Problem 76 - University Physics with Modern Physics

Question

A \(1.50-\mathrm{m}\) cylinder of radius 1.10 \(\mathrm{cm}\) is made of a complicated mixture of materials. Its resistivity depends on the distance \(x\) from the left end and obeys the formula \(\rho(x)=\) \(a+b x^{2},\) where \(a\) and \(b\) are constants. At the left end, the resistivity is \(2.25 \times 10^{-8} \Omega \cdot \mathrm{m},\) while at the right end it is \(8.50 \times\) \(10^{-8} \Omega \cdot \mathrm{m}\) . (a) What is the resistance of this rod? (b) What is the electric field at its midpoint if it carries a \(1.75-\) A current? (c) If we cut the rod into two 75.0 -cm halves, what is the resistance of each half?

Step-by-Step Solution

Answer

(a) The resistance is \(9.75 \times 10^{-7} \, \Omega\). (b) The electric field at midpoint is \(9.10 \times 10^{-7} \, V/m\). (c) Resistances are \(3.91 \times 10^{-7} \, \Omega\) and \(5.84 \times 10^{-7} \, \Omega\).

Step 1: Understand the Formula for Resistivity

The resistivity at any point along the cylinder depends on the position, given by \( \rho(x) = a + bx^2 \). We know the resistivity at both ends, so we can use their values to determine the constants \( a \) and \( b \).

Step 2: Determine Constants a and b

At \( x = 0 \) (left end), \( \rho(0) = a = 2.25 \times 10^{-8} \, \Omega \cdot \mathrm{m} \). At \( x = 1.50 \) m (right end), \( \rho(1.50) = a + b\cdot(1.50)^2 = 8.50 \times 10^{-8} \, \Omega \cdot \mathrm{m} \). Solving these equations gives us \( a = 2.25 \times 10^{-8} \, \Omega \cdot \mathrm{m} \) and \( b = \frac{8.50 \times 10^{-8} - 2.25 \times 10^{-8}}{1.50^2} = 2.78 \times 10^{-8} \, \Omega \cdot \mathrm{m} \).

Step 3: Calculate Resistance of the Entire Rod

The resistance \( R \) of the rod is given by the integral \[ R = \frac{1}{A} \int_0^{1.50} \rho(x) \, dx \] where \( A \) is the cross-sectional area, \( A = \pi \cdot (0.011)^2 = 3.80 \times 10^{-4} \, \mathrm{m^2} \). Substituting and integrating, \[ R = \frac{1}{3.80 \times 10^{-4}} \left[ 2.25 \times 10^{-8} x + 2.78 \times 10^{-8} \frac{x^3}{3} \right]_0^{1.50} = 9.75 \times 10^{-7} \, \Omega. \]

Step 4: Calculate Electric Field at Midpoint

The electric field \( E \) at the midpoint (0.75 m) can be found using \( E = \frac{V}{l} \), knowing \( V = I \cdot R_m \). \( R_m \) is the resistance from 0 to 0.75 m calculated similarly as in Step 3. The electric field is \( E = \frac{1.75 \, A \cdot 3.91 \times 10^{-7} \, \Omega}{0.75 \, m} = 9.10 \times 10^{-7} \, V/m \).

Step 5: Calculate Resistance of Each Half

For the left half (0 to 0.75 m) and the right half (0.75 to 1.50 m), we calculate separate resistances using the same method as in Step 3. The resistances are \( R_{left} = 3.91 \times 10^{-7} \, \Omega \) and \( R_{right} = 5.84 \times 10^{-7} \, \Omega \).

Key Concepts

Electric Field

Exploring the concept of the electric field is an essential part of understanding electromagnetic phenomena. In simple terms, the electric field is a vector field that represents the force experienced by a unit positive charge at a given point in space. In our example, the electric field at the midpoint of the cylinder can be calculated if we know the potential difference across that point and the length over which the potential is applied.

To find the electric field at the midpoint of the cylinder, we need to first determine the resistance between the endpoints from which the midpoint potential can be derived. The formula used here is \[ E = \frac{V}{l} \]where \( V \) represents the voltage across the midpoint, and \( l \) is the distance along the cylinder.

Depending on the current flowing through the cylinder and its resistance, using Ohm's Law (\( V = IR \)), we can find the electric field by rearranging these fundamental relationships. This integration of concepts allows us to see how electric fields manifest in conductive materials.

Resistance Calculation

Calculating resistance is key when dealing with circuits and conductive materials. Resistance measures how much a material opposes the flow of electric current. For a non-uniformly composed cylinder where resistivity varies with position, the concept of resistance depends heavily on its material's properties and geometry.

Resistance \( R \) can be calculated using the integral\[ R = \frac{1}{A} \int_{0}^{L} \rho(x) \, dx \]where \( A \) is the cross-sectional area, \( L \) is the length, and \( \rho(x) \) is the resistivity as a function of \( x \).

This calculation involves integrating the resistivity over the length of the cylinder to account for its variation. Once \( R \) is calculated, it's possible to predict how the entire cylinder will react under electrical influence, making resistance calculation a vital component in electrical engineering.

Cylinder Geometry

The cylinder's geometry plays a pivotal role in analyzing its electrical properties. Understanding this geometry involves knowing its length, radius, and how these relate to its volume and surface area.

A cylinder with a radius \( r \) and a length \( L \) has a cross-sectional area given by \[ A = \pi r^2 \]where \( \pi \) is a constant (approximately 3.14159). In our case, the radius of 1.10 cm needs to be converted into meters as \( 0.011 \, m \), allowing for proper units in calculations.

Understanding the cylindrical geometry enables us to calculate parameters like cross-sectional area, which is critical when computing resistance or electric fields. Additionally, it underscores how the physical structure affects the resistivity variation along the entire length.

Integral Calculus

Integral calculus is a mathematical tool used extensively to solve problems involving variables changing continuously, such as the resistivity in our rod example. This method allows us to handle and simplify the complex computation of variable-dependent quantities, integral to this problem.

The primary function of integral calculus in this context is to calculate the total resistance of the cylinder by summing up infinitely small resistive elements. The integral \[ \int \rho(x) \, dx \]computes the total resistivity over the entire length, respecting changes in resistivity at every point.

Through integrating, we obtain resistance values that reflect the physical reality of how current flows through the non-uniform material. Understanding integral calculus aids in visualizing how electrical properties manifest and change, representing them systematically and precisely.