/*! 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 79 A long, hollow, cylindrical cond... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A long, hollow, cylindrical conductor (with inner radius \(2.0\) \(\mathrm{mm}\) and outer radius \(4.0 \mathrm{~mm}\) ) carries a current of \(24 \mathrm{~A}\) distributed uniformly across its cross section. A long thin wire that is coaxial with the cylinder carries a current of \(24 \mathrm{~A}\) in the opposite direction. What is the magnitude of the magnetic field (a) \(1.0\) \(\mathrm{mm}\), (b) \(3.0 \mathrm{~mm}\), and (c) \(5.0 \mathrm{~mm}\) from the central axis of the wire and cylinder?

Short Answer

Expert verified
The magnetic fields are 0 T at 1 mm and 5 mm, and \(\mu_0 \times \frac{3}{2\pi \times 0.003} \) T at 3 mm.

Step by step solution

01

Understanding the System

We have a hollow cylindrical conductor with inner radius of 2.0 mm and outer radius of 4.0 mm, carrying a current of 24 A. A coaxial wire carries 24 A in the opposite direction. We need to find the magnetic field at specific distances from the central axis of this system.
02

Using Ampere's Law

Ampere's Law helps us find the magnetic field: \(\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{enc}\), where \(I_{enc}\) is the current enclosed by the path. We'll use this to find the magnetic field inside, between, and outside the cylindrical conductor.
03

Calculating Magnetic Field at 1 mm

At 1 mm from the axis, we're inside the wire, where no current from the cylinder is enclosed. Thus, the magnetic field depends only on the wire with \(I = 24\) A in the opposite direction. Since the wire's radius is 0, the enclosed current \(I_{enc} = 0\). This gives \(B = 0\) at 1 mm.
04

Calculating Magnetic Field at 3 mm

At 3 mm, we're inside the material of the cylindrical conductor. Here, the current enclosed, \(I_{enc}\), is proportional to the area enclosed: \(\frac{(3 \, \text{mm}^2 - 2 \, \text{mm}^2)}{(4 \, \text{mm}^2 - 2 \, \text{mm}^2)} \times 24 \, \text{A} = 3 \, \text{A}\). Using Ampere's Law: \(B \cdot 2\pi \cdot 0.003 = \mu_0 \times 3\), which gives \(B = \mu_0 \times \frac{3}{2\pi \cdot 0.003}\).
05

Calculating Magnetic Field at 5 mm

At 5 mm, the path encloses the whole cylinder (current = 24 A) and the wire (current = -24 A). The total enclosed current is 0, so by Ampere's Law, \(B = 0\) at this distance too.

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.

Ampere's Law
Ampere's Law is a fundamental principle in electromagnetism that relates the magnetic field circulating around a closed loop to the current passing through the area enclosed by the loop. It is mathematically expressed as:
  • \( \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{enc} \)
Here, \(B\) represents the magnetic field, \(dl\) is a differential element of the loop, \(\mu_0\) is the permeability of free space, and \(I_{enc}\) refers to the enclosed current. This law is particularly useful when dealing with symmetrically distributed currents, as in the exercise with coaxial conductors.
In practical terms, the law allows us to analyze the magnetic effects around conductors by summing the influence of distributed currents, simplifying the calculation of the magnetic field's intensity and direction.
Magnetic Field
The magnetic field, denoted as \( B \), is a vector field around a magnetic material or a moving electric charge, which exerts a force on other nearby moving charges or magnetic materials. Magnetic fields behave according to specific rules and principles, such as:
  • They have both magnitude and direction.
  • They circulate around current-carrying wires or magnets.
In the scenario of coaxial conductors, the use of Ampere’s Law illustrates how to calculate the magnetic field at different distances from the axis. For our exercise, the field changes depending on whether we are inside, outside, or between the wire and cylinder. Understanding how the magnetic field behaves in different regions is key to solving related electromagnetism problems.
Current Distribution
Current distribution refers to how electric current is spread across the surface or volume of a conductor. In this problem, the current distribution within the hollow cylindrical conductor and the coaxial wire can greatly affect the magnetic field. The current is assumed to be uniformly distributed across the cross-section of the cylindrical conductor. Thus:
  • This determines how much current is enclosed by any chosen Amperian loop.
  • The distribution profile directly contributes to the magnetic field calculated via Ampere's Law.
At different points within our system, like 1 mm, 3 mm, and 5 mm from the central axis, it is crucial to assess which portions of the conductor's current are enclosed within a certain distance to compute the local magnetic field's magnitude and direction.
Coaxial Conductors
Coaxial conductors consist of two or more coplanar conductors, typically organized with one within the other and sharing a common central axis. They are essential in this exercise as the arrangement defines the problem's symmetry and current interactions. In such setups:
  • The inner conductor might be a wire, and the outer one a cylindrical shell.
  • The configuration allows current to flow in opposing directions, creating useful properties in magnetic field interactions.
For coaxial conductors, the exercise demonstrates how current flowing in opposite directions in a wire and a cylindrical conductor results in specific field behaviors both inside and outside them. At certain points, these flows may cancel each other's field effects, crucial for understanding electromagnetic designs and practical applications, such as in cable shielding.

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

A conductor carries \(6.0 \mathrm{~A}\) along the closed path abcdefgha running along 8 of the 12 edges of a cube of edge length 10 \(\mathrm{cm}\). (a) Taking the path to be a combination of three square current loops \((b c f g b, a b g h a\), and \(c d e f c)\), find the net magnetic moment of the path in unit- vector notation. (b) What is the magnitude of the net magnetic field at the \(x y z\) coordinates of \((0,5.0 \mathrm{~m}, 0)\) ?

Four long straight wires are perpendicular to the page, and their cross sections form a square of edge length \(a=20 \mathrm{~cm}\). The currents are out of the page in wires 1 and 4 and into the page in wires 2 and 3 , and each wire carries 20 A. In unit-vector notation, what is the net magnetic field at the square's center?

Shows a snapshot of a proton moving at velocity \(\vec{v}=(-200 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) toward a long straight wire with current \(i=\) \(350 \mathrm{~m} \mathrm{~A}\). At the instant shown, the proton's distance from the wire is \(d=2.89 \mathrm{~cm} .\) In unit-vector nota- tion, what is the magnetic force on the proton due to the current?

Shows an arrangement known as a Helmholtz coil. It consists of two circular coaxial coils, each of 200 turns and radius \(R=25.0 \mathrm{~cm}\), separated by a distance \(s=R\). The two coils carry equal currents \(i=12.2 \mathrm{~mA}\) in the same direction. Find the magnitude of the net magnetic field at \(P\), midway between the coils.

A long wire is known to have a radius greater than \(4.0 \mathrm{~mm}\) and to carry a current that is uniformly distributed over its cross section. The magnitude of the magnetic field due to that current is \(0.28 \mathrm{mT}\) at a point \(4.0 \mathrm{~mm}\) from the axis of the wire, and \(0.20 \mathrm{mT}\) at a point 10 \(\mathrm{mm}\) from the axis of the wire. What is the radius of the wire?

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.