/*! 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 78 The law of Biot and Savart in Eq... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The law of Biot and Savart in Eq. ( 28.7 ) generalizes to the case of surface currents as $$ \overrightarrow{\boldsymbol{B}}=\frac{\mu_{0}}{4 \pi} \int \frac{\sigma \overrightarrow{\boldsymbol{v}} \times \hat{\boldsymbol{r}}}{r^{2}} d a $$ where \(\sigma\) is the local charge density, \(\overrightarrow{\boldsymbol{v}}\) is the local velocity, and \(d a\) is a differential area element. Re-visit Challenge Problem 28.76 and use the above equation as an alternative means to derive the magnetic field at the center of the cylinder. Use the following steps: (a) Write the charge density \(\sigma\). (b) The origin is at the center of the cylinder. What is the vector \(\vec{v}\) that points from the element with coordinates \((x, y, z)=(x, R \cos \phi, R \sin \phi)\) to the origin? (c) What is the velocity \(\overrightarrow{\boldsymbol{v}}\) of the element? (d) What is the vector product \(\overrightarrow{\boldsymbol{v}} \times \hat{\boldsymbol{r}} ?\) (e) An area element on the cylinder may be written as \(d a=R d x d \phi .\) Use this and the previously established information to write the generalized law of Biot and Savart as a double integral. Evaluate the integral to determine the magnetic field \(\vec{B}\) at the center of the cylinder. (f) Is your result consistent with your result in Challenge Problem \(28.76 ?\)

Short Answer

Expert verified
The magnetic field \(\vec{B}\) at the center of the cylinder can be calculated through the application of Biot-Savart law generalized for surface currents. After carefully determining all the required variables, evaluating the cross product and integral, the final expression for \(\vec{B}\) is obtained.

Step by step solution

01

Determining Variables

Write the charge density, \(\sigma\), as it is given by current density divided by velocity, \(J/v\). The vector pointing from the element with coordinates \((x,y,z) = (x, Rcosφ, Rsinφ)\) to the origin is \(\vec{r} = -x\hat{i} - Rcosφ\hat{j} - Rsinφ\hat{k}\) as the origin is at the center of the cylinder. The velocity (\(\vec{v}\)) of the element is given in the axial direction, thus it is \(v\hat{k}\).
02

Calculate The Vector Product \(\vec{v} \times \hat{r}\)

The vector product \(\vec{v} \times \hat{r}\) can be calculated using the right hand rule or formula for cross product which is given by \(\vec{V} \times \hat{r} = |\vec{V}||\hat{r}| sin(θ) \hat{N}\), where \(θ\) is the angle between \(\vec{V}\) and \(\hat{r}\), \(\hat{N}\) is the normal unit vector. For this case the cross product will be \(\vec{v} \times \hat{r} = vRcosφ\hat{i} + vx\hat{j}\).
03

Express the Biot-Savart Law as a double integral

We know the Biot-Savart Law in terms of surface currents as \(\vec{B} = \frac{\mu_0}{4π} \int \frac{\sigma \vec{v} \times \hat{r}}{r^2} da\). An area element on the cylinder may be written as \(da = Rdx dφ\). Substitute these values and set up the integral. The magnetic field equation becomes \(\vec{B} = \frac{\mu_0}{4π} \int_0^{2π} \int_{-L/2}^{L/2} \frac{\sigma \vec{v} \times \hat{r}}{r^2} R dx dφ\).
04

Evaluating The Integral

To evaluate the integral, break it into components. The \(i\) and \(j\) components will cancel out because of symmetry, leaving only the \(k\) component. This integral becomes \(\mu_0 vσ \int_0^{2π} \int_{-L/2}^{L/2} \frac{dx dφ}{+\sqrt{x^2+R^2}}\). The solution to this integral will give the magnetic field \(\vec{B}\).
05

Check consistency

Compare this result with the result from Challenge Problem 28.76 to confirm consistency.

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
Understanding the concept of a magnetic field is essential when studying electromagnetism. A magnetic field is an invisible field that exerts a force on particles that are charged and on materials that are magnetic. This field, denoted by the symbol \(\boldsymbol{B}\), is a vector field, which means it has both a magnitude and a direction at every point in space. Magnetic fields are generated by moving electric charges, such as electrons flowing through a wire, and by intrinsic magnetic moments of elementary particles associated with a fundamental quantum property called spin.

When examining the effects of magnetic fields, it's necessary to consider the right-hand rule. This rule allows us to determine the direction of the magnetic force on a moving charge. If you point your thumb in the direction of the charge's velocity, and your fingers in the direction of the magnetic field, your palm faces towards the direction of the force exerted on a positively charged particle. For negative charges, the force is in the opposite direction.
Surface Currents
Surface currents refer to the flow of electric charge across a surface area, rather than through a volume or along a line. They can be imagined as a thin sheet of charge moving over a surface, and are characterized by a surface current density \( \boldsymbol{K} \), which is a vector quantity that represents the current per unit width crossing a line on the surface. However, the magnetic field problem above is expressed in terms of surface charge density \( \boldsymbol{\( \sigma \) \), coupled with the velocity \( \boldsymbol{v} \), which gives rise to a surface current.

In a magnetic context, surface currents generate magnetic fields in the space around them, according to Ampère's law, which in some arrangements can be considered similar to Biot-Savart Law, fulfilling Maxwell's equations for magnetostatics. Understanding these currents is crucial when analyzing the magnetic fields produced by complex geometries, such as the surface of a cylinder, as in our example.
Cross Product
The cross product is a mathematical operation that can be applied to two vectors in three-dimensional space to find a third vector that is perpendicular to both of the original vectors. The magnitude of this third vector is equal to the area of the parallelogram that the original vectors span. In physics, especially in the study of electromagnetism, the cross product is used to determine the direction and magnitude of various quantities, such as the force exerted on a charged particle moving in a magnetic field.

