Problem 31 - University Physics with Modern Physics

Question

The wires in a household lamp cord are typically 3.0 mm apart center to center and carry equal currents in opposite directions. If the cord carries direct current to a 100-W light bulb connected across a 120-V potential difference, what force per meter does each wire of the cord exert on the other? Is the force attractive or repulsive? Is this force large enough so it should be considered in the design of the lamp cord? (Model the lamp cord as a very long straight wire.)

Step-by-Step Solution

Answer

The force is \(4.63 \times 10^{-5} \, \text{N/m}\) and attractive. It is too small to affect lamp cord design.

Step 1: Calculate the Current

First, we need to find the current flowing through the wires. Using the formula for power, \(P = IV\), where \(P = 100 \, \text{W}\) and \(V = 120 \, \text{V}\). Rearranging for current \(I\), we get \(I = \frac{P}{V} = \frac{100}{120} = 0.833 \, \text{A}\).

Step 2: Calculate the Magnetic Force per Unit Length

The force per unit length \(F/L\) between two parallel wires carrying currents in opposite directions can be calculated using the formula: \[ F/L = \frac{\mu_0}{2\pi} \cdot \frac{I_1 I_2}{d} \] where \(\mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}\) is the permeability of free space, \(I_1\) and \(I_2\) are the currents (0.833 A) and \(d\) is the distance between wires (0.003 m).

Step 3: Substitute and Calculate

Substituting the given values into the formula: \[ F/L = \frac{4\pi \times 10^{-7}}{2\pi} \cdot \frac{0.833 \times 0.833}{0.003} \] \[ F/L = 2 \times 10^{-7} \cdot \frac{0.693889}{0.003} = 4.63 \times 10^{-5} \, \text{N/m} \]

Step 4: Determine the Nature of Force

Since the currents are in opposite directions, the force between the wires is attractive according to the Ampère's force law.

Step 5: Consideration of Force in Design

The force calculated is \(4.63 \times 10^{-5} \, \text{N/m}\), which is relatively small. This force is not large enough to be a significant factor to consider in the design of a typical household lamp cord.

Key Concepts

Current Calculation

In this exercise, we are dealing with the concept of electrical current, which is a flow of electric charge carried by moving electrons in a wire. To find the current, we use the relationship between power, voltage, and current given by the equation \(P = IV\), where:

• \(P\) is the power in watts (W),
• \(I\) is the current in amperes (A),
• \(V\) is the voltage in volts (V).

For a 100-W light bulb connected to a 120-V source, the formula becomes \(I = \frac{P}{V}\). By substituting the given values, we calculate the current as \(I = \frac{100}{120} = 0.833 \text{ A}\).
This means each wire in the lamp cord carries an equal current of 0.833 A.

Ampère's Force Law

Ampère's force law is key to understanding how currents interact through magnetic forces. This law highlights that two parallel conductors carrying currents will exert forces on each other. Depending on the direction of the currents, the force can be attractive or repulsive.
- If the currents flow in the same direction, the force is attractive. - If they flow in opposite directions, the force is repulsive.
In our scenario, we have two currents of the same magnitude flowing in opposite directions, resulting in an attractive force. This means that the wires actually pull towards each other. Under practical conditions, Ampère’s force law explains many electromagnetic interactions we see in electrical circuits and devices.

Force per Unit Length

To calculate the force per unit length between two parallel wires, we use the specific formula for magnetic force per length \(\frac{F}{L}\):\[\frac{F}{L} = \frac{\mu_0}{2\pi} \cdot \frac{I_1 I_2}{d}\]where:

• \(\mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}\) (T = tesla), is the permeability of free space,
• \(I_1\) and \(I_2\) are the currents in amperes,
• \(d\) is the distance between the wires in meters.

With \(I_1 = I_2 = 0.833 \text{ A}\) and \(d = 0.003 \text{ m}\), substituting into the formula gives:\[\frac{F}{L} = \frac{4\pi \times 10^{-7}}{2\pi} \cdot \frac{0.833 \times 0.833}{0.003} = 4.63 \times 10^{-5} \text{ N/m}\]This force, although measurable, is quite small and does not significantly influence the design of everyday lamp cords, which are expected to handle small mechanical stresses.