/*! 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 87 Using Gauss's Law for Magnetism.... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 27: Problem 87

Using Gauss's Law for Magnetism. In a certain region of space, the magnetic field \(\overrightarrow{\boldsymbol{B}}\) is not uniform. The magnetic field has both a \(z\) -component and a component that points radially away from or toward the \(z\) -axis. The z-component is given by \(B_{z (z)=\beta z,\) where \(\beta\) is a positive constant. The radial component \(B_{x}\) depends only on \(r\) , the radial distance from the \(z\) -axis. (a) Use Gauss's law for magnetism, Eq. \((27.8),\) to find the radial component \(B_{r}\) as a function of \(r .\) (Hint Try a cylindrical Gaussian surface of radius \(r\) concentric with the \(z\) -axis, with one end at \(z=0\) and the other at \(z=L . )\) (b) Sketch the magnetic field lines.

Short Answer

Expert verified
(a) The radial component is \( B_r(r) = -\frac{\beta r}{2} \). (b) Field lines spiral around the z-axis.

Step by step solution

01

Understand Gauss's Law for Magnetism

Gauss's law for magnetism states that the net magnetic flux through any closed surface is zero, mathematically expressed as \( \oint \overrightarrow{B} \cdot d\overrightarrow{A} = 0 \). This implies that magnetic monopoles do not exist and magnetic field lines are closed loops.
02

Choose a Gaussian Surface

For this problem, choose a cylindrical Gaussian surface with radius \( r \), height \( L \), and concentric with the \( z \)-axis. The cylinder has a side surface and two end caps at \( z = 0 \) and \( z = L \).
03

Calculate the Flux through the End Caps

The magnetic flux through the end cap at \( z = 0 \) is zero because \( B_z(z=0) = \beta \times 0 = 0 \). At \( z = L \), the flux is \( \Phi_z = B_z(L) \cdot \pi r^2 = \beta L \cdot \pi r^2 \) because the magnetic field points along the normal to the surface.
04

Calculate the Flux through the Side Surface

The radial component contributes to the flux through the side surface of the cylinder. The differential area is \( dA = r d\theta dz \), so the magnetic flux through the side surface is \( \Phi_{r} = \int_0^{2\pi} \int_0^L B_r(r) r d\theta dz = B_r(r) \cdot 2\pi r L \) because \( B_r \) depends only on \( r \).
05

Apply Gauss's Law for Magnetism

Since the net flux through the entire closed cylindrical surface must be zero, equate the sum of the fluxes through the sides and the tops: \( \beta L \pi r^2 + B_r(r) \cdot 2\pi r L = 0 \). This simplifies to \( B_r(r) = -\frac{\beta r}{2} \).
06

Sketch the Magnetic Field Lines

The field lines for the \( B_z \) component are straight and parallel to the \( z \)-axis, increasing in strength as \( z \) increases, while the radial component causes the field lines to loop outward or inward with respect to the radial direction. The combination gives spiral-like lines in space.

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.

Magnetic Field Lines
In the realm of magnetism, magnetic field lines are a fundamental concept. They visualize how a magnetic field spreads out in space. These lines are hypothetical; we use them to help understand magnetic fields better.
Magnetic field lines have some key properties:
  • They form closed loops, indicating that there are no magnetic monopoles. Every line that enters an area must also leave it.
  • The direction of the field line shows the direction of the magnetic force.
  • Field lines never intersect each other.
Thus, in the problem discussed, the field lines appear as a combination of straight and spiral shapes. The straight lines are parallel to a certain axis and get more robust as we travel along that axis.
Meanwhile, the spiral lines represent the radial part of the field. It shows how the magnetic field grows radially outward, which depends on the radial distance away from the axis. This visualization helps us predict the behavior of magnetic forces in space.
Cylindrical Gaussian Surface
Choosing a cylindrical Gaussian surface is a smart strategic move in solving problems with radial and axial symmetry. A Gaussian surface is an imaginary closed surface where Gauss's laws are applied.
In this exercise, the Gaussian surface is cylindrical with a certain radius and height, concentric with a specific axis. This shape helps simplify calculations, especially when dealing with fields that have a radial and axial component.
Some features of the cylindrical Gaussian surface include:
  • The cylinder has a circular end at both top and bottom, and a curved side.
  • The surface helps identify areas where the magnetic fields can be easily calculated, especially when symmetry is present.
This approach is particularly helpful when you're considering fields that change in space, like the magnetic fields in this problem. It enables breaking down complex interactions into manageable calculations, focusing on different parts of the surface.
Magnetic Flux
Magnetic flux is a measure that describes how much magnetic field passes through a given area. It's an important value in Gauss's law for magnetism.
This law states that the total magnetic flux out of a closed surface is zero. This principle helps us understand that for every magnetic line that enters a space, it must exit as well.
Here's a closer look at the exercise:
  • The magnetic flux through each surface (end caps and sides of a cylinder) is calculated separately.
  • The radial component of the field affects the flux through the curved surface of the cylinder.
Using these calculations, it was shown that the net magnetic flux in this setup aligns with Gauss's law, providing insights into how the internal and external magnetic components interact. Recognizing changes in magnetic flux helps better understand dynamic magnetic fields and their influences.

