 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2018/paper-2/section-c/c12/a/solution

Distinguish electrical current density

J_{e}

J_{Q}

. Uniform current across the circular section gives

J_{e} = \frac{I}{π a ^{2}}, E = \frac{J _{e}}{σ}

H = J_{e} \cdot E = \frac{J _{e}^{2}}{σ} = \frac{I ^{2}}{σ π ^{2} a ^{4}} .

For a segment of length

ℓ

, integration gives

H π a^{2} ℓ = I^{2} ℓ / (σπ a^{2}) = I^{2} R

, confirming the usual resistor power. The model takes electrical conductivity constant and ignores temperature-induced current redistribution.

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