/*! 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 43 Shows a cross section across a d... [FREE SOLUTION] | 91影视

91影视

Shows a cross section across a diameter of a long cylindrical conductor of radius \(a=2.00 \mathrm{~cm}\) carrying uniform current \(170 \mathrm{~A}\). What is the magnitude of the current's magnetic field at radial distance (a) 0, (b) \(1.00 \mathrm{~cm},(\mathrm{c}) 2.00 \mathrm{~cm}\) (wire's surface), and (d) \(4.00 \mathrm{~cm} ?\)

Short Answer

Expert verified
(a) 0 T, (b) 鈮 2.68脳10鈦烩伌 T, (c) 鈮 1.70脳10鈦烩伌 T, (d) 鈮 8.50脳10鈦烩伒 T.

Step by step solution

01

Understand the Concept and Formula

The magnetic field inside a conductor carrying a current, defined by Ampere's Law, can be derived. For a current-carrying wire, Ampere's Circuital Law is given: \[ B \cdot 2 \pi r = \mu_0 I_{enclosed} \]where \( B \) is the magnetic field, \( r \) is the distance from the center, and \( I_{enclosed} \) is the current enclosed by radius \( r \). For a uniform current distribution, \( I_{enclosed} = I \times \frac{r^2}{a^2} \), where \( a \) is the radius of the wire and \( I \) is the total current.
02

Calculate the Magnetic Field at the Center (r = 0)

At the center of the wire, \( r = 0 \), the enclosed current \( I_{enclosed} \) is zero since no current flows across the center point. Thus, by applying Ampere's Law:\[ B = 0 \]
03

Calculate the Magnetic Field Inside the Wire (r = 1.00 cm)

Use the formula derived from Ampere's law for points inside the conductor:\[ I_{enclosed} = I \times \frac{r^2}{a^2} = 170 \times \frac{(1.00)^2}{(2.00)^2} \; \text{A} \]\[ I_{enclosed} = 42.5 \; \text{A} \]Using Ampere's Law: \[ B \cdot 2 \pi (0.01) = \mu_0 (42.5) \]\[ B = \frac{\mu_0 (42.5)}{2 \pi (0.01)} \]\[ B \approx 2.68 \times 10^{-4} \; \text{T} \]
04

Calculate the Magnetic Field on the Surface (r = 2.00 cm)

Here, we apply the law at the wire's surface (equal to radius):\[ I_{enclosed} = 170 \; \text{A} \]\[ B \cdot 2 \pi (0.02) = \mu_0 (170) \]\[ B = \frac{\mu_0 (170)}{2 \pi (0.02)} \]\[ B \approx 1.70 \times 10^{-4} \; \text{T} \]
05

Calculate the Magnetic Field Outside the Wire (r = 4.00 cm)

At a point outside the conductor, the total current enclosed is the full current of the wire.\[ I_{enclosed} = 170 \; \text{A} \]\[ B \cdot 2 \pi (0.04) = \mu_0 (170) \]\[ B = \frac{\mu_0 (170)}{2 \pi (0.04)} \]\[ B \approx 8.50 \times 10^{-5} \; \text{T} \]
06

Summarize the Results

The calculated magnetic field magnitudes are:(a) At the center \((r = 0)\), \( B = 0 \; \text{T} \).(b) At \( r = 1.00 \; \text{cm} \), \( B \approx 2.68 \times 10^{-4} \; \text{T} \).(c) At \( r = 2.00 \; \text{cm} \), \( B \approx 1.70 \times 10^{-4} \; \text{T} \).(d) At \( r = 4.00 \; \text{cm} \), \( B \approx 8.50 \times 10^{-5} \; \text{T} \).

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 91影视!

