 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2022/paper-1/section-c/c8/e/solution

E \cdot J = J^{2} / σ

, uniform in this wire. Multiplying by its volume gives the dissipated power

P_{loss} = \frac{1}{σ} (\frac{I}{π a ^{2}})^{2} π a^{2} L = \frac{I ^{2} L}{σπ a ^{2}} = I^{2} R_{wire},

R_{wire} = L / (σπ a^{2})

. The result is in watts; over a time interval

Δ t

the dissipated energy is

P_{loss} Δ t

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