Problem 72 - Physics for Scientists and Engineers with Modern Physics

Question

An unknown length of platinum wire \(1.22 \mathrm{~mm}\) in diameter is placed as the unknown resistance in a Wheatstone bridge (see Problem 71 , Fig. \(26-65\) ). Arms 1 and 2 have resistance of \(38.0 \Omega\) and \(29.2 \Omega,\) respectively. Balance is achieved when \(R_{3}\) is \(3.48 \Omega .\) How long is the platinum wire?

Step-by-Step Solution

Answer

The platinum wire is approximately 4.99 meters long.

Step 1: Understanding the Wheatstone Bridge

In a balanced Wheatstone bridge, the ratio of resistances in one pair of opposite arms equals the ratio in the other pair. The formula is \( \frac{R_1}{R_2} = \frac{R_x}{R_3} \), where \( R_1 = 38.0 \ \Omega \), \( R_2 = 29.2 \ \Omega \), \( R_x \) is the unknown resistance of the platinum wire, and \( R_3 = 3.48 \ \Omega \).

Step 2: Applying the Formula

Using the formula \( \frac{R_1}{R_2} = \frac{R_x}{R_3} \), substitute the known values to find \( R_x \). Thus, \( \frac{38.0}{29.2} = \frac{R_x}{3.48} \). Solve for \( R_x \) by calculating \( R_x = \frac{38.0}{29.2} \times 3.48 \).

Step 3: Solving for the Resistance of Platinum Wire

Calculate \( R_x = \frac{38.0}{29.2} \times 3.48 \). First, \( \frac{38.0}{29.2} \approx 1.3014 \). Then, \( R_x \approx 1.3014 \times 3.48 \approx 4.53 \ \Omega \).

Step 4: Relating Resistance to Wire Length

The resistance \( R \) of a wire is given by \( R = \rho \frac{L}{A} \), where \( \rho \) is the resistivity of platinum (approximately \( 10.6 \times 10^{-8} \ \Omega\cdot m \)), \( L \) is the length, and \( A \) is the cross-sectional area of the wire. The diameter \( d \) of the wire is \( 1.22 \ mm = 1.22 \times 10^{-3} \ m \). Thus, \( A = \pi \left(\frac{d}{2}\right)^2 \).

Step 5: Calculating the Cross-Sectional Area

Calculate the cross-sectional area \( A = \pi \left(\frac{1.22 \times 10^{-3}}{2}\right)^2 \). Thus, \( A \approx 1.166 \times 10^{-6} \ m^2 \).

Step 6: Solving for Length of the Wire

We have \( R = 4.53 \ \Omega \), \( \rho = 10.6 \times 10^{-8} \ \Omega\cdot m \), and \( A = 1.166 \times 10^{-6} \ m^2 \). Solve for \( L \) using \( L = \frac{R \cdot A}{\rho} = \frac{4.53 \times 1.166 \times 10^{-6}}{10.6 \times 10^{-8}} \). Thus, \( L \approx 4.99 \ m \).

Key Concepts

Platinum Wire Resistance

The resistance of a platinum wire is an essential measure in analyzing its behavior in electrical circuits, like those illustrated with a Wheatstone bridge. Resistance, denoted by \( R \), measures how much a wire opposes the flow of electric current. In our case, the platinum wire serves as an unknown resistance in a Wheatstone bridge setup.

A Wheatstone bridge is a circuit used to determine unknown resistances by balancing two legs of a bridge circuit. The formula \( \frac{R_1}{R_2} = \frac{R_x}{R_3} \) derives from this principle, where \( R_1 \) and \( R_2 \) are known resistances, \( R_x \) is the resistance of the platinum wire, and \( R_3 \) is a known adjustable resistance.

In our exercise, we're tasked with determining \( R_x \) by substituting the known values into this ratio and solving the equation. Once the platinum wire's resistance is calculated, it plays a critical role in understanding how long the wire must be when considering the resistivity and cross-sectional area calculations presented subsequently.

Resistivity of Materials

Resistivity is a fundamental property of materials that quantifies how strongly a material opposes the flow of electric current. It is denoted by \( \rho \) and is expressed in ohm-meters (\( \Omega\cdot m \)). Each material has a distinctive resistivity value, influencing how it conducts electricity.

In our exercise, the resistivity of platinum is given as approximately \( 10.6 \times 10^{-8} \Omega\cdot m \). This low resistivity makes platinum a great conductor, albeit not as conductive as some metals like copper. The resistivity value is crucial when calculating the total resistance of a wire, which further relates to its physical dimensions such as length and cross-sectional area.

Understanding the resistivity helps in predicting how a wire of a particular material will behave in an electrical circuit, influencing design decisions and ensuring effective circuit function. When calculating the wire's length, resistivity relates to resistance through the formula \( R = \rho \frac{L}{A} \), emphasizing the material's role in resisting current flow.

Cross-Sectional Area Calculation

Determining the cross-sectional area of a wire is vital in solving for its length when the resistance is known. The cross-sectional area \( A \) is a measure of the size of the wire's cross-section, typically calculated using the diameter of the wire, assuming a circular cross section.

The formula to find the cross-sectional area of a wire with diameter \( d \) is \( A = \pi \left(\frac{d}{2}\right)^2 \). In this exercise, the diameter of the platinum wire is \( 1.22 \ mm \), which is converted to meters as \( 1.22 \times 10^{-3} \ m \). By squaring half the diameter and multiplying by \( \pi \), we find \( A \approx 1.166 \times 10^{-6} \ m^2 \).

This calculation is a fundamental step to find the wire's length using the equation \( L = \frac{R \cdot A}{\rho} \). Here, \( L \), the length of the wire, directly depends on its cross-sectional area and shows the interdependence of these physical properties in determining electrical behavior.