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

In Fig. 28-30, a charged particle enters a uniform magnetic field with speedv0 , moves through a halfcirclein timeT0 , and then leaves the field

. (a) Is the charge positive or negative?

(b) Is the final speed of the particle greater than, less than, or equal tov0 ?

(c) If the initial speed had been0.5v0 , would the time spent in field have been greater than, less than, or equal toT0 ?

(d) Would the path have been ahalf-circle, more than a half-circle, or less than a half-circle?

Short Answer

Expert verified
  1. The charge on the particle is negative.
  2. The final speed of the particle is equal tov0.
  3. If the initial speed is0.5v0, then the time spent in the magnetic fieldBis equal toT0.
  4. The path would have been a half circle.

Step by step solution

01

Determine the formula for the radius

Consider the formula for the radius of the charge as:

r=mv|q|B

Here, r is radius, B is magnetic field, v is velocity, m is mass,q is charge on particle

02

(a) Determine the charge on the particle

The charge on the particle:

In order to get the force on the particle downwardaccording to the right hand rule, the charge on the particle must be negative.

Hence, the charge on the particle is negative.

03

(b) Determinewhether the final speed of the particle is greater than, less than, or equal to v0

The final speed of the particle greater than, less than, or equal tov0:

Magnetic field does not do any work on the charge particle. It only changes the direction of velocity in magnetic field region.

Hence, the speed of the particle does not change.

Hence, its final speed is the same as the initial speed.

04

(c) Determine if the initial speed is 0.5v0 , then whether the time spent in the magnetic field  B→ is greater than, less than, or equal to  T0

If the initial speed is0.5v0, then whether the time spent in the magnetic fieldBis greater than, less than, or equal toT0:

r=mv|q|B

If velocity is12v0,then usingtheabove relation,

r=12mvqB

Hence, radius will get halved.

Consider the formula:

T=2πrv

Since the speed is halved and the radius is halved, the period will not change. It will bethesame as the initial timeT0.

Hence, if the initial speed is 0.5v0, then the time spent in the magnetic field B is equal to T0.

05

(d) Determinewhether the path would have been a half circle, more than half circle, or less than a half circle

Path is half circle, more than half circle, or less than a half circle:

The path will be half circle. Speed does not change the direction; only the magnetic force changes the direction.

Hence, the path would have been a half circle.

Therefore, use the right hand rule to find the force on the particle and usethe relation of radius and magnetic field to find the radius of different particles.

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 strip of copper150 m thick and 4.5 mm wide is placed in a uniform magnetic fieldBof magnitude 0.65T, withB perpendicular to the strip. A currentlocalid="1663949700654" i=23 A is then sent through the strip such that a Hall potential difference Vappears across the width of the strip. Calculate V. (The number of charge carriers per unit volume for copper islocalid="1663949722414" 8.47×1028electrons/m3.)

A long, rigid conductor, lying along an xaxis, carries a current of5.0A in the negative direction. A magnetic field Bis present, given by B=(3.0i^+8.0x2j^)mTwith xin meters and Bin milliteslas. Find, in unit-vector notation, the force on the 2.0msegment of the conductor that lies between x=1.0mand x=3.0m.

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?

A single-turn current loop, carrying a current of4.00 A ,is in the shape of a right triangle with sides 50.0, 120, and 130cm. The loop is in a uniform magnetic field of magnitude 75.0mT whose direction is parallel to the current in the 130cm side of the loop. (a) What is the magnitude of the magnetic force on the 130cm side? (b) What is the magnitude of the magnetic force on the 50.0cm side? (c) What is the magnitude of the magnetic force on the 120cmside? (d)What is the magnitude of the net force on the loop?

A positron with kinetic energy2.00keV is projected into a uniform magnetic field Bof magnitude 0.100T, with its velocity vector making an angle of 89.0° with.

(a) Find the period.

(b) Find the pitch p.

(c) Find the radius rof its helical path.

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