Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Figure 28-31 gives snapshots for three situations in which a positively charged particle passes through a uniform magnetic field Bโ†’. The velocitiesVโ†’of the particle differ in orientation in the three snapshots but not in magnitude. Rank the situations according to (a) the period, (b) the frequency, and (c) the pitch of the particleโ€™s motion, greatest first.

Short Answer

Expert verified
  1. Ranking of situations according to the period is 3>2>1
  2. Ranking of situations according to frequency is1>2>3
  3. Ranking of situations according to the pitch of the particleโ€™s motion is3>2>1

Step by step solution

01

Step 1: Given

Mass, velocity, and charge of different particles are given in table.

02

Determining the concept

Equate centripetal force to magnetic force and find the relation for radius. Using radius, find the period, which depends upon the sine of the angle between magnetic field and velocity. From this, rank the situations according to the period. Then using the relation between period and frequency, rank the situations according to frequency. Substituting the period value in the formula for pitch, get the pitch value depending upon the cotangent of the angle between magnetic field and velocity. Using this, rank the situations according to the pitch of the particleโ€™s motion.

Formulae are as follow:

r=mvqBT=2ฯ€rVf=1TP=VparallelT

Where, r is radius, B is magnetic field, v is velocity, m is mass,q is charge on particle, T is time period, f is frequency, P is pitch.

03

(a) Determining the rank of the situations according to period

Rank the situations according to period:

In order to get circular motion, the centripetal force must be balanced by magnetic force.

mv2r=qVBsinฮธr=mvqBVsinฮธr=mqBsinฮธ

But,

T=2ฯ€rV

Substituting rin period,

T=2ฯ€vmqBsinฮธ

Velocity, magnetic field, and charge is constant. Hence, the period is inversely proportional to angleฮธbetween magnetic field and velocity vector. As angle decreases, period increases, and vice versa.

Hence, ranking of situations according to the period is 3>2>1.

04

 (b) Determining the rank of the situations according to frequency

Rank situations according to frequency:

f=1T

Hence, ranking of situations according to frequency is 1>2>3.

05

(c) Determining the rank of the situations according to the pitch of the particleโ€™s motion

Ranking according to pitch of particleโ€™s motion:

The parallel component of velocity to magnetic field is given by,

Vparallel=vcosฮธ

Where,ฮธis the angle between Vโ†’and Bโ†’.

Pitch P=VparallelT

Hence,

P=vcosฮธ2ฯ€mqBsinฮธP=2ฯ€mvqBcotฮธ

Hence, as ฦŸ decreases, the pitch value increases. Hence, situation 3 will has maximum value of pitch and situation 2 will have the smaller value of pitch. Situation 1 will have no pitch as it makes 90oangle.

Therefore, ranking of situations according to the pitch of the particleโ€™s motion is

3>2>1 .

Therefore, equating centripetal force and magnetic force, rank the situation of particleโ€™s motion in the magnetic field according to time period, frequency, and pitch of its motion.

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

Figure shows a wire ring of radiusa=1.8cmthat is perpendicular to the general direction of a radially symmetric, diverging magnetic field. The magnetic field at the ring is everywhere of the same magnitude B=3.4mT, and its direction at the ring everywhere makes an angle ฮธ=20ยฐwith a normal to the plane of the ring. The twisted lead wires have no effect on the problem. Find the magnitude of the force the field exerts on the ring if the ring carries a current i=4.6mA.

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).

Atom 1 of mass 35uand atom 2 of mass 37u are both singly ionized with a charge of +e. After being introduced into a mass spectrometer (Fig. 28-12) and accelerated from rest through a potential difference V=7.3kV, each ion follows a circular path in a uniform magnetic field of magnitude B=0.50T. What is the distanceฮ”xbetween the points where the ions strike the detector?

A wire50.0cmlong carries a current 0.500A in the positive direction of an xaxis through a magnetic field Bโ†’=(3.00mT)j^+(10.0mT)k^. In unit-vector notation, what is the magnetic force on the wire?

A beam of electrons whose kinetic energy is Kemerges from a thin-foil โ€œwindowโ€ at the end of an accelerator tube. A metal plate at distance dfrom this window is perpendicular to the direction of the emerging beam (Fig. 28-53). (a) Show that we can prevent the beam from hitting the plate if we apply a uniform magnetic field such that


Bโ‰ฅ2meK(e2d2)

in which mand eare the electron mass and charge. (b) How should Bโ†’be oriented?

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