A 2.0-mm-diameter wire formed from a composite material has a resistivity that decreases with distance along the wire as ρ=ρ₀e^−αx, where ρ₀=4.0×10⁻⁵ Ω m, x (in m) is measured from one end of the wire, and the constant α=4.0 m⁻¹. What is the resistance of a 50-cm-long length of this wire?

Verified step by step guidance

1. Step 1: Understand the problem. The wire has a resistivity that varies along its length according to the formula ρ = ρ₀e^−αx, where ρ₀ is the initial resistivity, α is a constant, and x is the distance from one end of the wire. We need to calculate the total resistance of a 50-cm-long wire, taking into account the varying resistivity.
2. Step 2: Recall the formula for resistance. Resistance R is given by the integral of resistivity ρ over the length of the wire, divided by the cross-sectional area A. Mathematically, R = ∫(ρ(x)/A) dx, where A = πr² is the cross-sectional area of the wire, and r is the radius of the wire.
3. Step 3: Calculate the cross-sectional area of the wire. The diameter of the wire is given as 2.0 mm, so the radius r = diameter/2 = 1.0 mm = 1.0 × 10^−3 m. Substitute this into A = πr² to find the area.
4. Step 4: Set up the integral for resistance. Substitute ρ(x) = ρ₀e^−αx and A = πr² into the formula for resistance: R = ∫(ρ₀e^−αx / πr²) dx. The limits of integration are from x = 0 to x = 0.50 m (since the wire is 50 cm long).
5. Step 5: Solve the integral. The integral of e^−αx with respect to x is (1/α)e^−αx. Apply the limits of integration (from 0 to 0.50 m) to find the total resistance R. Simplify the expression to get the final formula for R in terms of ρ₀, α, and the wire's dimensions.

Key Concepts

Resistivity

Resistivity is a material property that quantifies how strongly a given material opposes the flow of electric current. It is denoted by the symbol ρ and is measured in ohm-meters (Ω·m). In this problem, the resistivity of the wire varies with distance, which affects the overall resistance of the wire segment.

Resistance Calculation

Resistance (R) is calculated using the formula R = ρL/A, where ρ is the resistivity, L is the length of the conductor, and A is the cross-sectional area. For a cylindrical wire, the cross-sectional area can be determined using A = π(d/2)², where d is the diameter. In this case, the resistance must be integrated along the length of the wire due to the varying resistivity.

Integration in Physics

Integration is a mathematical technique used to find the total value of a quantity that varies over a certain range. In this context, it is necessary to integrate the resistance contributions along the length of the wire, as the resistivity changes with distance. This allows for the calculation of the total resistance of the wire segment by summing the infinitesimal resistances over its length.