{2}\mu_{1})( \lambda_{i}\mu_{1}+\gamma_{1})+(\gamma_{1}\mu_{2}-\gamma_{2}\mu_{1})(\lambda_{ i}\mu_{1}-\gamma_{1})}\). Thus, we obtain \(\widetilde{A}_{\rm HS}(\lambda_{i},x)=\sum\limits_{i=1}^{\infty}C_{g}e^{ \lambda_{i}x}\), \(\widetilde{A}_{\rm Href}(\lambda_{i},x)=\sum\limits_{i=1}^{\infty}D_{ref}e^{- \lambda_{i}x}\). Taking \(\lambda_{i}\) as the independent variable, the inverse Hankel transform is performed on \(\widetilde{A}_{\rm HS}(\lambda_{i},x)\) and \(\widetilde{A}_{\rm Href}(\lambda_{i},x)\) over the interval [0, h], yielding:  
\[\frac{\partial A_{\varphi}}{\partial r}+\frac{\partial^{2}A_{\varphi}}{\partial x ^{2}}-\frac{A_{\varphi}}{r^{2}}-\mu\sigma\frac{\partial A_{\varphi}}{\partial t }+\mu I\delta(r-r_{0})\delta(x-x_{0})=0 \tag{1}\]2  
\[\frac{\partial A_{\varphi}}{\partial r}+\frac{\partial^{2}A_{\varphi}}{\partial x ^{2}}-\frac{A_{\varphi}}{r^{2}}-\mu\sigma\frac{\partial A_{\varphi}}{\partial t }+\mu I\delta(r-r_{0})\delta(x-x_{0})=0 \tag{1}\]3  
Next, by considering the rectangular cross-section coaxial excitation and receiving coils with Hall sensors, we establish the truncated region eddy current field model as shown in Figure 3b. The current \(\widetilde{I}\) in the single-turn excitation coil is replaced with the current density in the excitation coil's rectangular cross-section, denoted as \(\widetilde{i}_{0}\), satisfying the condition \(\widetilde{i}_{0}=N_{1}\widetilde{I}/(r_{2d}-r_{1d})(x_{2d}-x_{1d})\). Since the radial dimension of the Hall sensor is small, its magnetic induction intensity in the \(z\)-axis direction can be neglected, and only its magnetic induction intensity in the \(x\)-axis direction is considered.