/*! 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 43 Vector potential inside a wire *... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Vector potential inside a wire ** A round wire of radius \(r_{0}\) carries a current \(I\) distributed uniformly over the cross section of the wire. Let the axis of the wire be the \(z\) axis, with \(\hat{z}\) the direction of the current. Show that a vector potential of the form \(\mathbf{A}=A_{0} \hat{\mathbf{z}}\left(x^{2}+y^{2}\right)\) will correctly give the magnetic field \(\mathbf{B}\) of this current at all points inside the wire. What is the value of the constant, \(A_{0}\) ?

Short Answer

Expert verified
The value of the constant \(A_{0}\) is \(\frac{\mu_{0} I}{4 \pi r_{0}^{2}}\).

Step by step solution

01

Determine the uniform current density

The uniform current density, \(J\), within the wire can be found by dividing the current \(I\) by the cross sectional area of the wire, \(\pi r_{0}^{2}\). So, \(J=\frac{I}{\pi r_{0}^{2}}\).
02

Obtain the Magnetic field inside the wire using Ampère's circuital Law

Ampère's circuital Law can be written as: \[\oint \mathbf{B} . d\mathbf{l} = \mu_{0} \iint \mathbf{J} . d\mathbf{S}\] Since the magnetic field is azimuthal and independent of \(z\) and \(\phi\), we can simplify the integral on the left hand side to: \[B 2 \pi r\]. The right hand side becomes: \[\mu_{0} J \pi r^{2}\]. Thus, setting these two equal gives us the magnitude of \(\mathbf{B}\) inside the wire as: \[B = \frac{\mu_{0} J r}{2} = \frac{\mu_{0} I r}{2 \pi r_{0}^{2}}\]
03

Determining the vector potential \(\mathbf{A}\)

The vector potential \(\mathbf{A}\) is given by the equation: \[\nabla \times \mathbf{A} = \mathbf{B}\] Taking the curl of the given vector potential \(\mathbf{A}=A_{0} \hat{\mathbf{z}}\left(x^{2}+y^{2}\right) = A_{0} \hat{\mathbf{z}} r^{2}\) yields \(\mathbf{B} = \frac{\mu_{0} I r}{2 \pi r_{0}^{2}}\).
04

Find the value of the constant \(A_{0}\)

Setting up these two expressions for the magnetic fields gives: \[\mu_{0} \frac{I r}{2 \pi r_{0}^{2}} = 2 A_{0} r\]. Solving for \(A_{0}\) gives: \[A_{0} = \frac{\mu_{0} I}{4 \pi r_{0}^{2}}\]

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ampère's Circuital Law
Ampère's Circuital Law is a fundamental principle in electromagnetism used to relate the integrated magnetic field around a closed loop to the electric current passing through the loop. This principle can be mathematically expressed as: \[ \oint \mathbf{B} \cdot d\mathbf{l} = \mu_{0} \iint \mathbf{J} \cdot d\mathbf{S} \] In this equation, the left side represents the line integral of the magnetic field \(\mathbf{B}\) along a closed path, and the right side involves the surface integral of the current density \(\mathbf{J}\) over any surface bounded by the closed path:
  • \( d\mathbf{l} \) is an infinitesimal element of the closed loop's path.
  • \( \mu_{0} \) is the permeability of free space, a constant important for magnetic calculations.
This law simplifies calculations when dealing with symmetrical situations, such as the magnetic field inside a wire. By choosing a circular loop within the wire, the integrals simplify greatly, allowing us to express the magnetic field \( B \) in terms of the current \( I \) and radius \( r \).
Uniform Current Density
The concept of uniform current density is crucial in understanding how current is distributed in a conductor. Uniform current density means that the current through the wire is distributed evenly over its cross-sectional area. To find it, we divide the total current \( I \) by the cross-sectional area \( \pi r_{0}^{2} \) of the wire:\[ J = \frac{I}{\pi r_{0}^{2}} \]This formula tells us that the current per unit area, \( J \), is the same at any point within the cross-section of the wire:
  • Provides a simplified model of current flow.
  • Allows for easier calculation of other electromagnetic properties.
Understanding this concept is vital for further calculations, such as finding the magnetic field generated by the current inside the wire.
Magnetic Field Inside a Wire
The magnetic field inside a current-carrying wire can be determined using Ampère's circuital law. For a wire of radius \( r_{0} \) with current \( I \), the magnetic field at a distance \( r \) from the center of the wire is calculated from:\[ B = \frac{\mu_{0} I r}{2 \pi r_{0}^{2}} \]This expression is derived by setting the magnetic field line integral equal to the current enclosed by the path chosen under Ampère’s law considerations:
  • The magnetic field \( B \) inside a uniform current-carrying wire is directly proportional to the radial distance \( r \).
  • The formula shows that \( B \) increases with \( r \) up to the surface of the wire.
This indicates that within the wire, as the distance from the center increases, the magnetic field strength also increases.Beyond the wire's surface, different rules apply.
Curl of Vector Potential
The vector potential \( \mathbf{A} \) is a vector field whose curl gives the magnetic field \( \mathbf{B} \). In our scenario, the vector potential inside a wire is given by \( \mathbf{A} = A_{0} \hat{\mathbf{z}}(x^{2} + y^{2}) \). Taking the curl of this vector potential should match with our expression for \( \mathbf{B} \). The curl is a vector operation that describes the rotation of a field:\[ abla \times \mathbf{A} = \mathbf{B} \]By applying this to our vector potential, we ensure it aligns with:
  • The resulting magnetic field must satisfy both the derived equation from Ampère's Law and the physical conditions of the problem.
  • \( A_{0} \) must be determined such that the derived magnetic field formula is consistent.
The calculated \( A_{0} = \frac{\mu_{0} I}{4 \pi r_{0}^{2}} \) is a specific condition fulfilling these criteria.

