/*! 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 69 Long, straight conductors with s... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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?

Short Answer

Expert verified
The magnitude and direction of the magnetic field a distance \(a\) below the current sheet is \(- \frac{\mu_0 n I}{2} \hat{k}\) and a distance \(a\) above the current sheet is \(\frac{\mu_0 n I}{2} \hat{k}\)

Step by step solution

01

Formulation of Ampere's law

Firstly, we use Ampere's Law as the foundational equation to solve this type of problem. Ampere's Law states that the closed line integral of the magnetic field around a current carrying wire is equal to the product of the current enclosed by the path and the permeability of free space (µ0). The law is mathematically expressed as: \[ \oint \vec{B} \cdot d \vec{l} = \mu_0 \times I_{enc} \]
02

Apply Ampere's law along the contour parallel to the x-axis

Next, apply Ampere's law along the contour that is parallel to x-axis. This way, all the terms on the right side are easy to calculate. The path of integration is put at a distance \(a\) from the sheet. This will help to get the magnitude of the magnetic field at that point. Then assume that B is uniform in magnitude along the path of integration and points to the +z or -z direction (this direction is determined by the right hand rule). So: \[ \oint B dl = B (2l) \] where l is the length of the Amperian loop. The equation is further simplified to: \[ B = \frac{\mu_0 n I}{2} \]
03

Solve for both cases (above and below the sheet)

Now let's solve for both cases. (a) For the case where the observation point is below the sheet: The right-hand rule indicates that the magnetic field lines will circle counter-clockwise when viewed from below and hence point in the -z direction. Therefore the magnitude and direction of the field is: \[ B = - \frac{\mu_0 n I}{2} \hat{k} \] (b) For the case where the observation point is above the sheet: The right-hand rule indicates that the magnetic field lines will circle clockwise when viewed from above and hence point in the +z direction. Therefore the magnitude and direction of the field is: \[ B = \frac{\mu_0 n I}{2} \hat{k} \]

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
When dealing with magnetic fields and electrical currents, Ampere's Law serves as a fundamental principle. Ampere's Law is part of Maxwell's equations, which are the cornerstones of classical electromagnetism. This law relates the integrated magnetic field around a closed loop to the electric current passing through the loop.

Ampere's Law is illustrated mathematically as \[\begin{equation}\oint \vec{B} \cdot d\vec{l} = \mu_0 \times I_{\text{enc}}\end{equation}\]where \( \vec{B} \) is the magnetic field, \( d\vec{l} \) is an infinitesimal vector element of the loop, \( \mu_0 \) symbolizes the permeability of free space, and \( I_{\text{enc}} \) is the current enclosed by the loop. Visualizing this concept can be challenging, but it essentially tells us that the magnetic field lines form closed loops, and they are directly proportional to the current producing them.

Applying Ampere’s Law requires choosing an appropriate Amperian loop. This is where we can make simplifications based on symmetry, such as in the case with the infinite current sheet, where the magnetic field can be assumed to be uniform along the path of the loop, simplifying calculations significantly.
Magnetic Field of Current Sheet
The idea of a 'current sheet' is a useful concept in electromagnetism, particularly when it comes to understanding the magnetic fields generated by large numbers of parallel current-carrying wires. Imagine an infinite plane where each point has a current flowing in the same direction — this is the current sheet. Think of it like a densely packed array of wires with current all moving together.

The magnetic field produced by such an infinite sheet is particularly simple due to the symmetry of the problem. Since there are multiple wires, the magnetic fields produced by them add up to create the net magnetic field. By considering a current sheet with currents running side by side, a consistent and well-ordered magnetic field is produced, which simplifies the analysis.

For an infinite sheet of current, the magnetic field at a distance \( a \) from the sheet is given by the remarkably simple expression \[\begin{equation}B = \frac{\mu_0 n I}{2}\end{equation}\]where \( n \) is the number of conductors per unit length, and \( I \) is the current in each conductor. The magnetic field's direction changes based on the side of the sheet you are on (above or below), which is where understanding the right hand rule becomes essential.
Right Hand Rule
The right hand rule is a mnemonic used to understand the direction of vectors in physics, especially for cross products in vector mathematics. In the context of magnetism, it helps in determining the direction of the magnetic field around a current-carrying conductor.

To apply the right hand rule for currents and magnetic fields, you extend your thumb in the direction of the current flow and curl your fingers. Your fingers will then mimic the direction of the magnetic field lines that encircle the current. It should be noted this applies to conventional current (positive to negative), and the actual electron flow would be in the opposite direction (negative to positive).

The right hand rule also comes into play when evaluating the magnetic field produced by a current sheet. If currents move upwards in the sheet, and you're looking at the sheet from below, using the right hand rule would show that the magnetic field points towards the observer. Conversely, from above the sheet, the magnetic field would be directed away. This phenomena further supports the solutions derived from the application of Ampere's Law for the magnetic field near a current sheet and why the direction changes from above to below the sheet.

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 short current element \(d \vec{l}=(0.500 \mathrm{~mm}) \hat{\jmath}\) carries a current of 5.40 A in the same direction as \(d \vec{l}\). Point \(P\) is located at \(\vec{r}=(-0.730 \mathrm{~m}) \hat{\imath}+(0.390 \mathrm{~m}) \hat{\boldsymbol{k}}\). Use unit vectors to express the mag- netic field at \(P\) produced by this current element.

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 but varies according to the relationship $$ \begin{array}{rlr} \overrightarrow{\boldsymbol{J}} & =\frac{2 I_{0}}{\pi a^{2}}\left[1-\left(\frac{r}{a}\right)^{2}\right] \hat{k} & \text { for } r \leq a \\ & =\mathbf{0} \quad & \text { for } r \geq a \end{array} $$ 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 \(+6.00 \mu\) 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 \(\overrightarrow{\boldsymbol{B}}\) it produces at the following points: (a) \(x=0.500 \mathrm{~m}, y=0, z=0\) (b) \(x=0, y=-0.500 \mathrm{~m}, z=0\) (c) \(x=0, \quad y=0, \quad z=+0.500 \mathrm{~m}\) (d) \(x=0, \quad y=-0.500 \mathrm{~m}\) \(z=+0.500 \mathrm{~m} ?\)

A closely wound coil has a radius of \(6.00 \mathrm{~cm}\) and carries a current of 2.50 A. How many turns must it have if, at a point on the coil axis \(6.00 \mathrm{~cm}\) from the center of the coil, the magnetic field is \(6.39 \times 10^{-4} \mathrm{~T} ?\)

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 \(55.0 \mathrm{~cm}\) long and \(2.80 \mathrm{~cm}\) in diameter. What current will you need to produce the necessary field?

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.