/*! 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} Q35P A proton circulates in a cyclotr... [FREE SOLUTION] | 91影视

91影视

A proton circulates in a cyclotron, beginning approximately at rest at the center. Whenever it passes through the gap between Dees, the electric potential difference between the Dees is 200 V.

(a)By how much does its kinetic energy increase with each passage through the gap?

(b)What is its kinetic energy as it completes 100passes through the gap? Let r100be the radius of the proton鈥檚 circular path as it completes those 100passes and enters a dee, and let r101be its next radius, as it enters a dee the next time.

(c)By what percentage does the radius increase when it changes from r100to r101? That is, what is Percentage increase =r101-r100r100100%?

Short Answer

Expert verified
  1. 200 eV is the increase in kinetic energy.
  2. 20.0 keV is the kinetic energy as it completes 100 passes through the gap.
  3. Percentage increase in the radius when it changes from r100 to r101 will be 0.499%

Step by step solution

01

Listing the given quantities

  • The potential difference between Dees = 200 eV
  • r = r100 when n = 100
  • r = r101 when n = 101
02

Understanding the concept of the law of conservation of energy

We are given the potential difference between Dees. According to the law of conservation of energy, when it passes through the gap, potential energy is converted into kinetic energy. From this, we can find the change in kinetic energy. We know the kinetic energy for completing one pass so that we can find the kinetic energy for 100 passes. Using the relationqvB=mv2r, we will find the relation between r and n. Once we get that relation, we can find the percentage increase in the radius.

Formula:

role="math" localid="1662725924089" K=nKqvB=mv2rKE1+PE1=KE2+PE2

03

(a) Calculation of increase in the kinetic energy

The total energy will always be conserved.

KE1+PE1=KE2+PE2KE2-KE1=PE2-PE1

Potential energy will convert into kinetic energy, then:

KE=PEKE=200eV

Thus, 200 eV is the increase in kinetic energy.

04

(b) Calculation of kinetic energy as it completes 100 passes through the gap

As we know, the kinetic energy increase in one pass is 200 eV.

Kinetic Energy increase in 100 passes will be:

K100=nKonepassK100=100200K100=20000eV=20.0keV

Thus, 20.0 keV is the kinetic energy as it completes 100 passes through the gap.

05

(c) Calculation of percentage increase in the radius when it changes

We know that:

K=12mv2v=2Km

We know that:

role="math" localid="1662726412882" qvB=mv2rqB=mvrqB=m2Kmr

We can write this in terms of n as:

qB=m2nKmrr=m2nKmqB

We can say that mqB2Kmis a constant term in the above equation.

So,

Rn

From this, we can say that:

r100n100r100n101

Percentageincrease=r101-r100r100100=n101-n100n100100=101-100100100=0.499%

Percentage increase in the radius when it changes from r100 to r101 will be 0.499 %.

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!

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

Fig. 28-49 shows a current loop ABCDEFAcarrying a current i= 5.00 A. The sides of the loop are parallel to the coordinate axes shown, with AB= 20.0 cm, BC= 30.0 cm, and FA= 10.0 cm. In unit-vector notation, what is the magnetic dipole moment of this loop? (Hint:Imagine equal and opposite currents iin the line segment AD; then treat the two rectangular loops ABCDA and ADEFA.)

An electron with kinetic energy 2.5keVmoving along the positive direction of an xaxis enters a region in which a uniform electric field Bof magnitude 10kV/mis in the negative direction of the yaxis. A uniform magnetic field is to be set up to keep the electron moving along the xaxis, and the direction of B is to be chosen to minimize the required magnitude of B. In unit-vector notation, whatBshould be set up?

(a)Find the frequency of revolution of an electron with an energy of 100 eV in a uniform magnetic field of magnitude 35.0T.

(b)Calculate the radius of the path of this electron if its velocity is perpendicular to the magnetic field.

A magnetic dipole with a dipole moment of magnitude 0.020 J/T is released from rest in a uniform magnetic field of magnitude 52 mT. The rotation of the dipole due to the magnetic force on it is unimpeded. When the dipole rotates through the orientation where its dipole moment is aligned with the magnetic field, its kinetic energy is 0.80mT. (a) What is the initial angle between the dipole moment and the magnetic field? (b) What is the angle when the dipole is next (momentarily) at rest?

An electron moves in a circle of radiusr=5.2910-11mwith speed 2.19106ms. Treat the circular path as a current loop with a constant current equal to the ratio of the electron鈥檚 charge magnitude to the period of the motion. If the circle lies in a uniform magnetic field of magnitude B=7.10mT, what is the maximum possible magnitude of the torque produced on the loop by the field?

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.