Past exam of the physics B course of the University of Cambridge / 2022 / Paper 1 / Section C / C8 / e / Solution

The Ohmic heating per unit volume is $\mathbf{E}\cdot \mathbf{J}=J^2/\sigma$, uniform in this wire. Multiplying by its volume gives the dissipated powerwhere the electrical resistance is $R_{\mathrm{wire}}=L/(\sigma \pi a^2)$. The result is in watts; over a time interval $\Delta t$ the dissipated energy is $P_{\mathrm{loss}}\Delta t$.