Chapter 26: Problem 19
What is the resistivity of a wire of \(1.0 \mathrm{~mm}\) diameter, \(2.0\) \(\mathrm{m}\) length, and \(50 \mathrm{~m} \Omega\) resistance?
Short Answer
Expert verified
The resistivity is approximately \(1.9625 \times 10^{-8} \Omega\cdot \text{m}\).
Step by step solution
01
Identify the given values
We have been given the diameter of the wire as \(1.0 \text{ mm}\), which needs to be converted into meters: \(1.0 \text{ mm} = 0.001 \text{ m}\). The length of the wire is \(2.0 \text{ m}\), and the resistance is \(50 \text{ m}\Omega = 50 \times 10^{-3} \text{ }\Omega\).
02
Calculate the cross-sectional area
The cross-section of the wire is circular. Use the formula for the area of a circle, \(A = \pi r^2\), where \(r\) is the radius. The radius \(r\) is half of the diameter, so \(r = 0.5 \times 0.001\, \text{m} = 0.0005\, \text{m}\). Now calculate the area:\[A = \pi (0.0005)^2 = \pi \times 0.00000025\, \text{m}^2\approx 7.85 \times 10^{-7}\, \text{m}^2\]
03
Use the resistivity formula
The formula that relates resistance \(R\), resistivity \(\rho\), length \(L\), and cross-sectional area \(A\) is:\[R = \rho \frac{L}{A}\]We can rearrange it to solve for resistivity \(\rho\):\[\rho = R \frac{A}{L}\]
04
Substitute values into the formula
Now substitute the known values into the formula from Step 3:\[\rho = (50 \times 10^{-3}) \Omega \cdot \frac{7.85 \times 10^{-7} \text{ m}^2}{2.0 \text{ m}}\]\[\rho = 50 \times 10^{-3} \times 3.925 \times 10^{-7} = 1.9625 \times 10^{-8} \Omega \cdot \text{m}\]
05
Double-check the calculation
Verify the calculation for accuracy:1. Calculate \(A = 7.85 \times 10^{-7}\, \text{m}^2\) (confirmed).2. Substitute into \(\rho = (50 \times 10^{-3})\Omega \cdot \frac{7.85 \times 10^{-7}\, \text{m}^2}{2.0\text{ m}}\) and simplify.3. Calculate \(\rho = 1.9625 \times 10^{-8} \Omega \cdot \text{m}\) (confirmed).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Ohm's Law
Ohm's Law is a fundamental principle in the field of electrical engineering and physics. It states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, provided the temperature remains constant. This can be mathematically represented by the formula:
- The formula is given by: \[ V = IR \] where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance of the conductor.
- A higher resistance implies that less current flows for a given voltage.
Circular Cross Section
A circular cross section refers to the shape of the area that a wire presents when viewed from its end. For wires manufactured in practical scenarios, most commonly, this shape is a circle. This is crucial particularly when calculating the cross-sectional area, which influences the wire’s resistance.To compute the area of a circle, use:
- The formula for the area of a circle is: \[ A = \pi r^2 \] where \( r \) is the radius of the circle.
- For the given exercise, the diameter of the wire is \(1.0 \text{ mm}\), which means the radius \( r \) is half of the diameter, i.e., \(0.5 \times 0.001 \text{ m}\ equals 0.0005 \text{ m}\).
- Using the formula, the area is calculated as \( A \approx 7.85 \times 10^{-7} \text{ m}^2 \).
Resistivity Formula
The resistivity formula plays a pivotal role in determining how materials resist current flow in an electrical circuit. It links together resistance, resistivity, the length of the conductor, and its cross-sectional area.The relationship is expressed by the formula:
- \[ R = \rho \frac{L}{A} \] Where \( R \) is resistance, \( \rho \) is resistivity, \( L \) is the length of the conductor, and \( A \) is the cross-sectional area.
- To find the resistivity \( \rho \) of a wire, rearrange the formula to: \[ \rho = R \frac{A}{L} \]
- A material with high resistivity will resist the flow of electric current more than a material with low resistivity.
- This resistivity depends on the material composition and its temperature.