/*! 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 34 A closely wound, circular coil w... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 28: Problem 34

A closely wound, circular coil with radius \(2.40 \mathrm{~cm}\) has 800 turns. (a) What must the current in the coil be if the magnetic field at the center of the coil is \(0.0770 \mathrm{~T}\) ? (b) At what distance \(x\) from the center of the coil, on the axis of the coil, is the magnetic field half its value at the center?

Short Answer

Expert verified
The current necessary in the coil is approximately 0.0402 A. The magnetic field is half of its center value at a distance about 2.72 cm from the center of the coil.

Step by step solution

01

Compute Current

The magnetic field at the center of current carrying coil is given by \(B = \frac{{\mu_0}}{2} * \frac{{NI}}{r}\), where \(\mu_0 = 4\pi * 10^{-7} T.m/A\) is the permeability of free space, \(N\) is the number of turns, \(I\) is the current and \(r\) is the radius. We are given \(B = 0.0770 T\), \(N = 800\) and \(r = 2.40 cm\). We can rearrange to find the current \(I\), thus \(I = \frac{{2Br}}{\mu_0N}.\)
02

Compute Distance

The magnetic field at a distance \(x\) along the central axis of a circular coil is given by \(B_x = \frac{{\mu_0}}{4\pi} * \frac{{2\pi NIA^2}}{(x^2 + A^2)^{3/2}}\), where \(A\) is the area of the circle and \(A = \pi r^2\). We also know from the problem that \(B_x = B/2\). Substituting the values and simplifying we get \(x = \sqrt{(8*A^2)-r^2}\).

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.

Biot-Savart Law
Understanding the Biot-Savart Law is fundamental for calculating the magnetic field produced by a steady current. This law relates the magnetic field \textbf{B} to the magnitude, direction, length, and proximity of the electric current. Essentially, it states that for a small segment of current-carrying wire, the magnetic field at a point in space is directly proportional to the current, inversely proportional to the square of the distance from the wire, and depends on the angle between the wire and the point in question.

The mathematical expression for Biot-Savart Law is \( dB = \frac{{\mu_0}}{{4\pi}} \frac{{Id\vec{s} \times \hat{r}}}{{r^2}} \) where \( Id\vec{s} \) is the vector representing the infinitesimal current element in both magnitude and direction, \( \hat{r} \) is the unit vector from the current element to the point of interest, \( r \) is the distance from the current element to the point, and \( \mu_0 \) is the permeability of free space.

This law provides the basis for the derivation of the magnetic field for various geometries, including a circular coil, as students must consider in this exercise.
Permeability of Free Space
The permeability of free space, denoted \( \mu_0 \), is a fundamental physical constant which characterizes the ability of a vacuum to support the formation of a magnetic field. Specifically, \( \mu_0 \) is the measure of the amount of resistance encountered when forming a magnetic field in a classical vacuum. The value of \( \mu_0 \) is approximately \( 4\pi \times 10^{-7} \) tesla meter per ampere (T路m/A).

The significance of \( \mu_0 \) is not only in its value but also in how it relates electric and magnetic field constants in the vacuum of space. In problems involving magnetic fields produced by currents, such as the magnetic field at a coil's center, it serves as a proportionality constant within the Biot-Savart Law and other electromagnetic equations.
Magnetic Field at Coil's Center
When dealing with the magnetic fields of coils, students need to pay particular attention to the field at the center of the coil, which is of interest in various practical applications, including electromagnets and induction coils. For a circular coil, the magnetic field at its center is given by a simplified version of the Biot-Savart Law.

The formula for the magnetic field at the center of a circular coil carrying a current \(I\) is given by:
\( B = \frac{{\mu_0}}{{2}} * \frac{{NI}}{{r}} \) where \(N\) is the number of turns, and \(r\) is the radius of the coil. The reliance on \(N\) and \(r\) indicates that the field strength is directly proportional to the current and number of turns, and inversely proportional to the coil's radius. Knowing this helps in understanding why, for instance, adding more turns to a coil or increasing the current leads to a stronger magnetic field at its center.
Circular Coil Current Calculation
In exercises such as the given problem, calculating the current in a circular coil involves rearranging the equation for the magnetic field at the coil's center to solve for the current \(I\). Using the given magnetic field \(B\), the number of turns \(N\), and the coil radius \(r\), we apply
\(I = \frac{{2Br}}{{\mu_0N}}\).

This formula indicates that the required current \(I\) is directly proportional to the desired magnetic field \(B\) and the radius of the coil \(r\), and inversely proportional to the permeability of free space \(\mu_0\) and the number of turns \(N\). Students can use this relationship to calculate the necessary current to produce a specific magnetic field at the center of a coil, which is a common exercise in both theoretical and practical contexts involving electromagnetism.

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

The magnetic field around the head has been measured to be approximately \(3.0 \times 10^{-8} \mathrm{G}\). 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 long, straight wire with a circular cross section of radius \(R\) carries a current \(I\). Assume that the current density is not constant across the cross section of the wire, but rather varies as \(J=\alpha r,\) where \(\alpha\) is a constant. (a) \(\mathrm{By}\) the requirement that \(J\) integrated over the cross section of the wire gives the total current \(I,\) calculate the constant \(\alpha\) in terms of \(I\) and \(R .\) (b) Use Ampere's law to calculate the magnetic field \(B(r)\) for (i) \(r \leq R\) and (ii) \(r \geq R .\) Express your answers in terms of \(I\).

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 \(r=a ?\)

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 \(28.76 ?\)

The solenoid is removed from the enclosure and then used in a location where the earth's magnetic field is \(50 \mu \mathrm{T}\) and points horizontally. A sample of bacteria is placed in the center of the solenoid, and the same current is applied that produced a magnetic field of \(150 \mu \mathrm{T}\) in the lab. Describe the field experienced by the bacteria: The field (a) is still \(150 \mu \mathrm{T} ;\) (b) is now \(200 \mu \mathrm{T} ;\) (c) is between 100 and \(200 \mu \mathrm{T},\) depending on how the solenoid is oriented; (d) is between 50 and \(150 \mu \mathrm{T}\), depending on how the solenoid is oriented.
See all solutions

Recommended explanations on Physics Textbooks

Work Energy and Power

Read Explanation

Dynamics

Read Explanation

Electricity

Read Explanation

Physics of Motion

Read Explanation

Electric Charge Field and Potential

Read Explanation

Kinematics Physics

Read Explanation
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.