 Solution

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

In a long wire away from its ends, the steady temperature is independent of axial position and azimuth. The heat equation reduces to

\frac{1}{ρ} \frac{d}{d ρ} (ρ \frac{d T}{d ρ}) = - \frac{H}{κ} .

Multiplying by

ρ

and integrating gives

ρ T^{'} = - \frac{H ρ ^{2}}{2 κ} + C_{1}, T = - \frac{H ρ ^{2}}{4 κ} + C_{1} ln ρ + C_{2} .

A finite temperature and nonsingular flux at the axis require

C_{1} = 0

. Applying

T (a) = T_{0}

determines

C_{2} = T_{0} + H a^{2} / (4 κ)

. Therefore

T (ρ) = T_{0} + \frac{H}{4 κ} (a^{2} - ρ^{2}) = T_{0} + \frac{I ^{2}}{4 π ^{2} a ^{4} κσ} (a^{2} - ρ^{2}) .

T (0) - T_{0} = I^{2} / (4 π^{2} a^{2} κσ)

. The outward heat flux is

J_{Q} (ρ) = - κ T^{'} (ρ) = H ρ /2

, so the power crossing a cylindrical surface per unit length,

2 π ρ J_{Q} (ρ) = H π ρ^{2}

, equals the heating generated inside it.

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