/*! 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 13 A long, straight wire lies along... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A long, straight wire lies along the \(z\) -axis and carries a 4.00 -A current in the \(+z\) -direction. Find the magnetic field (magnitude and direction) produced at the following points by a \(0.500-\mathrm{mm}\) segment of the wire centered at the origin: (a) \(x=2.00 \mathrm{m}, y=0\) , \(z=0 ;(\) b) \(x=0, y=2.00 \mathrm{m}, z=0 ;(\mathrm{c}) x=2.00 \mathrm{m}, y=2.00 \mathrm{m}\) \(z=0 ;(\mathrm{d}) x=0, y=0, z=2.00 \mathrm{m}\)

Short Answer

Expert verified
(a) \(5.0 \times 10^{-11} \mathbf{\hat{j}}\) T, (b) \(-5.0 \times 10^{-11} \mathbf{\hat{i}}\) T, (c) \(1.767 \times 10^{-11} [-\mathbf{\hat{i}} + \mathbf{\hat{j}}]\) T, (d) 0 T.

Step by step solution

01

Understand the Biot-Savart Law

The Biot-Savart Law helps us calculate the magnetic field generated by a small segment of a current-carrying wire. It is expressed as:\[d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I\,d\mathbf{l} \times \mathbf{\hat{r}}}{r^2}\]where \(d\mathbf{B}\) is the magnetic field contribution, \(\mu_0\) is the permeability of free space, \(I\) is the current, \(d\mathbf{l}\) is the vector length of the wire segment, and \(\mathbf{\hat{r}}\) is the unit vector from the wire to the point of interest.
02

Express the Position Vectors and Magnitudes

For each point of interest, determine the position vector \(\mathbf{r}\) from the middle of the wire segment to the point and its magnitude.- (a) \(\mathbf{r} = (2.00, 0, 0) \) and magnitude \(r = 2.00 \) m.- (b) \(\mathbf{r} = (0, 2.00, 0) \) and magnitude \( r = 2.00 \) m.- (c) \(\mathbf{r} = (2.00, 2.00, 0) \) and magnitude \(r = \sqrt{(2.00)^2 + (2.00)^2} = 2\sqrt{2} \) m.- (d) \(\mathbf{r} = (0, 0, 2.00) \) and magnitude \( r = 2.00 \) m.
03

Calculate the Magnetic Field

Apply Biot-Savart Law:### For Parts (a), (b), and (c):Since \(d\mathbf{l} = 0.500\times 10^{-3} \mathbf{k}\) (along z-axis), use cross product for direction:\[d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{4.00 \times 0.5\times 10^{-3} \mathbf{k} \times \mathbf{r}}{r^3} \]Plug in vectors and calculate:- (a) Cross \( (0, 0, 1) \) with \( (2, 0, 0) \) yields direction \( \mathbf{\hat{j}} \).- (b) Cross \( (0, 0, 1) \) with \( (0, 2, 0) \) yields direction \( -\mathbf{\hat{i}} \).- (c) Cross \( (0, 0, 1) \) with \( (2, 2, 0) \) yields direction \( -\mathbf{\hat{i}} + \mathbf{\hat{j}} \).Calculate respective magnitudes using the formula: \[B = \frac{\mu_0}{4\pi} \frac{4.00 \times 0.5\times 10^{-3}}{r^3}\]### For Part (d):No magnetic field because the point is on the axis of the wire segment.
04

Summarize Results

Conclusion from previous calculations:- (a) \(B = \frac{\mu_0}{4\pi} \frac{2 \times 10^{-3}}{(2)^3} \mathbf{\hat{j}} = 5.0 \times 10^{-11} \mathbf{\hat{j}}\) T.- (b) \(B = \frac{\mu_0}{4\pi} \frac{2 \times 10^{-3}}{(2)^3} (-\mathbf{\hat{i}}) = -5.0 \times 10^{-11} \mathbf{\hat{i}}\) T.- (c) \(B = \frac{\mu_0}{4\pi} \frac{2 \times 10^{-3}}{(2^{3/2})^3} [-\mathbf{\hat{i}} + \mathbf{\hat{j}}] = 1.767 \times 10^{-11} [-\mathbf{\hat{i}} + \mathbf{\hat{j}}]\) T.- (d) \(B = 0\) 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 91Ó°ÊÓ!

