Problem 70 - Fundamentals of Physics

Question

A caterpillar of length \(4.0 \mathrm{~cm}\) crawls in the direction of electron drift along a 5.2-mm-diameter bare copper wire that carries a uniform current of 12 A. (a) What is the potential difference between the two ends of the caterpillar? (b) Is its tail positive or negative relative to its head? (c) How much time does the caterpillar take to crawl \(1.0 \mathrm{~cm}\) if it crawls at the drift speed of the electrons in the wire? (The number of charge carriers per unit volume is \(8.49 \times 10^{28} \mathrm{~m}^{-3}\).)

Step-by-Step Solution

Answer

(a) 0.38 mV; (b) Tail is positive; (c) 28.57 s.

Step 1: Calculate the Resistance of the Wire Segment

First, find the resistance of the wire segment using the formula \( R = \frac{\rho L}{A} \), where \( \rho \) is the resistivity of copper (approximately \(1.68 \times 10^{-8} \ \Omega \cdot \mathrm{m}\)), \( L \) is the length of the caterpillar (0.04 m), and \( A \) is the cross-sectional area of the wire. The diameter of the wire is 5.2 mm, so the radius \( r = 2.6 \times 10^{-3} \ \mathrm{m} \). The area \( A = \pi r^2 = \pi (2.6 \times 10^{-3})^2 \). So, \( A \approx 2.123 \times 10^{-5} \ \mathrm{m}^2 \). Now, compute the resistance: \( R \approx \frac{1.68 \times 10^{-8} \times 0.04}{2.123 \times 10^{-5}} \approx 3.17 \times 10^{-5} \ \Omega \).

Step 2: Calculate the Potential Difference

Use Ohm’s Law \( V = IR \) to find the potential difference across the caterpillar. Here, \( I = 12 \ \mathrm{A} \). Therefore, \( V = 12 \times 3.17 \times 10^{-5} \approx 3.8 \times 10^{-4} \ \mathrm{V} \).

Step 3: Determine the Polarity of Tail Relative to Head

In a wire through which the current flows, the current direction is from the positive towards the negative. Since the caterpillar moves in the direction of electron drift, electrons move from negative to the positive end, meaning its head is more negative relative to its tail.

Step 4: Calculate Drift Speed of Electrons

The drift speed \( v_d \) can be calculated using the formula \( v_d = \frac{I}{nAe} \), where \( I = 12 \ \mathrm{A} \), \( n = 8.49 \times 10^{28} \ \mathrm{m}^{-3} \) is the number of charge carriers per unit volume, \( A = 2.123 \times 10^{-5} \ \mathrm{m}^2 \) is the cross-sectional area, and \( e = 1.6 \times 10^{-19} \ \mathrm{C} \) is the elementary charge. Substituting these, \( v_d \approx \frac{12}{8.49 \times 10^{28} \times 2.123 \times 10^{-5} \times 1.6 \times 10^{-19}} \approx 3.5 \times 10^{-4} \ \mathrm{m/s} \).

Step 5: Calculate Time to Crawl 1 cm

To find the time, use \( t = \frac{d}{v_d} \), where \( d = 0.01 \ \mathrm{m} \) is the distance. So, \( t = \frac{0.01}{3.5 \times 10^{-4}} \approx 28.57 \ \mathrm{s} \).

Key Concepts

Ohm's Law

Ohm's Law is a fundamental principle in the study of electricity and magnetism. It states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points. This relationship is governed by the formula: \[ V = IR \] where

• \( V \) is the voltage (potential difference) in volts (V).
• \( I \) is the current in amperes (A).
• \( R \) is the resistance in ohms (Ω).

In the context of our exercise, we calculate the voltage across a caterpillar crawling along a wire.
By knowing the current (12 A) and the resistance (calculated as approximately \( 3.17 \times 10^{-5} \ Omega \)), Ohm's Law allows us to determine the potential difference to be about \( 3.8 \times 10^{-4} \, \) V. This example helps illustrate how understanding Ohm's Law is crucial in solving various electrical circuit problems.

Drift Velocity

Drift velocity refers to the average velocity at which free charge carriers, such as electrons, move through a conductor under the influence of an electric field. It can be thought of as a slow, directed movement superimposed over the random thermal motion of the electrons.
Drift velocity is given by the formula: \[ v_d = \frac{I}{nAe} \] where

• \( I \) is the current in amperes (A).
• \( n \) is the number of charge carriers per unit volume, in our exercise \( 8.49 \times 10^{28} \, \) carriers/m³.
• \( A \) is the cross-sectional area in square meters (m²).
• \( e \) is the elementary charge, approximately \( 1.6 \times 10^{-19} \) C.

From the exercise, we calculated the drift velocity of electrons to be approximately \( 3.5 \times 10^{-4} \, \) m/s.
Understanding drift velocity is essential because it helps us determine how quickly electric charges are moving through a conductor.
In this case, it helps us estimate how long the caterpillar takes to travel along the wire.

Resistivity

Resistivity is a property of materials that measures how strongly a material opposes the flow of electric current. It is an intrinsic property that depends only on the material itself and not on its shape or size.
Copper, for instance, has a low resistivity, which makes it an excellent conductor of electricity.
The resistivity of a material is typically denoted by \( \rho \) and is expressed in ohm-meters (Ω·m).
The resistance \( R \) of a wire can be calculated using the formula: \[ R = \frac{\rho L}{A} \] where

• \( \rho \) is the resistivity.
• \( L \) is the length of the wire in meters (m).
• \( A \) is the cross-sectional area in square meters (m²).

In the caterpillar's exercise, we used copper's resistivity (approx. \( 1.68 \times 10^{-8} \, \Omega \cdot \mathrm{m} \)) to find the wire's resistance, helping us determine the voltage difference.
Grasping the concept of resistivity enables students to predict how different materials will behave as conductors or insulators.

Current

Electric current refers to the flow of electric charge through a conductor, such as a wire. Current is measured in amperes (A) and represents the rate at which charge is flowing.
It's crucial to remember that, in conventional terms, current flows from the positive to the negative terminal of a power source.
However, electrons, which carry the charge, move in the opposite direction, from negative to positive.

• Current in a circuit is sustained by a potential difference (voltage) across the circuit, and it's influenced by the circuit's resistance.
• The current can be visualized as the total number of charges passing through a wire's cross-section per unit time.

In our exercise, the wire carried a steady current of 12 A.
This stable flow of current made it possible to calculate both the drift velocity and potential difference accurately.
Knowing the current is vital for solving various electrical problems, as it ties directly into Ohm's Law and other fundamental concepts.