Problem 7 - Fundamentals of Physics

Question

A fuse in an electric circuit is a wire that is designed to melt, and thereby open the circuit, if the current exceeds a predetermined value. Suppose that the material to be used in a fuse melts when the current density rises to \(440 \mathrm{~A} / \mathrm{cm}^{2}\). What diameter of cylindrical wire should be used to make a fuse that will limit the current to \(0.50 \mathrm{~A} ?\)

Step-by-Step Solution

Answer

The wire diameter should be about 0.38 mm.

Step 1: Understand the Problem

We need to find the diameter of a cylindrical wire that will limit the current to 0.50 A such that the current density reaches a maximum of 440 A/cm².

Step 2: Recall the Formula for Current Density

The current density \(J\) is defined as the current \(I\) divided by the cross-sectional area \(A\) of the wire. Mathematically, it is expressed as: \[J = \frac{I}{A}\] Since we know the maximum current density \(J\) and the current \(I\), we can find the area \(A\) of the cross-section.

Step 3: Solve for the Cross-Sectional Area

Rearrange the formula for current density to find the area: \[A = \frac{I}{J}\] Substitute the given values: \[A = \frac{0.50}{440} = 0.001136 \, \text{cm}^2\].

Step 4: Relate Area to the Wire Diameter

The cross-sectional area \(A\) of a cylindrical wire is given by the formula for the area of a circle: \[A = \pi \left(\frac{d}{2}\right)^2\] where \(d\) is the diameter of the wire. Solve for \(d\) using the known area.

Step 5: Solve for the Diameter

Substitute the area obtained into the formula and solve for \(d\): \[0.001136 = \pi \left(\frac{d}{2}\right)^2\] Rearrange to solve for \(d\): \[d^2 = \frac{0.001136 \times 4}{\pi}\] \[d = \sqrt{\frac{0.001136 \times 4}{\pi}}\] \[d \approx \sqrt{0.00144663} \approx 0.03801 \, \text{cm}\]

Step 6: Convert Diameter to Millimeters

Since answers are often expected in millimeters, convert the diameter from cm to mm: \[d \approx 0.03801 \, \text{cm} = 0.3801 \, \text{mm}\].

Key Concepts

Understanding Current Density

Current density is a measure of how much electric current is flowing through a unit area of a material. It is denoted by the symbol \( J \) and is usually expressed in amperes per square centimeter \( (A/cm^2) \).
Current density can be thought of as how dense the current is within a certain area. If the current is like water, the current density tells us how tightly packed the water molecules are.
For any conductor, the current density can be calculated using the formula:

• \( J = \frac{I}{A} \)

Here, \( I \) is the current flowing through the conductor, and \( A \) is the cross-sectional area perpendicular to the flow of current.

This concept is crucial because the current density helps us understand the limits of materials in circuits. Ensure that a circuit component doesn't exceed a certain current density to avoid overheating or damage.
Using this knowledge, you can design components like fuses that are intentionally designed to melt if dangerous levels are reached.

Properties of a Cylindrical Wire

A cylindrical wire is a common shape used in many electrical applications because of its symmetry and ease of manufacturing.
For a cylindrical wire, the cross-sectional area is crucial as it affects the wire's resistance and how much current it can safely carry.
The area \( A \) is calculated from the diameter \( d \) using the formula for the area of a circle:

• \( A = \pi \left( \frac{d}{2} \right)^2 \)

This relationship is important for understanding how changes in diameter affect the current capacity of the wire. A larger diameter allows more current to flow through without increasing the current density.
The length of the wire does not affect the area but can influence the total resistance of the wire. While calculating current density, focus only on the cross-sectional area.
Understanding these geometric properties helps in designing safe and efficient electrical systems.

Designing Fuses for Safety

Fuses are safety devices in electric circuits designed to protect equipment by breaking the circuit when current becomes dangerously high.
This is typically achieved by making the fuse from a wire designed to melt, thus stopping the current.
The key to fuse design is selecting the wire with the appropriate current density limit.
- In the exercise, the material used in the fuse has a current density threshold of \( 440 \, A/cm^2 \). - When the current exceeds what the wire can carry without heating excessively, the wire's temperature rises until it melts, breaking the circuit.
To design a fuse effectively:

• Choose materials with predictable melting points.
• Calculate the diameter needed to ensure the wire will open the circuit at the desired current using the provided current density.

For example, to limit current to \( 0.50 \, A \), calculate the wire diameter that ensures this current leads to a current density of \( 440 \, A/cm^2 \).
This calculation avoids guessing and ensures the system responds correctly under overload conditions. By doing so, safety is maintained without continuous manual monitoring of electrical systems.