/*! 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 10 A short current element \(d \vec... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A short current element \(d \vec{l}=(0.500 \mathrm{mm}) \hat{\jmath}\) carries a current of 8.20 \(\mathrm{A}\) in the same direction as \(d \vec{l} .\) Point \(P\) is located at \(\vec{r}=(-0.730 \mathrm{m}) \hat{\imath}+(0.390 \mathrm{m}) \hat{k} .\) Use unit vectors to express the magnetic field at \(P\) produced by this current element.

Short Answer

Expert verified
The magnetic field at P is \(d\vec{B} = (0, 0, 1.23 \times 10^{-9})\,\mathrm{T}.\)

Step by step solution

01

Understand the Problem

We have a short current element described by \(d \vec{l}=(0.500 \mathrm{mm}) \hat{\jmath}\) carrying a current of 8.20 A in the direction of \(d \vec{l}\). Point \(P\) is at \(\vec{r}=(-0.730 \mathrm{m}) \hat{\imath}+(0.390 \mathrm{m}) \hat{k}\). We need to find the magnetic field at \(P\) produced by this current element using the Biot-Savart Law.
02

Convert Units

Convert the current element length from millimeters to meters: \(0.500 \mathrm{mm} = 0.500 \times 10^{-3} \mathrm{m}\).
03

Apply the Biot-Savart Law

According to the Biot-Savart Law, the magnetic field \(d\vec{B}\) at point \(P\) due to a current element \(I d\vec{l}\) is given by:\[d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \vec{r}}{r^3},\]where \(\vec{r}\) is the position vector from the current element to point \(P\), and \(r\) is its magnitude.
04

Calculate the Cross Product

First, note that \(d \vec{l} = (0, 0.500\times10^{-3}, 0) \ \mathrm{m}\) and \(\vec{r} = (-0.730, 0, 0.390) \ \mathrm{m}\). Calculate the cross product:\[d\vec{l} \times \vec{r} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \ 0 & 0.500 \times 10^{-3} & 0 \ -0.730 & 0 & 0.390 \end{vmatrix} = (0, 0, 0.365) \times 10^{-3} \, \widehat{\imath} + (0, 0, 0.365) \times 10^{-3} \, \hat{k}\]
05

Find the Magnitude of r

The magnitude \(r\) of \(\vec{r}\) is calculated as:\[r = \sqrt{(-0.730)^2 + (0)^2 + (0.390)^2} = \sqrt{0.5329 + 0.1521} = \sqrt{0.6850} \approx 0.828 \mathrm{m}.\]
06

Substitute into the Biot-Savart Law

Using the cross product and the magnitude of \(r\), substitute into the Biot-Savart Law:\[d\vec{B} = \frac{4\pi \times 10^{-7}}{4\pi} \frac{8.20 \times (0.365 \times 10^{-3})}{(0.828)^3} \, \hat{k},\]and simplify to find,\[d\vec{B} = 1.23 \times 10^{-9} \, \hat{k} \, \mathrm{T}.\]
07

Express the Result

Finally, express the magnetic field at \(P\) as:\[d\vec{B} = (0, 0, 1.23 \times 10^{-9}) \ \mathrm{T}.\]