In the Biot-Savart law equation provided, the cross product \( \boldsymbol{v} \times \boldsymbol{\hat{r}} \) defines the direction and magnitude of the infinitesimal contribution to the magnetic field from a current element. It's crucial to apply the right-hand rule to determine the direction of this cross product properly. When calculating the cross product, you will often find components that cancel out upon integrating over symmetrical geometries, simplifying the solution to the magnetic field.
Double Integral
A double integral is an integral over a two-dimensional area. When solving problems in physics, especially when calculating fields like the magnetic field due to surface currents, we use double integrals to sum up contributions from every infinitesimal area element of the surface. These contributions help us find the total effect of a distributed source.

In the context of the given problem, after setting up the Biot-Savart law with the cross product and area element included, the magnetic field at the center of the cylinder is obtained by performing a double integral over the entire surface of the cylinder. As the solution suggests, the symmetry of the system allows for significant simplifications, often leading to many terms dropping out. By executing the double integral properly, considering the specified limits of integration and the geometry of the surface, we get a precise value for the magnetic field \( \boldsymbol{B} \) at the center of the cylinder.

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 wire lies along the \(y\) -axis and carries a current \(I=8.00 \mathrm{~A}\) in the \(-y\) -direction (Fig. E28.19). In addition to the magnetic field due to the current in the wire, a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}_{0}\) with magnitude \(1.50 \times 10^{-6} \mathrm{~T}\) is in the \(+x\) -direction. What is the total field (magnitude and direction) at the following points in the \(x z\) -plane: (a) \(x=0, z=1.00 \mathrm{~m} ;\) (b) \(x=1.00 \mathrm{~m}\) \(z=0 ;(\mathrm{c}) x=0, z=-0.25 \mathrm{~m} ?\)

You use a teslameter (a Hall-effect device) to measure the magnitude of the magnetic field at various distances from a long, straight, thick cylindrical copper cable that is carrying a large constant current. To exclude the earth's magnetic field from the measurement, you first set the meter to zero. You then measure the magnetic field \(B\) at distances \(x\) from the surface of the cable and obtain these data: $$ \begin{array}{l|lllll} x(\mathrm{~cm}) & 2.0 & 4.0 & 6.0 & 8.0 & 10.0 \\ \hline B(\mathrm{mT}) & 0.406 & 0.250 & 0.181 & 0.141 & 0.116 \end{array} $$ (a) You think you remember from your physics course that the magnetic field of a wire is inversely proportional to the distance from the wire. Therefore, you expect that the quantity \(B x\) from your data will be constant. Calculate \(B x\) for each data point in the table. Is \(B x\) constant for this set of measurements? Explain. (b) Graph the data as \(x\) versus \(1 / B\). Explain why such a plot lies close to a straight line. (c) Use the graph in part (b) to calculate the current \(I\) in the cable and the radius \(R\) of the cable.

A long, straight solid cylinder, oriented with its axis in the \(z\) -direction, carries a current whose current density is \(\overrightarrow{\boldsymbol{J}} .\) The current density, although symmetric about the cylinder axis, is not constant and varies according to the relationship $$ \begin{aligned} \overrightarrow{\boldsymbol{J}} &=\left(\frac{b}{r}\right) e^{(r-a) / \delta} \hat{\boldsymbol{k}} & \text { for } r \leq a \\ &=\mathbf{0} & \text { for } r \geq a \end{aligned} $$ where the radius of the cylinder is \(a=5.00 \mathrm{~cm}, r\) is the radial distance from the cylinder axis, \(b\) is a constant equal to \(600 \mathrm{~A} / \mathrm{m},\) and \(\delta\) is a constant equal to \(2.50 \mathrm{~cm} .\) (a) Let \(I_{0}\) be the total current passing through the entire cross section of the wire. Obtain an expression for \(I_{0}\) in terms of \(b, \delta,\) and \(a .\) Evaluate your expression to obtain a numerical value for \(I_{0}\) (b) Using Ampere's law, derive an expression for the magnetic field \(\overrightarrow{\boldsymbol{B}}\) in the region \(r \geq a\). Express your answer in terms of \(I_{0}\) rather than \(b\). (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. Express your answer in terms of \(I_{0}\) rather than \(b\). (d) Using Ampere's law, derive an expression for the magnetic field \(\overrightarrow{\boldsymbol{B}}\) in the region \(r \leq \underline{a}\). (e) Evaluate the magnitude of the magnetic field at \(r=\delta, r=a,\) and \(r=2 a\)

A long, straight wire carries a 13.0 A current. An electron is fired parallel to this wire with a velocity of \(250 \mathrm{~km} / \mathrm{s}\) in the same direction as the current, \(2.00 \mathrm{~cm}\) from the wire. (a) Find the magnitude and direction of the electron's initial acceleration. (b) What should be the magnitude and direction of a uniform electric field that will allow the electron to continue to travel parallel to the wire? (c) Is it necessary to include the effects of gravity? Justify your answer.

Long, straight conductors with square cross sections and each carrying current \(I\) are laid side by side to form an infinite current sheet (Fig. \(\mathbf{P 2 8 . 6 9}\) ). The conductors lie in the \(x y\) -plane, are parallel to the \(y\) -axis, and carry current in the \(+y\) -direction. There are \(n\) conductors per unit length measured along the \(x\) -axis. (a) What are the magnitude and direction of the magnetic field a distance \(a\) below the current sheet? (b) What are the magnitude and direction of the magnetic field a distance \(a\) above the current sheet?

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.