Key Concepts

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

Magnetic Field Calculation
The magnetic field calculation is crucial when dealing with currents in conductors. According to Ampere's Law, we can find the magnetic field created by a current-carrying wire.
Ampere's Law is expressed as: \[ B \cdot 2 \pi r = \mu_0 I_{enclosed} \]where:
  • \( B \) is the magnetic field,
  • \( r \) is the radial distance from the center of the conductor,
  • \( \mu_0 \) is the permeability of free space (a constant),
  • \( I_{enclosed} \) is the current that is enclosed by the circular path of radius \( r \).

This equation helps derive the magnetic field within and around a cylindrical conductor by viewing the wire as a series of concentric rings of current.
This means the field strength depends on how much current is enclosed within each imaginary circular path.
The field's behavior also changes based on whether you are inside, on the surface, or outside the wire.
Current Distribution
When current flows through a cylindrical conductor, it is important to understand how it distributes inside the conductor. In this case, we consider that the wire carries a uniform current distribution.
This implies the current density remains constant across the cross-section of the conductor.### Uniform Current DistributionFor uniform distribution, the current at any point within the cylinder can be represented as:\[ I_{enclosed} = I \times \frac{r^2}{a^2} \],where:
  • \( I \) is the total current through the wire,
  • \( r \) is the radial distance from the center,
  • \( a \) is the radius of the conductor.
This formula helps determine what fraction of the total current is "seen" by a circular loop of radius \( r \) within the wire.
The uniform distribution means each small segment of the wire's cross-section carries an equal amount of current per unit area, important for calculating the magnetic field at various points.
Cylindrical Conductor
Cylindrical conductors are common in electrical applications because they provide a simple and effective way to transport electricity. Understanding the properties of these conductors is vital for solving problems involving magnetic fields.### Characteristics of Cylindrical Conductors
  • A cylindrical conductor has a consistent circular cross-section.
  • The radius \( a \) defines the size of the conductor.
  • The conductor in this exercise is long compared to its radius, implying the ends have negligible effects on magnetic field calculations near the center.
A key aspect for calculations is recognizing that the field patterns differ between the inside and outside of the conductor.
Within the conductor, the magnetic field results from current uniformly distributed across its cross-section.
Outside the conductor's surface, the magnetic field depends on the total current flowing through the conductor because all current is "enclosed" by paths further away from the center.
This simplifies calculations as if all the current is concentrated in the center of the wire, though in reality, it spreads across the entire cross-section.

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

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?

Three long wires are parallel to a \(z\) axis, and each carries a current of \(10 \mathrm{~A}\) in the positive \(z\) direction. Thein points of intersection with the \(x y\) plane form an equilateral triangle with sides of \(50 \mathrm{~cm}\), as shown in Fig. \(29-77\). A fourth wire (wire \(b\) ) passes through the midpoint of the base of the triangle and is parallel to the other three wires. If the net magnetic force on wire \(a\) is zero, what are the (a) size and (b) direction \((+z\) or \(-z)\) of the current in wire \(b ?\)

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 \((0,5.0 \mathrm{~m}, 0)\) ?

A long solenoid with \(10.0\) turns \(/ \mathrm{cm}\) and a radius of \(7.00 \mathrm{~cm}\) carries a current of \(20.0 \mathrm{~mA}\). A current of \(6.00 \mathrm{~A}\) exists in a straight conductor located along the central axis of the solenoid. (a) At what radial distance from the axis will the direction of the resulting magnetic field be at \(45.0^{\circ}\) to the axial direction? (b) What is the magnitude of the magnetic field there?

Shows wire 1 in cross section; the wire is long and straight, carries a current of \(4.00 \mathrm{~mA}\) out of the page, and is at distance \(d_{1}=2.40 \mathrm{~cm}\) from a surface. Wire 2, which is parallel to wire 1 and also long, is at horizontal distance \(d_{2}=5.00 \mathrm{~cm}\) from wire 1 and carries a current of \(6.80\) \(\mathrm{mA}\) into the page. What is the \(x\) component of the magnetic force per unit length on wire 2 due to wire \(1 ?\)

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.