Question

Two conductors are made of the same material and have the same length \(L\). Conductor \(A\) is a hollow tube with inside diameter \(2.00 \mathrm{~mm}\) and outside diameter \(3.00 \mathrm{~mm}\); conductor \(\mathrm{B}\) is a solid wire with radius \(R_{\mathrm{B}}\). What value of \(R_{n}\) is required for the two conductors to have the same resistance measured between their ends?

Short answer

Answer: \(R_{B} \approx 1.50 \, \mathrm{mm}\)

Step-by-step solution

Write down the resistance formula for both conductors A and B

We have the resistance formula \(R = \frac{\rho L}{A}\). For conductor A, the cross-sectional area is the area of the hollow tube, and for conductor B, it's the area of the solid wire.

Find the cross-sectional area of conductor A

Conductor A is a hollow tube with inside diameter \(2.00 \mathrm{~mm}\) and outside diameter \(3.00 \mathrm{~mm}\). To find its cross-sectional area, subtract the area of the inside circle from the area of the outside circle. The area of a circle is given by \(A = \pi r^2\): \(A_{\mathrm{A}} = \pi (r_{\text{outside}}^2 - r_{\text{inside}}^2) = \pi \left(\left(\frac{3.00}{2}\right)^2 - \left(\frac{2.00}{2}\right)^2\right) \mathrm{mm}^2\)

Write down the cross-sectional area of conductor B

Conductor B is a solid wire with radius \(R_{B}\). Its cross-sectional area is given by: \(A_{\mathrm{B}} = \pi R_{\mathrm{B}}^2\)

Set up the equation for equal resistances and solve for \(R_{B}\)

Since both conductors have the same resistance, we can equate their resistances using the resistance formula: \(\frac{\rho L}{A_{\mathrm{A}}} = \frac{\rho L}{A_{\mathrm{B}}}\) Cancel out \(\rho\) and \(L\), as they are the same for both conductors: \(\frac{1}{A_{\mathrm{A}}} = \frac{1}{A_{\mathrm{B}}}\) Now, substitute the cross-sectional area expressions for \(A\) and \(B\) from Steps 2 and 3: \(\frac{1}{\pi \left(\left(\frac{3.00}{2}\right)^2 - \left(\frac{2.00}{2}\right)^2\right)} = \frac{1}{\pi R_{\mathrm{B}}^2}\) We can cancel out \(\pi\) on both sides of the equation: \(\frac{1}{\left(\frac{3.00}{2}\right)^2 - \left(\frac{2.00}{2}\right)^2} = \frac{1}{R_{\mathrm{B}}^2}\) Now, solve for \(R_{\mathrm{B}}\): \(R_{\mathrm{B}}^2 = \left(\frac{3.00}{2}\right)^2 - \left(\frac{2.00}{2}\right)^2\) \(R_{\mathrm{B}} = \sqrt{\left(\frac{3.00}{2}\right)^2 - \left(\frac{2.00}{2}\right)^2} \approx 1.50 \mathrm{~mm}\) Therefore, the required value of \(R_{\mathrm{B}}\) for the two conductors to have the same resistance measured between their ends is approximately \(1.50 \mathrm{~mm}\).