A hollow cylindrical tube with a rectangular cross section has external dimensions of \(0.5 \mathrm{in}\). by 1 in. and a wall thickness of \(0.05 \mathrm{in}\). Assume that the material is brass, for which \(\sigma=1.5 \times 10^{7} \mathrm{~S} / \mathrm{m}\). A current of \(200 \mathrm{~A} \mathrm{dc}\) is flowing down the tube. \((a)\) What voltage drop is present across a \(1 \mathrm{~m}\) length of the tube? (b) Find the voltage drop if the interior of the tube is filled with conducting material for which \(\sigma=1.5 \times 10^{5} \mathrm{~S} / \mathrm{m}\).

First, we'll need to determine the tube's resistance. The formula for resistance R in a rectangular wire is given by: $$R = \frac{\rho L}{A}$$ where $$\rho$$ is the resistivity, L is the length of the wire, and A is the cross-sectional area. The conductivity $$\sigma$$ is the inverse of the resistivity, so $$\rho = \frac{1}{\sigma}$$. We are given the dimensions of the tube and its conductivity, so we can compute the resistivity and find the cross-sectional area. The cross-sectional area of the hollow tube is the difference between the outer and inner rectangular areas. Given the wall thickness $$t = 0.05\,\mathrm{in}$$, the inner rectangle has dimensions of $$0.4$$ and $$0.9$$ inches. Converting the dimensions to meters, and calculating the areas: $$A_\text{outer} = (0.5\,\mathrm{in} \times 1\,\mathrm{in}) \times (0.0254\,\mathrm{m}/\mathrm{in})^2$$ $$A_\text{inner} = (0.4\,\mathrm{in} \times 0.9 \mathrm{in}) \times (0.0254\,\mathrm{m}/\mathrm{in})^2$$ The cross-sectional area of the tube, $$A = A_\text{outer} - A_\text{inner}$$.

Now we have the cross-sectional area, we can calculate the resistance of the tube using its resistivity and length. $$R_\text{tube} = \frac{\rho L}{A}$$ Given the current $$I=200\,\mathrm{A}$$, we can find the voltage drop across the tube, $$V_\text{tube}$$, using Ohm's law: $$V_\text{tube} = I \times R_\text{tube}$$ This will give us the answer to part (a).

Now, we need to calculate the resistance of the conducting material that fills the inner part of the tube. We can use the same formula for resistance as before, but this time, we'll use the inner rectangular cross-sectional area, $$A_\text{inner}$$, and the conductivity of the conducting material, $$\sigma_\text{material}$$. $$R_\text{material} = \frac{\rho_\text{material} L}{A_\text{inner}}$$

When the tube is filled with the conducting material, the two resistances - tube and material - are effectively in parallel. Therefore, the total resistance, $$R_\text{total}$$, can be calculated using the formula for resistances in parallel: $$\frac{1}{R_\text{total}} = \frac{1}{R_\text{tube}} + \frac{1}{R_\text{material}}$$

Finally, with the total resistance, we can find the voltage drop across the tube when it is filled with the conducting material, $$V_\text{filled}$$: $$V_\text{filled} = I \times R_\text{total}$$ This will give us the answer to part (b).