## 3 Numerical Validation  
To verify the validity of the new presented formula for calculating, the following set of examples is presented. In each calculation one begins with the exact basic formula in the form of the double integral. The verification of this integral is verified with the expressions after the first and second integration. Thus, these analytical and semi-analytical expressions validate the presented new formulas. The special case is when the given configuration, thin conical sheet coil and the circular loop degenerates to the combination, thin cylindrical solenoid, and the circular loop. **Example 1**. : _Calculate the mutual inductance between the thin conical sheet inductor and the circular loop for which \(R_{1}\) = 10 m, \(r_{1}\) = 2 m, \(z_{1}\) =\(-\)1 m, \(z_{2}\) = 2 m, \(N\) = 1000, \(R\) = 5 m, and \(z_{Q}\) = 0 m._  
This is the case \(R_{1}>r_{1}\) (Figure 1). Let us begin with the basic Formula (5) where the mutual inductance is given by the double integration. The mutual inductance is the following. \[M(Double)=7.401731104798464\ {\rm mH}\]  
Using Formula (9) with (10), the mutual inductance is obtained by the single integration. \[M(Single)=7.401731104798464\ {\rm mH}\]  
Let us use the semi-analytical Formula (22) with (23). \[M(Semi-analytical)=7.401731104798464\ {\rm mH}\]  
All results are in an excellent agreement. **Example 2**. : _Calculate the mutual inductance between the thin conical sheet inductor and the circular loop for which \(R_{1}\) = 2 m, \(r_{1}\) = 10 m, \(z_{1}\) = \(-\)1 m, \(z_{2}\) = 2 m, \(N\) = 1000, \(R\) = 5 m, and \(z_{Q}\) = 0 m._  
This is the case \(R_{1}<r_{1}\) (Figure 2). Using the basic Formula (38) for the double integration, the mutual inductance is the following. \[M(Double)=8.607861541512988\ \ {\rm mH}\]  
Using Formula (39) with (40), the mutual inductance is obtained by the single integration. \[M(Single)=8.607861541512988\ \ {\rm mH}\]  
Finally, let us use the semi-analytical Formula (41) with (42) the mutual inductance as follows. \[M(Semi-analytical)=8.607861541512988\ \ {\rm mH}\]  
All results are in an excellent agreement. **Example 3**. : _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 \(z_{Q}\) = 0.2 m._This is the case \(R_{1}>r_{1}\) (Figure 1). The basic formula for the mutual inductance is given by the double integration (5). The mutual inductance is the following. \[M(Double)=0.8559140919190895\;\mathrm{mH}\]  
Using Formula (9) with (10) the mutual inductance is obtained by the single integration. \[M(Single)=0.8559140919190892\;\mathrm{mH}\]  
Finally, let us use the semi-analytical Formula (22) with (23) for the mutual inductance which gives the following. \[M(Semi-analytical)=0.8559140919190874\;\mathrm{mH}\]  
All results are in an excellent agreement. The figures that agree are in bold. **Example 4**. : _Calculate the mutual inductance between the thin conical sheet inductor and the circular loop for which \(R_{1}=2\;m\), \(r_{1}=3\;m\), \(z_{1}=0\;m\), \(z_{2}=0.2\;m\), \(N=1000\), \(R=1\;m\), and \(z_{Q}=0.2\;m\)._  
This is the case \(r_{1}>R_{1}\) (Figure 2). The basic formula for the mutual inductance is given by the double integration (38) which gives the following. \[M(Double)=0.8529571443235432\;\mathrm{mH}\]  
Using Formula (39) with (40), the self-inductance is obtained by the single integration. \[M(Single)=7.401731104798464\ {\rm mH}\]0  
Let us use the semi-analytical Formula (41) with (42). \[M(Single)=7.401731104798464\ {\rm mH}\]1  
All results are in an excellent agreement. The figures that agree are in bold. **Example 5**.