a) In Checkpoint 5, if the dipole moment is rotated from orientation 2 to orientation 1 by an external agent, is the work done on the dipole by the agent positive, negative, or zero?

(b) Rank the work done on the dipole by the agent for these three rotations, greatest first.21,24,23

Short Answer

Expert verified
  1. Work that agent did is positive.
  2. .W21=W24>W23

Step by step solution

01

Step 1: Given

Four orientations of magnetic dipole moments in the given magnetic field.

02

Determining the concept

Use the relation of work done with the potential energy of a magnetic dipole and find the work done for different orientations of the dipole. From its sign, determine whether it is positive or negative.

The work done is given as-

W=ΔU=μ.B

Where W is work done, U is potential energy,B is magnetic field, and 𝜇 is magnetic dipole moment.

03

(a) Determining the work done on the dipole by the agent is positive, negative or zero

The work done on the dipole appears as the change in potential energy. The work done in rotating magnetic dipole of magnitudeμ,due to magnetic field of magnitude, from initial orientationθitoθfis given by,

Wa=ΔU=UfUiWa=μBcosθf(μBcosθi)

Wa=μBcosθiμBcosθf

Wa=μB(cosθicosθf).(1)

In orientation 2,θi=θand in orientation 1

Substituting this values in 1),θf=(180°θi)

Wa=μB(cosθcos(180θ))

Wa=μB(cosθ+cosθ)

Wa=2μBcosθ

Since θis less than 90°, cosθis greater than zero.

So, the work done is positive.

Hence,the work done is positive.

04

(b) Determining the rank of the work done on the dipole by the agent for three rotations 2→1,2→4,2→3

Work done for three rotations 21,24,23:

Work done in rotation21isW21=2μBcosθ,

Work done in rotation from24is,

θi=θandθf=(180+θ)

Substituting in equation 1),

Wa=μB(cosθcos(180+θ))

Wa=μB(cosθ(cosθ))

Wa=2μBcosθ

Work done in rotation from 23is,

θi=θandθf=(360θ)

Wa=μB(cosθcos(360θ))

cos(360θ)=cos360*cosθ+sin360*sinθ

cos(360θ)=cosθ

Substituting in 1),

Wa=μB(cosθcos(360θ))

Wa=μB(cosθcosθ)

Wa=0

Hence,the ranking for the three cases isW21=W24>W23.

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 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 13.0g wire of length L = 62.0 cm is suspended by a pair of flexible leads in a uniform magnetic field of magnitude 0.440T (Fig. 28-41). What are the (a) magnitude and (b) direction (left or right) of the current required to remove the tension in the supporting leads?

A mass spectrometer (Figure) is used to separate uranium ions of mass3.92×1025kg and charge3.20×1019C from related species. The ions are accelerated through a potential difference of 100 kV and then pass into a uniform magnetic field, where they are bent in a path of radius 1.00 m. After traveling through 180° and passing through a slit of width 1.00 mm and height 1.00 cm, they are collected in a cup.

(a)What is the magnitude of the (perpendicular) magnetic field in the separator?If the machine is used to separate out 100 mg of material per hour

(b)Calculate the current of the desired ions in the machine.

(c)Calculate the thermal energy produced in the cup in 1.00 h.

Bainbridge’s mass spectrometer, shown in Fig. 28-54, separates ions having the same velocity. The ions, after entering through slits, S1and S2, pass through a velocity selector composed of an electric field produced by the charged plates Pand P', and a magnetic field Bperpendicular to the electric field and the ion path. The ions that then pass undeviated through the crossedEand Bfields enter into a region where a second magnetic field Bexists, where they are made to follow circular paths. A photographic plate (or a modern detector) registers their arrival. Show that, for the ions, q/m=E/rBB,where ris the radius of the circular orbit.

The coil in Figure carries current i=2.00Ain the direction indicated, is parallel to an xz plane, has 3.00turns and an area of 4.00×10-3m2, and lies in a uniform magnetic field B=(2.00i^-3.00j^-4.00k^)mT(a) What are the orientation energy of the coil in the magnetic field (b)What are the torque (in unit-vector notation) on the coil due to the magnetic field?

Question: In Fig. 28-57, the two ends of a U-shaped wire of mass m=10.0gand length L=20.0cmare immersed in mercury (which is a conductor).The wire is in a uniform field of magnitude B=0.100T. A switch (unshown) is rapidly closed and then reopened, sending a pulse of current through the wire, which causes the wire to jump upward. If jump height h=3.00m, how much charge was in the pulse? Assume that the duration of the pulse is much less than the time of flight. Consider the definition of impulse (Eq. 9-30) and its relationship with momentum (Eq. 9-31). Also consider the relationship between charge and current (Eq. 26-2).

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.

Sign-up for free