## 2 Basic Formulas  
Let us consider a thin conical sheet and a circular loop as shown in Figure 1. The thin conical sheet has the radii of basis \(R_{1}\) and \(r_{1}\left(R_{1}>r_{1}\right).\) and the axial positions \(z_{1}\) and \(z_{2}\), with the number of sheets turns \(N.\) The circular loop has the radius \(R\) and axial position \(z_{Q}.\)  
Let us begin the complete analysis for the first case \(\left(R_{1}>r_{1}\right)\), Figure 1.  
From Figure 1. one has the following. \[\frac{R_{1}-r_{1}}{z_{2}-z_{1}}=\tan(\alpha)=\eta\,\ \ R_{1}=r_{1}+\eta(z_{2}-z_{1} )\ \ \text{or}\ \ \ r_{1}=R_{1}-\eta(z_{2}-z_{1}) \tag{1}\]  
\[\frac{z_{2}-z}{\rho-r_{1}}=\frac{z_{2}-z_{1}}{R_{1}-r_{1}}=\frac{1}{\eta}\,\ \rho=r_{1}-\eta(z-z_{2})=R_{1}-\eta(z-z_{1}) \tag{2}\]  
The mutual inductance between the thin conical sheet inductor and the circular loop can be calculated by the following. \[M=\frac{\mu_{0}N}{z_{2}-z_{1}}\int\limits_{0}^{\pi}\int\limits_{z_{1}}^{z_{2} }\frac{R\rho\cos(\theta)}{r_{0}}dzd\theta \tag{3}\]  
with  
\[r_{0}=\sqrt{\left(R_{1}-\eta z+\eta z_{1}\right)^{2}-2\left(R_{1}-\eta z+\eta z _{1}\right)R\cos(\theta)+R^{2}+\left(z-z_{Q}\right)^{2}} \tag{4}\]  
Figure 1: Thin conical sheet inductor and circular loop (\(R_{1}>r_{1}\)). Equation (3) is obtained from the Neumann formula for the mutual inductance, derived from the magnetic vector potential [7] which is applicable for two coils. By inspection for \(R_{1}=r_{1}\), Equation (3) becomes [8]. Even though \(R_{1}>r_{1}\) (\(\eta>0\)) one must use \(\rho=R_{1}-\eta(z-z_{1})\). By the simple verification for \(R_{1}=r_{1}\) (\(\eta=0\)) the thin conical sheet degenerates to the thin cylindrical solenoid. For this case, the Formula (3) is the formula for calculating the mutual inductance between the thin cylindrical solenoid and the circular loop [8]. Introducing the substitution, \(\theta=\pi-2\beta\) in (3) one has the following. \[M=-\frac{2\mu_{0}NR}{z_{2}-z_{1}}\int\limits_{0}^{\pi/2}\int\limits_{z_{1}}^{ z_{2}}\frac{(R_{1}-\eta z+\eta z_{1})\cos(2\beta)}{r_{0}}dzd\beta \tag{5}\]  
where  
\[r_{0}=\sqrt{\left(R_{1}-\eta z+\eta z_{1}\right)^{2}+2(R_{1}-\eta z+\eta z_{1 })R\cos(2\beta)+R^{2}+\left(z-z_{Q}\right)^{2}}\]  
Let us calculate the first integral with respect to \(z\) in (5). Figure 1.0  
with  
Figure 1.1  
Figure 1.2  
Figure 1.3  
Figure 1.4  
Figure 1.5  
Figure 1.6  
Figure 1.7  
Using [25],  
\[ \begin{array}{l}I_{1}=\Big{\{}\frac{\sqrt{c_{1}R_{1}^{2}+b_{1}R_{1}+a_{1}}} {\eta^{2}+1}-\frac{\sqrt{c_{1}r_{1}^{2}+b_{1}r_{1}+a_{1}}}{\eta^{2}+1}-\frac{ \eta^{2}R\cos(2\beta)-R_{1}+\eta\big{(}z_{1}-z_{Q}\big{)}}{(\eta^{2}+1)^{3/2}} \times\\ \Big{[}arsinh\frac{\eta R\cos(2\beta)+\eta R_{1}+z_{Q}-z_{1}}{\sqrt{D_{10}}} -arsinh\frac{\eta R\cos(2\beta)+\eta r_{1}+z_{Q}-z_{2}}{\sqrt{D_{10}}}\Big{]} \Big{\}}\end{array}  \tag{8}\]  
Thus, the solution of this integral is obtained in the close form. Now, the Formula (5) can be given as follows. \[M=-\frac{2\mu_{0}NR}{(z_{2}-z_{1})(\eta^{2}+1)^{3/2}}\int\limits_{0}^{\pi/2}V \cos(2\beta)d\beta \tag{9}\]where  
\[V=\sqrt{\eta^{2}+1}\left[\sqrt{c_{1}R_{1}^{2}+b_{1}R_{1}+a_{1}}-\sqrt{c_{1}r_{1}^{2 }+b_{1}r_{1}+a_{1}}\right]-\]  
\[\left[\eta^{2}R\cos(2\beta)-R_{1}+\eta\big{(}z_{Q}-z_{1}\big{)}\right]\times \tag{10}\]  
\[\left[arsinh\frac{\eta R\cos(2\beta)+\eta R_{1}+z_{Q}-z_{1}}{\sqrt{D_{10}}} \right.-\left.arsinh\frac{\eta R\cos(2\beta)+\eta r_{1}+z_{Q}-z_{2}}{\sqrt{D_{1 0}}}\right]\right\}\]  
Let us solve the following integrals in (9) and (10).