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 shunt-wound de motor with the field coils and rotor connected in parallel (Fig. 27.56 ) operates from a \(120-\mathrm{V}\) dc power linc. The resistance of the ficld windings, \(R_{f},\) is 218\(\Omega\) . The resistance of the rotor, \(R_{r}\) is 5.9 . When the motor is running, the rotor develops an emf \(\mathcal{E}\) . The motor draws a current of 4.82 A from the line. Friction losses amount to 45.0 W. Compute (a) the field current; \((b)\) the rotor current; \((c)\) the emf \(\mathcal{E} ;\) (d) the rate of development of thermal energy in the field windings; (e) the rate of development of thermal energy in the rotor; \((f)\) the power input to the motor; (g) the efficiency of the motor.

A circular loop of wire with area \(A\) lies in the \(x y\) -plane. As viewed along the \(z\) -axis looking in the \(-z\) -direction toward the origin, a current \(I\) is circulating clockwise around the loop. The torque produced by an extemal magnetic field \(\vec{B}\) is given by \(\vec{\tau}=D(4 \hat{z}-3 \hat{y}),\) where \(D\) is a positive constant, and for this orientation of the loop the magnetic potential energy \(U=-\vec{\mu} \cdot \vec{B}\) is negative. The magnitude of the magnetic field is \(B_{0}=13 D / L A\) (a) Determine the vector magnetic moment of the current loop. (b) Determine the components \(B_{x}, B_{y}\) , and \(B_{z}\) of \(\vec{B}\) .

A particle of mass 0.195 g carries a charge of \(-2.50 \times\) \(10^{-8} \mathrm{C}\) . The particle is given an initial horizontal velocity that is due north and has magnitude \(4.00 \times 10^{4} \mathrm{m} / \mathrm{s}\) . What are the magnitude and direction of the minimum magnetic field that will keep the particle moving in the earth's gravitational field in the same horizontal, northward direction?

A particle with a charge of \(-1.24 \times 10^{-8} \mathrm{C}\) is moving with instantancous \(\quad\) velocity \(\quad \vec{v}=\left(4.19 \times 10^{4} \mathrm{m} / \mathrm{s}\right) \hat{\imath}+(-3.85 \times\) \(10^{4} \mathrm{m} / \mathrm{s}\) ) \(\hat{\mathrm{v}}\) . What is the force exerted on this particle by a magnetic field \((a) \vec{B}=(1.40 \mathrm{T}) \hat{\imath}\) and \((b) \vec{B}=(1.40 \mathrm{T}) \hat{k} ?\)

When a particle of charge \(q>0\) moves with a velocity of \(\overrightarrow{\boldsymbol{v}}_{1}\) at \(45.0^{\circ}\) from the \(+x\) -axis in the \(x y\) -plane, a uniform magnetic field exerts a force \(\overrightarrow{\boldsymbol{F}}_{1}\) along the \(-z\) -axis (Fig. \(27.58 ) .\) When the same particle moves with a velocity \(\overrightarrow{\boldsymbol{v}}_{2}\) with the same magnitude as \(\overrightarrow{\boldsymbol{v}}_{1}\) but along the \(+z\) -axis, a force \(\overrightarrow{\boldsymbol{F}}_{2}\) of magnitude \(\boldsymbol{F}_{2}\) is exerted on it along the \(+x\) -axis. (a) What are the magnitude (in terms of \(q\) , \(v_{1},\) and \(F_{2} )\) and direction of the magnetic field? (b) What is the magnitude of \(\overrightarrow{\boldsymbol{F}}_{1}\) in terms of \(F_{2} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Dynamics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Turning Points in Physics

Read Explanation

Radiation

Read Explanation

Quantum Physics

Read Explanation

Torque and Rotational Motion

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.