Induction stove

If a circle with radius $r$ is selected as the integration area $A$ for Lenz's law, the integrals can be simplified $\begin{aligned} -\frac{\partial \Phi}{\partial t} &= - \int \frac{\partial \vec B}{\partial t} \cdot d\vec A = -\frac{\partial B}{\partial t} \cdot A = \omega \pi r^2 B_0 \sin(\omega t) \\ &= \oint \vec E \cdot d \vec l = E \cdot l = 2 \pi r E(r) \sin(\omega t) \\ \Rightarrow \quad E(r) &= \frac{1}{2} B_0 \omega r \end{aligned}$ with the magnetic field amplitude $B_0 = \mu_r \mu_0 H_0$ inside the metal.

A volume element $dV = w \cdot r d\phi \cdot dr$ of the disk has a voltage drop $dU$ and a current flow $dI$ and with $\begin{aligned} dU &= E r d\phi \\ dI &= j w dr \\ \Rightarrow \quad dP &= dU dI = \sigma E^2 w r dr d\phi \end{aligned}$ Integration over the whole bottom plate yields the total power $\begin{aligned} P &= \frac{1}{4} \omega^2 \sigma B_0^2 \sin^2(\omega t) w \int_0^R r^3 dr \int_0^{2\pi} d\pi \\ &= \frac{1}{8} \pi \omega^2 \sigma B_0^2 \sin^2(\omega t) w R^4 \\ \Rightarrow \quad \langle P \rangle &= \frac{1}{8} \pi \omega^2 \sigma B_0^2 w R^4 \end{aligned}$ with the time average over one period $\begin{aligned} \langle \sin^2(\omega t) \rangle &= \frac{1}{T} \int_0^T \sin^2(\omega t) dt = \frac{1}{T} \int_0^T \cos^2(\omega t) dt \\ &= \frac{1}{2T} \int_0^T (\sin^2(\omega t) + \cos^2(\omega t)) dt = \frac{1}{2 T} \int_0^T dt = \frac{1}{2} \end{aligned}$ Numerical evaluation results $\langle P \rangle = 3.876 \,\text{kW}$