/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 83 Two parallel rods are each 0.50 ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two parallel rods are each 0.50 m in length. They are attached at their centers to either end of a spring (spring constant \(=150 \mathrm{N} / \mathrm{m} )\) that is initially neither stretched nor compressed. When 950 \(\mathrm{A}\) of current is in each rod in the same direction, the spring is observed to be compressed by 2.0 \(\mathrm{cm}\) . Treat the rods as long, straight wires and find the separation between them when the current is present.

Short Answer

Expert verified
The separation between the rods is approximately 2.1 cm.

Step by step solution

01

Identify the Forces at Play

The exercise involves determining the separation distance between two wires carrying currents. The current produces a magnetic force that influences the spring's length. Here, the spring's force and the magnetic force between the currents create equilibrium when the spring is compressed.
02

Calculate the Magnetic Force

The magnetic force per unit length between two parallel currents is given by \[ F/L = \frac{\mu_0 I_1 I_2}{2\pi d} \]where \( F \) is the force, \( L \) is the length of the wires, \( \mu_0 = 4\pi \times 10^{-7} \, \mathrm{T\cdot m/A} \) is the permeability of free space, \( I_1 \) and \( I_2 \) are the currents, and \( d \) is the distance between the wires. Here, \( I_1 = I_2 = 950 \, \mathrm{A} \).
03

Relate Magnetic Force to Spring Force

The magnetic force \( F \) when the spring is compressed should equal the spring force exerted by the compression. The spring force \( F_s \) is given by \[ F_s = k \times \Delta x \]where \( k = 150 \, \mathrm{N/m}\) is the spring constant and \( \Delta x = 0.02 \, \mathrm{m} \) is the change in length of the spring.
04

Equate Forces and Solve for Distance

Setting the magnetic force equal to the spring force,\[ \frac{\mu_0 I_1 I_2}{2\pi d} \times L = k \times \Delta x \]Solving for \( d \), we get:\[ d = \frac{\mu_0 I^2 L}{2\pi k \Delta x} \]Substitute the known values into the equation:\[ d = \frac{(4\pi \times 10^{-7})(950)^2(0.50)}{2\pi (150)(0.02)} \]Perform the calculations to find \( d \).
05

Calculate and Present the Result

After plugging in the values and performing the arithmetic,\[ d = \frac{4\pi \times 10^{-7} \times 902500 \times 0.5}{2\pi \times 150 \times 0.02} \]\[ d \approx 0.021 \, \mathrm{m} = 2.1 \, \mathrm{cm} \]Thus, the separation between the rods, when the current is present, is approximately 2.1 cm.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Magnetic Field Calculation
Understanding the calculation of the magnetic field is key in problems involving current-carrying wires. When two parallel wires carry current, they interact through magnetic forces. The force that each wire exerts on the other can be calculated per unit length using the formula:
  • \( F/L = \frac{\mu_0 I_1 I_2}{2\pi d} \)
where \( F \) is the magnetic force, \( L \) is the wire length, \( I_1 \) and \( I_2 \) are the currents in the wires, and \( d \) is the separation between them. The formula involves the permeability of free space, \( \mu_0 \), which is a constant, equal to \( 4\pi \times 10^{-7} \, \mathrm{T\cdot m/A} \). Understanding this equation helps you determine how the magnetic forces change with varying current levels or distances between wires. A greater current or a smaller distance will result in a larger magnetic force.
Spring Force Equilibrium
In situations where a spring changes its length, it is often due to equilibrium between different forces acting on it. Here, we are balancing the magnetic force generated by the currents in the wires with the spring force that comes into play when the spring is compressed or stretched. The spring force \( F_s \) can be calculated using Hooke's Law:
  • \( F_s = k \times \Delta x \)
where \( k \) is the spring constant, and \( \Delta x \) is the change in the spring's length. In this problem, the spring is compressed by 2 cm, creating a spring force that equals the magnetic force between the currents, allowing us to solve for the separation between wires. The spring's force must counterbalance the magnetic attraction to maintain the system in equilibrium.
Permeability of Free Space
The permeability of free space, denoted as \( \mu_0 \), is a fundamental physical constant necessary in the calculation of magnetic fields. It is a measure of how much resistance the vacuum of space poses against magnetic field lines. In the context of current-carrying wires, it plays a critical role in quantifying the strength of the magnetic interaction between the wires. This constant has a value of \( 4\pi \times 10^{-7} \, \mathrm{T\cdot m/A} \).
  • In our case, \( \mu_0 \) appears in the formula for the magnetic force, where it dictates how strongly the wires' magnetic fields interact given their currents.
Understanding \( \mu_0 \) helps in appreciating how magnetic fields propagate in environments devoid of matter, which remains consistent across various applications of electromagnetism.
Current-Carrying Conductors
Wires that carry current, known as current-carrying conductors, exhibit interesting behaviors due to their magnetic properties. When a current flows through a conductor, it creates a magnetic field around it. This field can exert forces on other nearby conductors carrying currents.
  • The direction of the force between two parallel wires depends on whether the currents are in the same or opposite directions.
  • If currents flow in the same direction, they attract each other; if in opposite directions, they repel.
In the problem, both wires carry currents in the same direction, resulting in an attractive magnetic force. This fundamental principle is exploited in many practical applications, such as magnetic levitation and the functioning of electric motors. Observing how these forces manifest in conductors is critical for their use in designing electrical and magnetic systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Particle 1 and particle 2 have masses of \(m_{1}=2.3 \times 10^{-8} \mathrm{kg}\) and \(m_{2}=5.9 \times 10^{-8} \mathrm{kg},\) but they carry the same charge \(q .\) The two particles accelerate from rest through the same electric potential difference \(V\) and enter the same magnetic field, which has a magnitude \(B\). The particles travel perpendicular to the magnetic field on circular paths. The radius of the circular path for particle 1 is \(r_{1}=12 \mathrm{cm} .\) What is the radius (in cm) of the circular path for particle 2\(?\)

Two charged particles move in the same direction with respect to the same magnetic field. Particle 1 travels three times faster than particle 2. However, each particle experiences a magnetic force of the same magnitude. Find the ratio \(\left|q_{1}\right| /\left|q_{2}\right|\) of the magnitudes of the charges.

ssm In New England, the horizontal component of the earth's magnetic field has a magnitude of \(1.6 \times 10^{-5}\) T. An electron is shot vertically straight up from the ground with a speed of \(2.1 \times 10^{6} \mathrm{m} / \mathrm{s}\) . What is the magnitude of the acceleration caused by the magnetic force? Ignore the gravitational force acting on the electron.

An ionized helium atom has a mass of \(6.6 \times 10^{-27} \mathrm{kg}\) and a speed of \(4.4 \times 10^{5} \mathrm{m} / \mathrm{s}\) . It moves perpendicular to a \(0.75-\mathrm{T}\) magnetic field on a circular path that has a 0.012 -m radius. Determine whether the charge of the ionized atom is \(+e\) or \(+2 e .\)

A proton is projected perpendicularly into a magnetic field that has a magnitude of 0.50 T. The ficld is then adjusted so that an electron will follow a circular path of the same radius when it is projected perpendicularly into the field with the same velocity that the proton had. What is the magnitude of the field used for the electron?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.