Question

A wire 50.0 m long and 2.00 mm in diameter is connected to a source with a potential difference of 9.11 V, and the current is found to be 36.0 A. Find the resistivity of the material.

Answer 1

Answer:

ρ = 1.6*10⁻⁸ Ω/m.

Explanation:

• Applying Ohm's Law to the wire, assuming that it can be treated as a pure resistance, the resistance of the wire can be obtained as follows:

       $R = \frac{V}{I} = \frac{9.11V}{36.0A} = 0.253 \Omega (1)$

• At the same time, we know that there exists a relationship between the resistance, the resistivity ρ, the length L and the area A of the wire, that is given for the following expression:

       $R = \rho* \frac{L}{A} (2)$

• The area of the circular section of the wire, can be expressed as a function of the diameter d, as follows:

      $A = \frac{\pi*d^{2} }{4} = \frac{\pi*(0.002m)^{2}}{4} = \pi*10e-6 (3)$

• Replacing  the left side of (2) by (1), and (3) on the right side, we can solve for the resistivity ρ as follows:

       $\rho = \frac{R*A}{L} = \frac{0.253\Omega*\pi*10e-6}{50.0m} = 1.6e-8 \Omega/m$

• ρ = 1.6*10⁻⁸ Ω/m