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My question is how do i calculate the total energy flux through the walls of the torus in this situation? The Poynting flux is given by $\vec{N} = \frac{1}{\mu} ( \vec{E} \times \vec{B})$ but the $\vec{E}$ field is zero inside the torus no?

Also, instead of the current flowing in the copper, the current is switched off completely such that $\vec{j} = 0$, then how to we calculate the magnetic field and electric field decay in here? How can $\vec{j}$ even be equal to 0; I thought the idea was that you can't just switch off the current/magnetic field because the reduction in current produces a change a flux which induces an E-field that re-enforces the current to keep going?

Any help appreciated :)

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So basically, I missed out an important symmetry in the question which is that the $B \propto \frac{1}{r}$ meaning that the flux integrated around the loop circumference $ = 2\pi r$ results in the $r$-dependence cancelling out simplifying the integral greatly. The result is that: $\vec{N} = \frac{ \vec{E} \times \vec{B} } {\mu_{0}} = \frac{10 I }{\pi r} \vec{E} \times \vec{\hat{n}}$. When this flux is integrated over the closed surface:

$ \oint \vec{N} \cdot \vec{dS} = \oint \frac{10I}{\pi r} E_{||} \cdot 2\pi r \cdot dl = 20 I \oint \vec{E} \cdot \vec{dl} = 20I (IR) = 20 I^2 R$ = Ohmic heating in the system.

Hence the energy flux out of the torus is equal to the Ohmic heating in the coils.

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