Problem 26 - University Physics with Modern Physics

Question

Two long, parallel wires are separated by a distance of \(2.50 \mathrm{cm} .\) The force per unit length that each wire exerts on the other is 4.00 \(\times 10^{-5} \mathrm{N} / \mathrm{m}\) , and the wires repel each other. The current in one wire is 0.600 \(\mathrm{A}\) . (a) What is the current in the second wire? (b) Are the two currents in the same direction or in opposite directions?

Step-by-Step Solution

Answer

(a) The current in the second wire is 0.333 A. (b) The currents are in opposite directions.

Step 1: Understand the Problem

We have two parallel wires carrying currents that repel each other. We need to find the current in the second wire and determine the direction of the currents. The force per unit length between the wires is given, along with the current in one wire.

Step 2: Use the Formula for Magnetic Force Between Two Wires

The formula for the magnetic force per unit length between two parallel wires is given by: \[ F/L = \frac{\mu_0 I_1 I_2}{2\pi d} \] where \( F/L \) is the force per unit length (\(4.00 \times 10^{-5}\, \mathrm{N/m}\)), \( \mu_0 = 4\pi \times 10^{-7}\, \mathrm{T\cdot m/A}\) is the permeability of free space, \( I_1 = 0.600\, \mathrm{A} \) is the current in the first wire, \( I_2 \) is the current in the second wire, and \( d = 2.50\, \mathrm{cm} = 0.025\, \mathrm{m} \) is the distance between the wires.

Step 3: Solve for the Current in the Second Wire

Rearrange the formula to solve for \( I_2 \):\[ I_2 = \frac{2\pi d (F/L)}{\mu_0 I_1} \]Substitute the given values into the equation:\[ I_2 = \frac{2\pi (0.025\, \mathrm{m})(4.00 \times 10^{-5}\, \mathrm{N/m})}{(4\pi \times 10^{-7}\, \mathrm{T\cdot m/A})(0.600\, \mathrm{A})} \]\[ I_2 = 0.333\, \mathrm{A} \]

Step 4: Determine the Direction of the Currents

The problem states that the wires repel each other. According to the right-hand rule, if two wires repel, the currents must be in opposite directions. Hence, the current directions in the wires are opposite.

Key Concepts

Current Direction

In electromagnetism, the direction of current in a wire plays a crucial role in determining the interaction between two parallel wires carrying current. When currents flow in the same direction, the wires are attracted to each other. On the other hand, when currents flow in opposite directions, the wires experience a repulsive force.
This is because the magnetic fields created by the currents interact with each other, and the resulting force depends on the relative direction of the currents.

• If the currents are parallel and in the same direction, they create magnetic fields that attract one another.
• Conversely, if the currents are in opposite directions, the magnetic fields repel one another.

In this particular problem, it was stated that the parallel wires repel each other, indicating that the currents are flowing in opposite directions. This concept relies on the basic principles of magnetic field interactions and can be easily visualized using the right-hand rule.

Permeability of Free Space

The permeability of free space, commonly denoted as \( \mu_0 \), is a fundamental physical constant that describes the ability of a vacuum to support a magnetic field. It plays a crucial role in the equations of electromagnetism, providing a link between the magnetic force and the properties of the space around the wires.The value of the permeability of free space is \( \mu_0 = 4\pi \times 10^{-7} \, \mathrm{T\cdot m/A} \). This constant appears in the formula for calculating the magnetic force per unit length between two parallel current-carrying wires: \[ F/L = \frac{\mu_0 I_1 I_2}{2\pi d} \]

• This formula shows that the force per unit length is directly proportional to the product of the currents in the two wires (\(I_1\) and \(I_2\)).
• It is inversely proportional to the distance \(d\) between the wires.

Understanding the permeability of free space helps in grasping how magnetic forces operate in different environments and how alterations in distance or current affect the resultant forces.

Force Per Unit Length

The force per unit length between two parallel wires is a measure of the magnetic interaction between them. It's expressed as the force acting over a certain length of the wire. This concept is useful because it simplifies the analysis of magnetic forces in problems involving long wires.
Given by the formula:\[ F/L = \frac{\mu_0 I_1 I_2}{2\pi d} \]where \( F/L \) is the force per unit length, \( \mu_0 \) is the permeability of free space, \( I_1 \) and \( I_2 \) are the currents in the wires, and \( d \) is the separation between them.

• In this problem, the force per unit length was given as \( 4.00 \times 10^{-5} \, \mathrm{N/m} \).
• The force calculation involves understanding how different factors, such as the magnitude of current and the distance, affect this force.

By using this formula, you can determine how changes in current or the distance between wires influences the force, making it easier to design and predict outcomes in electrical circuits.