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Chapter 25: Problem 65

The resistivity of a conductor is \(\rho=1.00 \cdot 10^{-5} \Omega \mathrm{m}\). If a cylindrical wire is made of this conductor, with a crosssectional area of \(1.00 \cdot 10^{-6} \mathrm{~m}^{2},\) what should the length of the wire be for its resistance to be \(10.0 \Omega ?\)

Short Answer

Expert verified
Answer: The required length of the conductor wire is 10.0 meters.

Step by step solution

01

Recall the formula for the resistance of a wire

We will use the formula for the resistance of a wire: $$ R = \frac{\rho L}{A} $$ where \(R\) is the resistance, \(\rho\) is the resistivity, \(L\) is the length of the wire, and \(A\) is the cross-sectional area.
02

Plug the given values

We are given the resistivity \(\rho = 1.00 \cdot 10^{-5} \Omega \mathrm{m}\), the cross-sectional area \(A = 1.00 \cdot 10^{-6} \mathrm{m}^2\), and the desired resistance \(R = 10.0 \Omega\). We plug these values into the formula: $$ 10.0 \Omega = \frac{(1.00 \times 10^{-5} \Omega \mathrm{m}) L}{(1.00 \times 10^{-6} \mathrm{m}^2)} $$
03

Solve for the length, L

Now, we will solve for the length \(L\): $$ L = \frac{10.0 \Omega \times 1.00 \times 10^{-6} \mathrm{m}^2}{1.00 \times 10^{-5} \Omega \mathrm{m}} $$
04

Simplify the expression

By dividing the numbers and canceling out the units, we get: $$ L = 10.0 \mathrm{m} $$
05

State the final answer

The length of the wire should be \(10.0\) meters for its resistance to be \(10.0 \Omega\).

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