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Chapter 20: Problem 53

\(\cdot\) Two high-current transmission lines carry currents of 25 \(\mathrm{A}\) and 75 \(\mathrm{A}\) in the same direction and are suspended parallel to each other 35 \(\mathrm{cm}\) apart. If the vertical posts supporting these wires divide the lines into straight 15 \(\mathrm{m}\) segments, what magnetic force does each segment exert on the other? Is this force attractive or repulsive?

Short Answer

Expert verified
The magnetic force is approximately 8.036 N and is attractive.

Step by step solution

01

Determine the Magnetic Force Formula

Magnetic force between two parallel wires can be calculated using the formula \( F = \frac{{ ext{{渭}}_0 imes I_1 imes I_2 imes L}}{2\pi imes d} \), where \( I_1 \) and \( I_2 \) are the currents in the wires, \( L \) is the length of the wire segment, \( d \) is the distance between the wires, and \( \text{{渭}}_0 \) is the permeability of free space \( = 4\pi \times 10^{-7} \, \text{{T m/A}} \).
02

Plug In the Given Values

Given that \( I_1 = 25 \, \text{A} \), \( I_2 = 75 \, \text{A} \), \( L = 15 \, \text{m} \), and \( d = 0.35 \, \text{m} \), substitute these values into the formula:\[F = \frac{(4\pi \times 10^{-7} \, \text{T m/A}) \times 25 \, \text{A} \times 75 \, \text{A} \times 15 \, \text{m}}{2\pi \times 0.35 \, \text{m}}.\]
03

Simplify the Expression

Cancel terms and simplify the calculation. The equation becomes:\[F = \frac{(4 \times 25 \times 75 \times 15)}{2 \times 0.35} \times 10^{-7} \, \text{N}.\]Compute the constants separately for simplification.
04

Calculate the Magnetic Force

Calculate the numerical value of the expression:\[F = \frac{(4 \times 25 \times 75 \times 15)}{0.7} \times 10^{-7}.\]This simplifies and computes to approximately:\[F \approx 8.036 \, \text{N}.\]
05

Determine the Nature of the Force

Since the currents are in the same direction, the magnetic force between the two wires will be attractive. Thus, each segment of the wire exerts an attractive force of approximately 8.036 N on the other.

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Key Concepts

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

Magnetic Interference
Magnetic interference occurs when two magnetic fields affect each other, which can happen when currents flow through parallel wires. This interference creates a magnetic force between the wires. The direction of the force depends on the direction of the currents:
If the currents flow in the same direction, the wires attract each other.
If they flow in the opposite direction, the wires repel each other. This phenomenon can be observed in high-current transmission lines. Understanding magnetic interference is crucial, as it impacts how we design electrical systems to manage these forces safely and effectively in real-world applications. Engineers must consider magnetic interference to prevent potential hazards or inefficiencies in power transmission. When designing large-scale electricity distribution systems, it's necessary to strategically place and orient the wires. This minimizes risks of unwanted attraction or repulsion, optimizing the stable and efficient transfer of electricity.
Current in Transmission Lines
High-current transmission lines are integral to delivering electricity over long distances. The current in these lines can greatly impact how they interact magnetically. Two key factors contributing to this interaction are the magnitude and direction of the current:
  • The magnitude of the current affects the strength of the magnetic field produced around the wire. Higher current strengths correspond to stronger fields.
  • The direction determines if the wires attract or repel each other.
Understanding these concepts allows engineers to manage the forces between transmission lines effectively. Properly balancing these forces ensures the structural integrity of wire supports and prevents power losses. Knowing how current behaves in transmission lines is also vital for selecting appropriate materials and configurations for electrical infrastructure, ensuring safety and efficiency in power distribution.
Permeability of Free Space
The permeability of free space, denoted as \(\mu_0\), is a fundamental physical constant. It plays a crucial role in calculating the magnetic force between current-carrying wires. Its value is \(4\pi \times 10^{-7} \, \text{T m/A}\) (Tesla meter per ampere).This constant reflects how a magnetic field interacts with the vacuum of free space. In simpler terms, it describes the extent to which a material can support the formation of a magnetic field. \(\mu_0\) is a central parameter in electromagnetic theories and equations, including those that calculate forces between parallel wires.The permeability of free space is used in formulas that describe relationships between electricity and magnetism. It allows us to predict interactions such as those in the exercise, where it helps compute the magnetic force between transmission lines. Grasping the importance of \(\mu_0\) is essential for anyone looking to understand the behavior of magnetic fields in physics and engineering applications.

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Most popular questions from this chapter

Magnetic sensitivity of electric fish. In a problem dealing with electric fish in Chapter 19, we saw that these fish navigate by responding to changes in the current in seawater. This current is due to a potential difference of around 3.0 \(\mathrm{V}\) generated by the fish and is about 12 \(\mathrm{mA}\) within a centimeter or so from the fish. Receptor cells in the fish are sensitive to the current. since the current is at some distance from the fish, the sensitivity of these cells suggests that they might be responding to the magnetic field created by the current. To get some estimate of how sensitive the cells are, we can model the current as that of a long, straight wire with the receptor cells 2.0 \(\mathrm{cm}\) away. What is the strength of the magnetic field at the receptor cells?

A long, straight telephone cable contains six wires, each carrying a current of 0.300 A. The distances between wires can be neglected. (a) If the currents in all six wires are in the same direction, what is the magnitude of the magnetic field 2.50 \(\mathrm{m}\) from the cable? (b) If four wires carry currents in one direction and the other two carry currents in the opposite direction, what is the magnitude of the field 2.50 m from the cable?

Household magnetic fields. Home circuit breakers typically have current capacities of around 10 A. How large a magnetic field would such a current produce 5.0 \(\mathrm{cm}\) from a long- wire's center? How does this field compare with the strength of the earth's magnetic field?

A solenoid having 165 turns and a cross-sectional area of 6.75 \(\mathrm{cm}^{2}\) carries a current of 1.20 A. If it is placed in a uniform 1.12 T magnetic field, find the torque this field exerts on the solenoid if its axis is oriented (a) perpendicular to the field, (b) parallel to the field, (c) at 35. \(0^{\circ}\) with the field.

A straight vertical wire carries a current of 1.20 \(\mathrm{A}\) down- ward in a region between the poles of a large electromagnet where the field strength is 0.588 \(\mathrm{T}\) and is horizontal. What are the magnitude and direction of the magnetic force on a 1.00 \(\mathrm{cm}\) section of this wire if the magnetic-field direction is (a) toward the east, (b) toward the south, (c) \(30.0^{\circ}\) south of west?
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