Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 26: Problem 75

A solid conducting wire of radius \(R\) runs parallel to the \(z\) -axis and carries a current density given by \(\vec{J}=J_{0}(1-r / R) \hat{k},\) where \(J_{0}\) is a constant and \(r\) is the distance from the wire axis. Find expressions for (a) the total current in the wire and (b) the magnetic field for \(r>R\) and (c) \(r

Short Answer

Expert verified
The total current in the wire is \( I = \pi R^{2}J_{0}/2 \). The magnetic field for \( r > R \) is \( B = \mu_{0} I/(2 \pi r) \) and for \( r < R \) is \( B = \mu_{0} J_{0} r / 2 \).

Step by step solution

01

Current Density Expression

First, observe that the current density is radially symmetric, and in the negative \( z \) direction, expressed as \( \vec{J} = J_{0}(1 - r/R) \hat{k} \). This indicates that the current density decreases linearly with distance from the center of the wire.
02

Total Current Calculation

Find the total current through the wire by integrating the current density over the area \( A = \pi R^2 \), which is a circular cross section of the wire. The expression becomes \( I = \int \vec{J} \cdot d\vec{A} = \int J_{0}(1-r/R) dA \). Solving this results in \( I = \pi R^{2}J_{0}/2 \) as the total current.
03

Magnetic Field Calculation for r > R

To find the magnetic field outside the wire (when \( r > R \)), use Ampère's circuital law which in integral form states \( \oint \vec{B} \cdot d\vec{l} = \mu_{0} I_{encl} \). The magnetic field is circular and in the azimuthal direction. The enclosed current is the total current. Solving the integral gives \( B = \mu_{0} I/(2 \pi r) \) when \( r > R \).
04

Magnetic Field Calculation for r < R

Now, consider when \( r < R \). Ampère's circuital law still applies, but the enclosed current within radial distance \( r \) is given by integrating the current density over the area \( A = \pi r^2 \), resulting in \( I_{encl} = \pi r^{2}J_{0}/2 \). Substituting this in Ampère's law results in \( B = \mu_{0} J_{0} r/(2) \) when \( r < R \).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 rectangular copper strip measures \(1.0 \mathrm{mm}\) in the direction of a uniform 2.4 -T magnetic field. When the strip carries a \(6.8-\) A current perpendicular to the field, a 1.2 - \(\mu\) V Hall potential develops across the strip. Find the number density of free electrons in the copper.

An electron and a proton moving at the same speed enter a region containing a uniform magnetic field. Which is deflected more from its original path?

A disk of radius \(a\) carries uniform surface charge density \(\sigma\) and rotates with angular speed \(\omega\) about the disk axis. Show that the magnetic field at the disk's center is \(\frac{1}{2} \mu_{0} \sigma \omega a\)

If a current is passed through an unstretched spring, will the spring contract or expand? Explain.

The Biot-Savart law shows that the magnetic field of a current element decreases as \(1 / r^{2} .\) Could you put together a complete circuit whose field exhibits this decrease? Why or why not?
See all solutions

Recommended explanations on Physics Textbooks

Translational Dynamics

Read Explanation

Kinematics Physics

Read Explanation

Astrophysics

Read Explanation

Mechanics and Materials

Read Explanation

Fluids

Read Explanation

Geometrical and Physical Optics

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.

Sign-up for free