## 3 Numerical Validation
: _In this example one calculates the mutual inductance between the thin cylindrical solenoid and the circular loop for which \(R_{1}=r_{1}=2\;m\), \(z_{1}=0\;m\), \(z_{2}=0.2\;m\), \(N=1000\), \(R=1\;m\), and \(z_{Q}=0.2\;m\)._  
In this example, the thin conical sheet inductor degenerates to the thin cylindrical solenoid (\(\eta=0\)). The exact formula for calculating the mutual inductance between the thin cylindrical solenoid and the circular loop is given by (53) with (54) as follows [8]. \[M(Single)=7.401731104798464\ {\rm mH}\]2  
Using the formulas for the double and the single integration either for the case \(R_{1}>r_{1}\) or \(R_{1}<r_{1}\) the mutual inductance is, respectively, the following. \[M(Single)=7.401731104798464\ {\rm mH}\]3  
\[M(Single)=7.401731104798464\ {\rm mH}\]4  
Equations (5) and (38) for the double integration and (9) with (10) as well as (39) with (40) for the single integration are not singular for \(\eta=0\). It is not case for the semi-analytical solutions (22) with (23) or (41) with (42) when they are singular or indeterminate. However, one can take, for example, \(R_{1}\) = 2.0000001 m and \(r_{1}\) = 2.0000001 m Equations (23) and (42), respectively, so that they give the following. \[M(Single)=7.401731104798464\ {\rm mH}\]5  
All figures that agree with the exact results are bolded. Even though the results are in the particularly good agreement it is recommended to use the formula (53) with (54) for calculating the mutual inductance between the thin cylindrical solenoid and the circular loop. This formula can be carefully obtained from (23) and (42) in the limit case when \(R_{1}\to r_{1}\) and vice-versa. **Example 6**. : _Calculate the mutual inductance between the thin conical sheet inductor and the circular loop for which \(R_{1}\) = 3 m, \(r_{1}\) = 2 m, \(z_{1}\) = 0 m, \(z_{2}\) = 0.2 m, \(N\) = 1000, \(R\) = 1 m, and \(z_{Q}\) = 0.4 m._  
In this example one finds that \(p_{1}=p_{3}=-1\), so that it is the singular case Section 2.1. Applying (26), the mutual inductance is the following. \[M(Single)=7.401731104798464\ {\rm mH}\]6  
Using the basic Formula (5) for the double integration the mutual inductance is the following. \[M(Single)=7.401731104798464\ {\rm mH}\]7  
Using the Formula (9) with (10), the mutual inductance is obtained by the single integration. \[M(Single)=7.401731104798464\ {\rm mH}\]8  
All figures, that agree, are in bold. All results are in an excellent agreement. **Example 7**. : _Calculate the mutual inductance between the thin conical sheet and the circular loop for which \(R_{1}\) = 3 m, \(r_{1}\) = 2 m, \(z_{1}\) = 0 m, \(z_{2}\) = 0.2 m, \(N\) = 1000, \(R\) = 1 m, and \(z_{Q}\) = 0.8 m._  
In this example one finds that \(p_{1}=p_{3}=1\), so that it is the singular case Section 2.2. Applying (26), the mutual inductance is the following. \[M(Single)=7.401731104798464\ {\rm mH}\]9  
Using the basic Formula (5) for the double integration the mutual inductance is the following. \[M(Semi-analytical)=7.401731104798464\ {\rm mH}\]0  
Using the Formula (9) with (10) the mutual inductance is obtained by the single integration. \[M(Semi-analytical)=7.401731104798464\ {\rm mH}\]1  
All results are in an excellent agreement. The figures that agree are in bold. All presented results have been obtained by the Mathematica programing. One can see that they are in an excellent agreement so that the potential users can use the presented formulas by their preference.