## 3 Electromagnetic Field (EMF) Distribution in Region \(R_{ii}\)  
In region \(R_{II}\), the system of Equation (4) can be reduced to the following equation for the magnetic flux density:  
\[\frac{1}{r}\frac{\partial}{\partial r}\Bigg{(}r\frac{\partial H_{n,\varphi}^{ (II)}}{\partial r}\Bigg{)}+\frac{\partial^{2}H_{n,\varphi}^{(II)}}{\partial z ^{2}}-\frac{H_{n,\varphi}^{(II)}}{r^{2}}-jn\omega\mu\sigma H_{n,\varphi}^{( II)}=0. \tag{10}\]Equation (10) can be solved using the separation of variables technique [24]. As a result, the following expression is obtained:  
\[H_{n,\varphi}^{(II)}(r,z)=\Big{(}\frac{A_{n}}{r}+B_{n}r\Big{)}\cosh(\beta_{n}z)+ \sum\limits_{m=0}^{\infty}\{C_{n,m}I_{1}(\lambda_{n,m}r)+D_{n,m}K_{1}(\lambda_ {n,m}r)\}\cos(k_{m}z), \tag{11}\]  
where \(I_{1}(\lambda_{n,m}r)\) and \(K_{1}(\lambda_{n,m}r)\) are the modified Bessel functions of the first order, \(\beta_{n}^{2}=jn\omega\mu\), \(k_{m}=\frac{\pi(2m+1)}{h}\), and \(\lambda_{n,m}^{2}=k_{m}^{2}+\beta_{n}^{2}\), where \(m=0,\ 1,\ 2,\ldots\). In order to find the unknown constants \(C_{n,m}\) and \(D_{n,m}\), boundary conditions (6) and (7) need to be applied. Therefore, we get:  
\[\left\{\begin{array}{ll}\beta_{n}\Big{(}\frac{A_{n}}{a}+B_{n}a\Big{)}\sinh( \beta_{n}z)-\sum\limits_{m=0}^{\infty}k_{m}[C_{n,m}I_{1}(\lambda_{n,m}a)+D_{n,m}K_{1}(\lambda_{n,m}a)]\sin(k_{m}z)=0,\\ \beta_{n}\Big{(}\frac{A_{n}}{b}+B_{n}b\Big{)}\sinh(\beta_{n}z)-\sum\limits_{m= 0}^{\infty}k_{m}[C_{n,m}I_{1}(\lambda_{n,m}b)+D_{n,m}K_{1}(\lambda_{n,m}b)] \sin(k_{m}z)=0,\end{array}\right. \tag{12}\]  
where \(\sinh(\beta_{n}z)\) can be expanded in a Fourier series within the segment \(z\in\left[-\frac{h}{2},\frac{h}{2}\right]\) as follows (see [9]):  
\[\sinh(\beta_{n}z)=\sum\limits_{m=0}^{\infty}\frac{4\beta_{n}(-1)^{m}}{h \lambda_{n,m}^{2}}\cosh(\frac{\beta_{n}h}{2})\sin(k_{m}z). \tag{13}\]  
The substitution of (13) into the system of Equation (12) yields:  
\[\left\{\begin{array}{ll}k_{m}[C_{n,m}I_{1}(\lambda_{n,m}a)+D_{n,m}K_{1}( \lambda_{n,m}a)]=\beta_{n}\Big{(}\frac{A_{n}}{a}+B_{n}a\Big{)}\frac{4\beta_{n} (-1)^{m}}{h\lambda_{n,m}^{2}}\cosh(\frac{\beta_{n}h}{2}),\\ k_{m}[C_{n,m}I_{1}(\lambda_{n,m}b)+D_{n,m}K_{1}(\lambda_{n,m}b)]=\beta_{n}\Big{(} \frac{A_{n}}{b}+B_{n}b\Big{)}\frac{4\beta_{n}(-1)^{m}}{h\lambda_{n,m}^{2}} \cosh(\frac{\beta_{n}h}{2}).\end{array}\right.\]  
The solution to this system of equations (with respect to the constants \(C_{n,m}\) and \(D_{n,m}\)) should be substituted into Equation (9), resulting in the following solution to Equation (10):  
\[\begin{array}{ll}H_{n,\varphi}^{(II)}(r,z)&=\Big{(}\frac{A_{n}}{r}+B_{n}r \Big{)}\cosh(\beta_{n}z)\\ &+\sum\limits_{m=0}^{\infty}\frac{4\beta_{n}^{2}(-1)^{m}}{h\lambda_{n,m}^{2} }\frac{\cosh(\frac{\beta_{n}h}{2})\cos(k_{m}z)}{l_{1}(\lambda_{n,m}a)K_{1}( \lambda_{n,m}b)-I_{1}(\lambda_{n,m}b)K_{1}(\lambda_{n,m}a)}\\ &\times\Big{(}\frac{A_{n}}{a}+B_{n}a\Big{)}\big{(}I_{1}(\lambda_{n,m}r)K_{1}( \lambda_{n,m}b)-I_{1}(\lambda_{n,m}b)K_{1}(\lambda_{n,m}r)\big{)}\\ &+\Big{(}\frac{A_{n}}{b}+B_{n}b\Big{)}\big{(}I_{1}(\lambda_{n,m}a)K_{1}( \lambda_{n,m}r)-I_{1}(\lambda_{n,m}r)K_{1}(\lambda_{n,m}a)\big{)}\Big{\}},\end{array} \tag{14}\]  
where constants \(A_{n}\) and \(B_{n}\) are to be found in the next section.