taking into account the boundary conditions, are readily derived from Eqs. (5)-(7). As a result, we get for the total energy $E=\int edV$, to leading order in $Rm$ [@Priede_etal07; @Willis_Barenghi02a] $$\frac{dE}{dt}=-\int u_ru_{\phi}r\frac{d\Omega}{dr}dV+D,$$ where $D$ is the dissipation function $$D=-\frac{1}{Re}\int [(\nabla\times {\bf u})^2+Ha^2(\nabla \times
{\bf b})^2]dV.$$ The first term in Eq. (8) is the Reynolds stress $u_ru_{\phi}$ multiplied by the shear, $rd\Omega/dr$, and describes energy exchange between the basic flow and perturbations, which is therefore due to the shear and is not affected by the magnetic field. The dissipation function $D$ is always negative definite and consists of two parts. The first part describes the usual viscous dissipation of the kinetic energy, while the second part describes the Joule losses and in general is comparable to the former. Since the magnetic energy is negligible in this case, the Joule losses act as an additional effective dissipation of special kind for the kinetic energy through the kinetic-magnetic exchange term proportional to $\Lambda$ in Eq. (5) [@Zikanov_Thess98]. Note that a net contribution from nonlinear terms has canceled out in Eq. (8) after averaging over the domain due to the boundary conditions. Thus, only Reynolds stress, when positive, can act as a source of 