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

Far field from a square loop ** Consider a square loop with current \(I\) and side length \(a\). The goal of this problem is to determine the magnetic field at a point a large, distance \(r\) (with \(r \gg a\) ) from the loop. (a) At the distant point \(P\) in Fig. 6.36, the two vertical sides give essentially zero Biot-Savart contributions to the field, because they are essentially parallel to the radius vector to \(P\). What are the Biot-Savart contributions from the two horizontal sides? These are easy to calculate because every little interval in these sides is essentially perpendicular to the radius vector to \(P\). Show that the sum (or difference) of these contributions equals \(\mu_{0} I a^{2} / 2 \pi r^{3}\), to leading order in \(a\). (b) This result of \(\mu_{0} I a^{2} / 2 \pi r^{3}\) is not the correct field from the loop at point \(P\). The correct field is half of this, or \(\mu_{0} I a^{2} / 4 \pi r^{3} .\) We will eventually derive this in Chapter 11, where we will show that the general result is \(\mu_{0} I A / 4 \pi r^{3}\), where \(A\) is the area of a loop with arbitrary shape. But we should be able to calculate it via the Biot-Savart law. Where is the error in the reasoning in part (a), and how do you go about fixing it? This is a nice one - don't peek at the answer too soon!

Hall voltage \(* *\) A Hall probe for measuring magnetic fields is made from arsenic-doped silicon, which has \(2 \cdot 10^{21}\) conduction electrons per \(\mathrm{m}^{3}\) and a resistivity of \(0.016 \mathrm{ohm}-\mathrm{m}\). The Hall voltage is measured across a ribbon of this \(n\)-type silicon that is \(0.2 \mathrm{~cm}\) wide, \(0.005\) \(\mathrm{cm}\) thick, and \(0.5 \mathrm{~cm}\) long between thicker ends at which it is connected into a \(1 \mathrm{~V}\) battery circuit. What voltage will be measured across the \(0.2 \mathrm{~cm}\) dimension of the ribbon when the probe is inserted into a field of 1 kilogauss?

Magnetic scalar "potential" \(* *\) (a) Consider an infinite straight wire carrying current \(I\). We know that the magnetic field outside the wire is \(\mathbf{B}=\left(\mu_{0} I / 2 \pi r\right) \hat{\boldsymbol{\theta}}\) There are no currents outside the wire, so \(\nabla \times \mathbf{B}=0 ;\) verify this by explicitly calculating the curl. (b) Since \(\nabla \times \mathbf{B}=0\), we should be able to write \(\mathbf{B}\) as the gradient of a function, \(\mathbf{B}=\nabla \psi .\) Find \(\psi\), but then explain why the usefulness of \(\psi\) as a potential function is limited.

Scaled-up ring \(*\) Consider two circular rings of copper wire. One ring is a scaled-up version of the other, twice as large in all regards (radius, crosssectional radius). If currents around the rings are driven by equal voltage sources, how do the magnetic fields at the centers compare?

Constant magnitude of \(B\) ** How should the current density inside a thick cylindrical wire depend on \(r\) so that the magnetic field has constant magnitude inside the wire?

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.