Temperature difference in slot copper strips

Assumption: each strip is 8 mm wide (slot width) and 10 mm deep, so the three strips fill a depth of 30 mm. Heat in the slot portion flows (i) along the copper to the cooler overhang, and (ii) across the slot insulation to the iron.

Heat produced per unit volume of copper:

$$q = J^2\rho = (4\times10^6)^2 \times 0.021\times10^{-6} = 3.36\times10^5\ \text{W/m}^3$$

i) Centre of embedded portion to overhang

Take $x$ from the centre of the slot length $L$. Heat crossing a section at $x$ is $qAx$, and over $dx$ the temperature drop is $d\theta = qAx \cdot \rho_t\,dx/A = q\rho_t x\,dx$. Integrating from 0 to $L/2$:

$$\begin{aligned} \theta_1 &= \frac{q\rho_t}{2}\left(\frac{L}{2}\right)^2 = \frac{q\rho_t L^2}{8}\\ &= \frac{3.36\times10^5 \times 0.0025 \times 0.5^2}{8} = 26.25\ ^\circ\text{C} \end{aligned}$$

ii) Conductor to slot walls

Loss in the three strips (slot portion):

$$\begin{aligned} V &= 3 \times (8\times10\times10^{-6}) \times 0.5 = 1.2\times10^{-4}\ \text{m}^3\\ P &= qV = 3.36\times10^5 \times 1.2\times10^{-4} = 40.32\ \text{W} \end{aligned}$$

Heat leaves through the two sides and the bottom of the slot (the open top is assumed not to dissipate):

$$S = (2 \times 30 + 8)\times10^{-3} \times 0.5 = 0.034\ \text{m}^2$$

Thermal resistance of insulation:

$$R = \frac{\rho_t\,t}{S} = \frac{3 \times 0.004}{0.034} = 0.353\ ^\circ\text{C/W}$$ $$\theta_2 = PR = 40.32 \times 0.353 = 14.23\ ^\circ\text{C}$$

   open top (wedge)
  |  [ strip 1 ]  |
  |  [ strip 2 ]  |  <- 4 mm insulation
  |  [ strip 3 ]  |     on sides and bottom
  |_______________|
     slot wall

Answer: (i) ≈ 26.25 °C between the centre of the embedded part and the overhang; (ii) ≈ 14.2 °C between conductors and slot walls.