Problem 64 - Fundamentals of Physics

Question

A cylindrical resistor of radius \(5.0 \mathrm{~mm}\) and length \(2.0 \mathrm{~cm}\) is made of material that has a resistivity of \(3.5 \times 10^{-5} \Omega \cdot \mathrm{m} .\) What are (a) the magnitude of the current density and (b) the potential difference when the energy dissipation rate in the resistor is \(1.0 \mathrm{~W} ?\)

Step-by-Step Solution

Answer

(a) Current density is \(1.35 \times 10^5 \text{ A/m}^2\). (b) Potential difference is \(0.0945 \text{ V}\).

Step 1: Understand the Given Information

We are given a cylindrical resistor with a radius of \(5.0 \text{ mm}\), a length of \(2.0 \text{ cm}\), and a resistivity of \(3.5 \times 10^{-5} \Omega \cdot \text{m}\). The energy dissipation rate, or power, is \(1.0 \text{ W}\). We need to find (a) the current density and (b) the potential difference.

Step 2: Calculate the Cross-Sectional Area

First, convert the radius from millimeters to meters: \(5.0 \text{ mm} = 0.005 \text{ m}\). The cross-sectional area \(A\) of the cylinder is given by the formula \(A = \pi r^2\). Substituting \(r = 0.005 \text{ m}\), we get:\[A = \pi (0.005)^2 = 7.85 \times 10^{-5} \text{ m}^2\]

Step 3: Calculate the Resistance

The resistance \(R\) of the cylindrical resistor can be calculated using the formula:\[R = \rho \frac{L}{A}\]where \(\rho = 3.5 \times 10^{-5} \Omega \cdot \text{m}\) is the resistivity and \(L = 0.02 \text{ m}\) is the length. Substituting the values, we have:\[R = 3.5 \times 10^{-5} \cdot \frac{0.02}{7.85 \times 10^{-5}} = 8.92 \times 10^{-3} \Omega\]

Step 4: Determine the Current Using Power Formula

The power \(P\) dissipated is given by \(P = I^2 R\). We can rearrange this to find the current \(I\):\[I = \sqrt{\frac{P}{R}} = \sqrt{\frac{1.0}{8.92 \times 10^{-3}}} = 10.6 \text{ A}\]

Step 5: Calculate the Current Density

The current density \(J\) is found using: \(J = \frac{I}{A}\). Substituting \(I = 10.6 \text{ A}\) and \(A = 7.85 \times 10^{-5} \text{ m}^2\), we find:\[J = \frac{10.6}{7.85 \times 10^{-5}} = 1.35 \times 10^{5} \text{ A/m}^2\]

Step 6: Calculate the Potential Difference

The potential difference \(V\) is given by Ohm's Law: \(V = IR\). Using \(I = 10.6 \text{ A}\) and \(R = 8.92 \times 10^{-3} \Omega\), we have:\[V = 10.6 \times 8.92 \times 10^{-3} = 0.0945 \text{ V}\]

Key Concepts

Resistivity

Resistivity is a fundamental property of materials that quantifies how strongly a material opposes the flow of electric current. It is determined by the material's intrinsic characteristics and is denoted by the symbol \( \rho \). The unit of resistivity is ohm-meter (\( \Omega \cdot \mathrm{m} \)). In the context of a cylindrical resistor, resistivity plays a crucial role in calculating how much resistance the material provides against current flow. The formula to determine resistance \( R \) for a cylindrical shape is \[ R = \rho \frac{L}{A} \] where \( L \) is the length of the cylinder, and \( A \) is its cross-sectional area.
For this particular resistor, with a resistivity of \( 3.5 \times 10^{-5} \ \Omega \cdot \mathrm{m} \), the resistance is calculated to be approximately \( 8.92 \times 10^{-3} \ \Omega \). This low resistivity suggests the material conducts electricity well, which is typical for conductive materials used in resistors.

Ohm's Law

Ohm's Law is a fundamental principle in electrical engineering and physics, defining the relationship between voltage, current, and resistance within an electrical circuit. The law is elegantly simple and is expressed by the formula \( V = IR \), where \( V \) represents voltage (potential difference), \( I \) is the current flowing through the resistor, and \( R \) is the resistance.
In this exercise, we utilize Ohm's Law to calculate the potential difference across the resistor, given the current of \( 10.6 \) A and the calculated resistance of \( 8.92 \times 10^{-3} \ \Omega \). Applying the formula, the potential difference is \( 0.0945 \) V.
Ohm's Law is foundational because it allows us to predict how a circuit will behave given two known quantities, making it a versatile tool in circuit analysis.

Power Dissipation

Power dissipation in an electrical component refers to the rate at which it converts electrical energy into heat or other forms of energy. It is a key consideration in circuit design, as excess heat can damage or reduce the efficiency of electronic elements. The formula for power dissipation in resistors leverages the relationship \( P = I^2 R \), where \( P \) is the power in watts, \( I \) is the current, and \( R \) is the resistance.
For this particular cylindrical resistor, the power dissipation is given as \( 1.0 \) W. Using this information and known resistance, we rearrange the power formula to solve for current \( I \), which leads us to a current of \( 10.6 \) A.
Understanding power dissipation helps in designing resistors and circuits that operate efficiently and safely, preventing overheating and ensuring components function within their specified limits.