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Chapter 28: Problem 26

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?

Short Answer

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

Step by step solution

01

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.
02

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.
03

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:\[ I_2 = \frac{2\pi (0.025\, \mathrm{m})(4.00 \times 10^{-5}\, \mathrm{N/m})}{(4\pi \times 10^{-7}\, \mathrm{T\cdot m/A})(0.600\, \mathrm{A})} \]\[ I_2 = 0.333\, \mathrm{A} \]
04

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A long, straight wire carries a current of 2.50 A. An electron is traveling in the vicinity of the wire. At the instant when the electron is 4.50 \(\mathrm{cm}\) from the wire and traveling with a speed of \(6.00 \times 10^{4} \mathrm{m} / \mathrm{s}\) directly toward the wire, what are the magnitude and direction (relative to the direction of the current) of the force that the magnetic field of the current exerts on the electron?

Lamp Cord Wires. 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 current to a 100 -W light bulb connected across a \(120-\mathrm{V}\) 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 lamp cord? (Model the lamp cord as a very long straight wire.)

A long, straight wire lies along the \(y\) -axis and carries a current \(I=8.00\) A in the \(-y\) -direction (Fig. 28.39\()\) . In addition to the magnetic field due to the current in the wire, a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}_{0}\) with magnitude \(1.50 \times 10^{-6} \mathrm{T}\) is in the \(+x\) -direction \(\mathrm{What}\) is the total field (magnitude and direction) at the following points in the \(x z\) -plane: \((a) x=0, z=1.00 \mathrm{m}\) (b) \(x=1.00 \mathrm{m}, z=0 ;(\mathrm{c}) x=0\) \(z=-0.25 \mathrm{m} ?\)

A long, horizontal wire \(A B\) rests on the surface of a table and carries a current \(I\) . Horizontal wire \(C D\) is vertically above wire \(A B\) and is free to slide up and down on the two vertical metal guides \(C\) and \(D\) (Fig. 28.45\()\) . Wire \(C D\) is connected through the sliding contacts to another wire that also carries a current \(I,\) opposite in direction to the current in wire \(A B .\) The mass per unit length of the wire \(C D\) is \(\lambda\) . To what equilibrium height \(h\) will the wire \(C D\) rise, assuming that the magnetic force on it is due entirely to the current in the wire \(A B ?\)

A long solenoid with 60 turns of wire per centimeter carries a current of 0.15 \(\mathrm{A}\) . The wire that makes up the solenoid is wrapped around a solid core of silicon steel \(\left(K_{\mathrm{m}}=5200\right) .\) (The wire of the solenoid is jacketed with an insulator so that none of the current flows into the core.) (a) For a point inside the core, find the magnitudes of (i) the magnetic field \(\overrightarrow{\boldsymbol{B}}_{0}\) due to the solenoid current; (ii) the magnetization \(\vec{M} ;\) (iii) the total magnetic field \(\overrightarrow{\boldsymbol{B}}\) . (b) In a sketch of the solenoid and core, show the directions of the vectors \(\overrightarrow{\boldsymbol{B}}, \overrightarrow{\boldsymbol{B}}_{0}\) , and \(\overrightarrow{\boldsymbol{M}}\) inside the core.
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