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
The magnetic field calculation in this context relies upon the Biot-Savart Law, an important principle in electromagnetism. For a small current element, this law provides a way to determine how a magnetic field is produced at a point due to the flow of electric current. The formula for the Biot-Savart Law is:\[d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \vec{r}}{r^3}\]Let's break it down:
  • \(d\vec{B}\) is the differential magnetic field at a point \(P\).
  • \(\mu_0\) is the permeability of free space, which has a constant value.
  • \(I\) is the current through the wire segment.
  • \(d\vec{l}\) is the infinitesimal length of the current element.
  • \(\vec{r}\) is the position vector from the current element to the point \(P\).
  • \(r\) is the magnitude of the position vector \(\vec{r}\).
In the example problem, the goal is to express the magnetic field at point \(P\) produced by a short current element. This requires precise calculation using the parameters given and applying vector analysis.
Current Element
A current element is a segment of a wire carrying an electric current. In our exercise, we analyzed a very short piece of the wire, denoted as \(d\vec{l} = (0.500 \times 10^{-3}) \hat{\jmath}\), which is expressed in meters after unit conversion. The significance of considering such a small segment lies in accurately determining its contribution to the magnetic field at a particular point.The current element is aligned in the direction of the current flow, which in our case is along the \(\hat{\jmath}\) axis. Here, \(d\vec{l}\) is paired with a current of \(8.20 \text{ A}\), indicating that a steady flow of electric charge exists in that direction. This alignment and magnitude create a specific magnetic field influence at any given point \(P\).When solving magnetic field problems using the Biot-Savart Law, distinguishing the components of the current segment becomes crucial. It allows the identification of how the electric current interacts with surrounding space to generate a magnetic field, and directs us in using the vectors' cross product for accurate calculations.
Cross Product Calculation
The cross product calculation is a key step when applying the Biot-Savart Law, as it determines the direction and relative magnitude of the magnetic field produced. The cross product of the current element vector \(d\vec{l}\) and the position vector \(\vec{r}\) is used to establish the perpendicular influence of the current on the magnetic field.In the exercise, the vectors are:
  • \(d\vec{l} = (0, 0.500 \times 10^{-3}, 0) \ \mathrm{m}\)
  • \(\vec{r} = (-0.730, 0, 0.390) \ \mathrm{m}\)
The cross product, \(d\vec{l} \times \vec{r}\), is calculated using the determinant:\[d\vec{l} \times \vec{r} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \ 0 & 0.500 \times 10^{-3} & 0 \ -0.730 & 0 & 0.390 \end{vmatrix} = (0, 0, 0.365 \times 10^{-3}) \ \mathrm{m^2}\]Points to remember:
  • The cross product results in a vector that is orthogonal to both \(d\vec{l}\) and \(\vec{r}\).
  • This orthogonal nature determines the direction of the magnetic field around the current element.
  • A zero component in the cross product indicates that some vector directions do not contribute to the magnetic field at \(P\).
Effectively grasping how to compute the cross product in vector calculations is fundamental in the study of electromagnetic field theory.

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 long, straight, solid cylinder, oriented with its axis in the \(z\) -direction, carries a current whose current density is \(\vec{J}\) . The current density, although symmetric about the cylinder axis, is not constant but varies according to the relationship $$ \begin{aligned} \vec{J} &=\frac{2 I_{0}}{\pi a^{2}}\left[1-\left(\frac{r}{a}\right)^{2}\right] \hat{k} & \text { for } r \leq a \\ &=0 \quad \text { for } r \geq a \end{aligned} $$ where \(a\) is the radius of the cylinder, \(r\) is the radial distance from the cylinder axis, and \(I_{0}\) is a constant having units of amperes. (a) Show that \(I_{0}\) is the total current passing through the entire cross section of the wire. (b) Using Ampere's law, derive an expression for the magnitude of the magnetic field \(\vec{B}\) in the region \(r \geq a\) . (c) Obtain an expression for the current \(I\) contained in a circular cross section of radius \(r \leq a\) and centered at the cylinder axis. (d) Using Ampere's law, derive an expression for the magnitude of the magnetic field \(\vec{B}\) in the region \(r \leq a .\) How do your results in parts (b) and (d) compare for \(r=a ?\)

A closely wound, circular coil with a diameter of 4.00 \(\mathrm{cm}\) has 600 turns and carries a current of 0.500 A. What is the magnitude of the magnetic field (a) at the center of the coil and (b) at a point on the axis of the coil 8.00 \(\mathrm{cm}\) from its center?

At a particular instant, charge \(q_{1}=+4.80 \times 10^{-6} \mathrm{C}\) is at the point \((0,0.250 \mathrm{m}, 0)\) and has velocity \(\vec{\boldsymbol{v}}_{1}=\left(9.20 \times 10^{5} \mathrm{m} / \mathrm{s}\right) \hat{\imath}\) Charge \(q_{2}=-2.90 \times 10^{-6} \mathrm{C}\) is at the point \((0.150 \mathrm{m}, 0,0)\) and has velocity \(\vec{v}_{2}=\left(-5.30 \times 10^{5} \mathrm{m} / \mathrm{s}\right) \hat{\boldsymbol{J}}\) . At this instant, what are the magnitude and direction of the magnetic force that \(q_{1}\) exerts on \(q_{2} ?\)

A flat, round iron ring 5.00 \(\mathrm{cm}\) in diameter has a current running through it that produces a magnetic field of 75.4\(\mu \mathrm{T}\) at its center. This ring is placed in a uniform external magnetic field of 0.375 T. What is the maximum torque the external field can exert on the ring? Show how the ring should be oriented relative to the field for the torque to have its maximum value.

BIO Currents in the Heart. The body contains many small currents caused by the motion of ions in the organs and cells. Measurements of the magnetic field around the chest due to currents in the heart give values of about 10\(\mu \mathrm{G}\) . Although the actual currents are rather complicated, we can gain a rough understanding of their magnitude if we model them as a long, straight wire. If the surface of the chest is 5.0 \(\mathrm{cm}\) from this current, how large is the current in the heart?

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.