/*! 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 30 Two long, parallel wires are sep... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two long, parallel wires are separated by a distance of 2.50 cm. The force per unit length that each wire exerts on the other is 4.00 \(\times\) 10\(^{-5}\) N/m, and the wires repel each other. The current in one wire is 0.600 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 0.833 A. (b) The currents are in opposite directions.

Step by step solution

01

Understand the Problem

We have two parallel wires separated by a certain distance, experiencing a mutual repulsion force per unit length. We know the current in one wire and need to find the current in the second wire. Additionally, we need to determine if the currents are in the same or opposite directions.
02

Recognize the Formula

The formula for the force per unit length between two parallel wires is given by:\[ F = \frac{\mu_0}{2\pi} \frac{I_1 I_2}{d} \]where \( F \) is the force per unit length, \( \mu_0 \) is the permeability of free space \((4\pi \times 10^{-7} \text{ N/A}^2)\), \( I_1 \) and \( I_2 \) are the currents in the wires, and \( d \) is the distance between the wires.
03

Plug in Known Values and Solve for Current

We know:- \( F = 4.00 \times 10^{-5} \text{ N/m} \)- \( I_1 = 0.600 \text{ A} \)- \( d = 0.0250 \text{ m} \)Plug these values into the formula:\[ 4.00 \times 10^{-5} = \frac{4\pi \times 10^{-7}}{2\pi} \frac{0.600 \times I_2}{0.0250} \]Simplify and solve for \( I_2 \):\[ I_2 = \frac{4.00 \times 10^{-5} \times 2 \pi \times 0.0250}{4\pi \times 10^{-7} \times 0.600} \approx 0.833 \text{ A} \]
04

Analyze the Force Direction

The problem states the wires repel each other. For parallel wires, currents in opposite directions repel, while currents in the same direction attract. Since these wires repel, the currents must be in opposite directions.

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 Fields
To understand the magnetic force between two parallel wires, we must first explore the concept of magnetic fields. Magnetic fields are invisible areas around a magnetic source that exert force on other magnetic materials and moving charges. When electric current flows through a wire, it generates a magnetic field around it. The direction of this field is determined by the Right-Hand Rule: if you point your thumb in the direction of the current, the curl of your fingers shows the magnetic field's direction.

This field interacts with other magnetic fields, creating forces that can either attract or repel the involved objects. Parallel wires carrying currents create magnetic fields that interact with each other. The strength and direction of these interactions are fundamental in calculating the force per unit length between the wires. Understanding this interaction is crucial because it allows us to apply mathematical laws, such as Ampere's Law, to predict and explain real-world phenomena.
Ampere's Law
Ampere's Law offers a powerful tool to relate the magnetic field around a conductor to the electric current passing through it. The law states: \[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc} \]where \( \vec{B} \) is the magnetic field, \( d\vec{l} \) is a differential length element through which the field is calculated, and \( I_{enc} \) is the current enclosed by the path. Here, \( \mu_0 \) is the permeability of free space, a constant value of \( 4\pi \times 10^{-7} \text{ N/A}^2 \).

In the context of two parallel wires, Ampere's Law helps us derive the formula for the magnetic force per unit length. By recognizing that each wire creates a magnetic field, we can calculate the force experienced by the other via:\[ F = \frac{\mu_0}{2\pi} \frac{I_1 I_2}{d} \]where \( F \) is the given force per length, and \( I_1 \), \( I_2 \) are the currents in the two wires. This relationship shows the direct dependency on both the current carried and the distance between the wires. Understanding how this equation is derived from Ampere’s Law is key to solving problems related to magnetic force between wires.
Current Direction Analysis
Analyzing the direction of current is vital to determine whether two wires will attract or repel each other. The direction of the current affects the direction of the magnetic field produced around the wire. When analyzing forces between wires:
  • If currents flow in the same direction, the magnetic fields will attract
  • If currents flow in opposite directions, the magnetic fields will repel

Given that the problem statement mentions repulsion between the wires, this inherently tells us the currents must flow in opposite directions. This principle helps not only in academic exercises but also in practical applications, such as designing electrical circuits to minimize interference or in industrial settings where strong magnetic interactions might occur. Correctly analyzing current directions can avert possible system malfunctions and optimize circuit performance.

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

Lightning bolts can carry currents up to approximately 20 kA. We can model such a current as the equivalent of a very long, straight wire. (a) If you were unfortunate enough to be 5.0 m away from such a lightning bolt, how large a magnetic field would you experience? (b) How does this field compare to one you would experience by being 5.0 cm from a long, straight household current of 10 A?

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.)

A wooden ring whose mean diameter is 14.0 cm is wound with a closely spaced toroidal winding of 600 turns. Compute the magnitude of the magnetic field at the center of the cross section of the windings when the current in the windings is 0.650 A.

A short current element \(\overrightarrow{dl}\) S = (0.500 mm)\(\hat{\imath}\) carries a current of 5.40 A in the same direction as \(\overrightarrow{dl}\). Point P is located at \(\overrightarrow{r} =\) (-0.730 m)\(\hat{\imath}\) + (0.390 m)\(\hat{k}\) . Use unit vectors to express the magnetic field at \(P\) produced by this current element.

The current in the windings of a toroidal solenoid is 2.400 A. There are 500 turns, and the mean radius is 25.00 cm. The toroidal solenoid is filled with a magnetic material. The magnetic field inside the windings is found to be 1.940 T. Calculate (a) the relative permeability and (b) the magnetic susceptibility of the material that fills the toroid.

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.