A copper wire of diameter [tex][tex]$1.58 \text{ mm}$[/tex][/tex] and resistivity [tex][tex]$1.7 \times 10^{-8} \Omega \cdot \text{m}$[/tex][/tex] has a total resistance of [tex][tex]$0.50 \Omega$[/tex][/tex]. How long is the wire?



Answer :

Certainly! Let's go through the steps to determine the length of the copper wire given the diameter, resistivity, and resistance.

### Step 1: Convert Diameter to Radius
First, convert the diameter of the wire into its radius. The given diameter is [tex]\(1.58 \, \text{mm}\)[/tex].

[tex]\[ \text{Radius} = \frac{\text{Diameter}}{2} = \frac{1.58 \, \text{mm}}{2} = 0.79 \, \text{mm} \][/tex]

Since the resistivity is given in meters, we need to convert the radius from millimeters to meters.

[tex]\[ 0.79 \, \text{mm} = 0.79 \times 10^{-3} \, \text{m} = 0.00079 \, \text{m} \][/tex]

### Step 2: Compute the Cross-Sectional Area
Next, calculate the cross-sectional area (A) of the wire. The cross-sectional area of a wire with a circular cross-section is given by the formula:

[tex]\[ A = \pi r^2 \][/tex]

Substitute the radius into the formula:

[tex]\[ A = \pi (0.00079 \, \text{m})^2 \approx 1.9606679751053896 \times 10^{-6} \, \text{m}^2 \][/tex]

### Step 3: Use the Resistivity Formula to Find Length
The relationship between resistance (R), resistivity ([tex]\(\rho\)[/tex]), length (L), and cross-sectional area (A) of a wire is given by the formula:

[tex]\[ R = \rho \frac{L}{A} \][/tex]

Rearrange the formula to solve for length (L):

[tex]\[ L = \frac{R \cdot A}{\rho} \][/tex]

Substitute the known values into the equation. Here, [tex]\(R = 0.50 \, \Omega\)[/tex], [tex]\(\rho = 1.7 \times 10^{-8} \, \Omega \cdot \text{m}\)[/tex], and [tex]\(A = 1.9606679751053896 \times 10^{-6} \, \text{m}^2\)[/tex]:

[tex]\[ L = \frac{0.50 \, \Omega \times 1.9606679751053896 \times 10^{-6} \, \text{m}^2}{1.7 \times 10^{-8} \, \Omega \cdot \text{m}} \][/tex]

Perform the calculation:

[tex]\[ L \approx 57.66670515015852 \, \text{m} \][/tex]

### Conclusion

The length of the copper wire is approximately [tex]\(57.67 \, \text{m}\)[/tex].