## 5 Electric Field (EF) and Eddy Current (EC) Losses in the Tank Wall  
Due to the Ohmic nature of EC losses, the average power density of losses can be expressed in the form:  
\[P(\mathbf{r})=\frac{1}{T}\int\limits_{0}^{T}\sigma\mathbf{E}^{2}(\mathbf{r},t )dt, \tag{23}\]  
where \(\mathbf{E}(\mathbf{r},t)\) is the EF in region \(R_{II}\), which can be obtained from [PERSON]'s Equation (4) in the form:  
\[\mathbf{E}=\frac{\
abla\times H}{\sigma}=\sum\limits_{n=-\infty}^{+\infty} \mathbf{E}_{n}e^{jn\alpha t}=\sum\limits_{n=-\infty}^{+\infty}(\mathbf{e}_{r} E_{n,r}(r,z)+\mathbf{e}_{z}E_{n,z}(r,z))e^{jn\alpha t}. \tag{24}\]  
The substitution of (24) into (23) and accurate calculation of the respective integrals yields:  
\[P(r,z)=\sum\limits_{n=-\infty}^{+\infty}P_{n}(r,z),\]  
where  
\[P_{n}(r,z)=\frac{1}{2}\sigma\bigg{(}\big{|}E_{n,r}(r,z)\big{|}^{2}+\big{|}E_{ n,z}(r,z)\big{|}^{2}\bigg{)}. \tag{25}\]  
Here, the radial component of the EF in the tank wall is:  
\[ \begin{array}{c}E_{n,r}(r,z)\!=\!-\frac{1}{\sigma}\frac{\partial H_{n,\rho} }{\partial z}\\ =-\frac{\beta_{n}l_{n}}{2\pi n\
u}\frac{\sinh(\beta_{n}z)}{\cosh\left(\frac{ \beta_{n}z}{2}\right)}+\frac{l_{n}}{2\pi n}\sum\limits_{m=0}^{\infty}\frac{4 \beta_{n}^{2}(-1)^{m}}{hh_{n,m}^{2}}\frac{\sin(\kappa_{m}z)}{I_{1}(\lambda_{n, m}a)K_{1}(\lambda_{n,m}b)-I_{1}(\lambda_{n,m}b)K_{1}(\lambda_{n,m}a)}\\ \times\Big{\{}\frac{1}{a}\big{[}I_{1}(\lambda_{n,m}r)K_{1}(\lambda_{n,m}b)-I_ {1}(\lambda_{n,m}b)K_{1}(\lambda_{n,m}r)\big{]}\\ +\frac{1}{b}\big{[}I_{1}(\lambda_{n,m}a)K_{1}(\lambda_{n,m}r)-I_{1}(\lambda_{ n,m}r)K_{1}(\lambda_{n,m}a)\big{]}\big{\}},\end{array} \]  
whereas the axial component takes the form:  
\[ \begin{array}{ccl}E_{n,z}(r,z)&=\frac{1}{\sigma}\frac{1}{r}\frac{\partial( rH_{n,\rho})}{\partial r}=\frac{l_{n}}{2\pi n\sigma}\sum\limits_{m=0}^{ \infty}\frac{4\beta_{n}^{2}(-1)^{m}}{hh_{n,m}^{2}}\frac{\cos(\kappa_{m}z)}{I_{1 }(\lambda_{n,m}a)K_{1}(\lambda_{n,m}b)-I_{1}(\lambda_{n,m}b)K_{1}(\lambda_{n, m}a)}\\ &&\times\Big{\{}\frac{1}{a}\big{[}I_{0}(\lambda_{n,m}r)K_{1}(\lambda_{n,m}b)+I _{1}(\lambda_{n,m}b)K_{0}(\lambda_{n,m}r)\big{]}\\ &&-\frac{1}{b}\big{[}I_{1}(\lambda_{n,m}a)K_{0}(\lambda_{n,m}r)+I_{0}(\lambda _{n,m}r)K_{1}(\lambda_{n,m}a)\big{]}\big{]}.\end{array} \]  
Using the same approximation as in (22), these components of the EF can be represented as follows:  
\[ \begin{array}{ccl}E_{n,r}(r,z)&=-\frac{\beta_{n}l_{n}}{2\pi n\sigma}\frac{ \sinh(\beta_{n}z)}{\cosh\left(\frac{\beta_{n}z}{2}\right)}+\frac{2\beta_{n}^ {2}l_{n}}{\pi n\alpha}\sum\limits_{m=0}^{\infty}\frac{(-1)^{m}}{\lambda_{n,m}^ {2}}\frac{K_{1}(\lambda_{n,m}r)}{K_{1}(\lambda_{n,m}a)}\sin(k_{m}z)\\ &=\frac{2\beta_{n}^{2}l_{n}}{\pi n\alpha h\alpha}\sum\limits_{m=0}^{\infty} \frac{(-1)^{m}}{\lambda_{n,m}^{2}}\bigg{(}\sqrt{\frac{\alpha}{r}}e^{-\lambda_{ n,m}(r-a)}-\frac{a}{r}\bigg{)}\sin(k_{m}z),\\ &&\\ E_{n,z}(r,z)&=-\frac{2\beta_{n}^{2}l_{n}}{\pi n\alpha h\alpha}\sum\limits_{m= 0}^{\infty}\frac{(-1)^{m}}{k_{m}\lambda_{n,m}}\sqrt{\frac{a}{r}}e^{-\lambda_{ n,m}(r-a)}\cos(k_{m}z),\end{array}  \tag{26}\]  
where expansion (13) of the function \(\sinh(\beta_{n}z)\) in a Fourier series has been used. By taking the integral of power density losses (23) over the whole conducting disk, the total losses in the tank wall can be computed. Thus, it can be written as:  
\[ \begin{array}{rl}P_{\text{tot}}&=\int\limits_{a}^{2\pi}d\varrho\int\limits_{a}^{ b}rdr\int\limits_{-h/2}^{h/2}dzP(r,z)=2\pi\sum\limits_{n=-\infty}^{+\infty}\int \limits_{a}^{b}rdr\int\limits_{-h/2}^{h/2}dzP_{n}(r,z)\\ &=\pi\alpha\sum\limits_{n=-\infty}^{+\infty}\int\limits_{a}^{b}rdr\int\limits_ {-h/2}^{h/2}dz(\big{|}E_{n,r}(r,z)\big{|}^{2}+\big{|}E_{n,z}(r,z)\big{|}^{2}). \end{array}  \tag{28}\]  
Results (26) and (27) for \(E_{n,r}(r,z)\) and \(E_{n,z}(r,z)\) should be substituted into Equation (28), where the integration over the variable \(z\) can be performed using the orthogonality of the system of functions \(\{\sin(k_{m}2),\cos(k_{n}z)\}\).