Key Concepts

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

Magnetic Field Calculation
To find the magnetic field around a current-carrying wire, we can use the Biot-Savart Law. This law is a fundamental tool in electromagnetism for calculating the magnetic field produced by electric currents. It involves evaluating the contribution to the magnetic field from a small segment of the wire at various points in space.
The equation for the Biot-Savart Law is:\[d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I\,d\mathbf{l} \times \mathbf{\hat{r}}}{r^2}\]
This states that the differential magnetic field \(d\mathbf{B}\) is proportional to the cross product of the current element \(d\mathbf{l}\) and the unit vector \(\mathbf{\hat{r}}\), which points from the wire to the field point. The equation is divided by \(r^2\), emphasizing that the magnetic field weakens with the square of the distance from the wire segment.
For each specific point of interest, we perform the following steps:
  • Identify the three-dimensional coordinates of the point.
  • Determine the position vector \(\mathbf{r}\) pointing from the wire to the point.
  • Calculate the magnitude of \(\mathbf{r}\) to find \(r\).
Understanding these steps accurately allows us to simplify the complex nature of magnetic field calculations around current-carrying wires.
Current-Carrying Wire
Current-carrying wires create magnetic fields that circulate around the wire. The direction of this magnetic field can be determined using the right-hand rule. If the thumb of your right hand points in the direction of the current, your fingers curl in the direction of the magnetic field.
In the given exercise, we are dealing with a wire aligned along the \(z\)-axis carrying a current of 4.00 A in the \(+z\) direction. This configuration simplifies our analysis since we can treat the wire segment \(d\mathbf{l}\) as oriented in the \(\mathbf{k}\)-direction.
In practical scenarios:
  • The magnetic effect is prominent in the regions perpendicular to the wire, diminishing as you go further away.
  • It is essential to comprehend how the magnetic field interacts with other materials and influences charges within its vicinity.
    This understanding is crucial, especially when using devices such as solenoids and motors where current-carrying wires form essential parts.
Vector Operations
Vector operations play a critical role in electromagnetism, especially when dealing with current and magnetic fields. In the context of the Biot-Savart Law, the cross product operation \(d\mathbf{l} \times \mathbf{r}\) has a vital effect.
The Cross Product:
  • Given two vectors, the cross product results in a third vector perpendicular to the plane formed by the original two.
  • The magnitude of the cross product is given by \(|d\mathbf{l}||\mathbf{r}|\sin\theta\), where \(\theta\) is the angle between the vectors.
The Biot-Savart Law uses this operation to determine the direction of the magnetic field generated. In our exercise:
  • We evaluate \(d\mathbf{l} \times \mathbf{r}\) to pinpoint the orientation of the resultant magnetic field at each observation point.
  • Considering unit vectors simplifies this calculation and highlights the directionality without solving for the full vector magnitude.
Understanding vector operations is pivotal in navigating through electromagnetics, helping us visualize and solve complex problems with ease.
Electromagnetism
Electromagnetism is a branch of physics that studies the electromagnetic forces between charged particles. It unifies electricity and magnetism into a single coherent theory, governed by Maxwell's equations. The Biot-Savart Law is an integral part of this framework.
Key Points in Electromagnetism:
  • Electric currents generate magnetic fields, and changing magnetic fields can induce electric currents.
  • Understanding electromagnetism is crucial for various technological applications, including power generation, electric motors, and telecommunications.
In real-world applications, wires and currents create intricate electromagnetic environments. We use equations like Biot-Savart to predict the impact of these environments. Solutions must often consider field strengths, directions, and the effects on nearby materials. Electromagnetism not only explains natural phenomena but is also foundational in the design and functioning of modern technological devices.

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

BIO Currents in the Brain. The magnetic field around the head has been measured to be approximately \(3.0 \times 10^{-8}\) G. Although the currents that cause this field are quite complicated, we can get a rough estimate of their size by modeling them as a single circular current loop 16 \(\mathrm{cm}\) (the width of a typical head) in diameter. What is the current needed to produce such a field at the center of the loop?

A \(+6.00-\mu \mathrm{C}\) point charge is moving at a constant \(8.00 \times 10^{6} \mathrm{m} / \mathrm{s}\) in the \(+y\) -direction, relative to a reference frame. At the instant when the point charge is at the origin of this reference frame, what is the magnetic-field vector \(\vec{B}\) it produces at the following points: (a) \(x=0.500 \mathrm{m}, y=0, \quad z=0 ;\) (b) \(x=0\) \(y=-0.500 \mathrm{m}, z=0 ; \quad(\mathrm{c}) x=0, \quad y=0, z=+0.500 \mathrm{m} ;\) (d) \(x=0, y=-0.500 \mathrm{m}, z=+0.500 \mathrm{m} ?\)

As a new electrical technician, you are designing a large solenoid to produce a uniform \(0.150-\) T magnetic field near the center of the solenoid. You have enough wire for 4000 circular turns. This solenoid must be 1.40 \(\mathrm{m}\) long and 2.80 \(\mathrm{cm}\) in diameter. What current will you need to produce the necessary field?

Coaxial Cable. A solid conductor with radius \(a\) is supported by insulating disks on the axis of a conducting tube with inner radius \(b\) and outer radius \(c\) (Fig. \(\mathrm{E} 28.45\) ). The central conductor and tube carry equal currents \(I\) in opposite directions. The currents are distributed uniformly over the cross sections of each conductor. Derive an expression for the magnitude of the magnetic field (a) at points outside the central, solid conductor but inside the tube \((ac) .\)

A toroidal solenoid with 400 turns of wire and a mean radius of 6.0 \(\mathrm{cm}\) carries a current of 0.25 A. The relative permeability of the core is \(80 .\) (a) What is the magnetic field in the core? (b) What part of the magnetic field is due to atomic currents?

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.