/*! 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} Q12P At time t1, an electron is sent ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

At time t1, an electron is sent along the positive direction of an x-axis, through both an electric fieldand a magnetic fieldB→, withE→directed parallel to the y-axis. Figure 28-33 gives the ycomponent Fnet, yof the net force on the electron due to the two fields, as a function of theelectron’s speed vat time t1.The scale of the velocity axis is set byvx=100.0 m/s. The xand zcomponents of the net force are zero at t1. AssumingBx=0, find

(a)the magnitude E and

(b B→)in unit-vector notation.

Short Answer

Expert verified

a.E=1.25V/m

b.B⇶Ä=0.025TK∧

Step by step solution

01

Step 1: Given

When,v=0,F=−2×10-19N

02

Determining the concept

The direction of the magnetic forceisperpendicular to the plane formed byv¯andB¯as determined by the right-hand rule.

Right Hand Rule states that if we arrange our thumb, forefinger, and middle finger of the right-hand perpendicular to each other, then the thumb points towards the direction of the motion of the conductor relative to the magnetic field, and the forefinger points towards the direction of the magnetic field and the middle finger points towards the direction of the induced current.

Formulae are as follows:

role="math" localid="1663013317461" E=Fq=q(V⇶Ä×B⇶Ä)

Where F is a magnetic force, v is velocity, E is the electric field, B is the magnetic field, and q is the charge of the particle.

03

(a) Determining the magnitude  

To find the magnitude of E:

Here,

F=qEE=Fq=−2×10−19 N−1.6×10−19 C=1.25 N/C

Hence, the magnitude of E is1.25 N/C

04

(b) Determining the B→ in-unit vector notation

To find a magnetic field(B⇶Ä):

B=Ev=1.25 N/C50 m/s=0.025 T

To find the direction ofrole="math" localid="1663013448881" B⇶Ä,

F⇶Ä=q(V⇶Ä×B⇶Ä)

The net force is directed indirection and velocity in+xdirection, so by applying the right-hand rule,B⇶Ämust be directed in+zdirection.

Hence,

role="math" localid="1663013638135" B⇶Ä=0.025TK∧

Hence, the magnetic field isB⇶Ä=0.025TK∧.

Therefore, the magnitude of the electric field and magnetic field can be determined by using the respective formulae. The direction of the magnetic field can be found by using the right-hand rule.

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 91Ó°ÊÓ!

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

At time t=0, an electron with kinetic energy 12KeVmoves through x=0in the positive direction of an xaxis that is parallel to the horizontal component of Earth’s magnetic field B⇶Ä. The field’s vertical component is downward and has magnitude 55.0μT. (a) What is the magnitude of the electron’s acceleration due toB⇶Ä? (b) What is the electron’s distance from the xaxis when the electron reaches coordinate x=20cm?

Figure 28-52 gives the orientation energy Uof a magnetic dipole in an external magnetic field B→, as a function of angle ϕ between the directions B→, of and the dipole moment. The vertical axis scale is set by Us=2.0×10-4J. The dipole can be rotated about an axle with negligible friction in order that to change ϕ. Counterclockwise rotation from ϕ=0yields positive values of ϕ, and clockwise rotations yield negative values. The dipole is to be released at angle ϕ=0with a rotational kinetic energy of 6.7×10-4J, so that it rotates counterclockwise. To what maximum value of ϕwill it rotate? (What valueis the turning point in the potential well of Fig 28-52?)

A cyclotron with dee radius 53.0 cm is operated at an oscillator frequency of 12.0 MHz to accelerate protons.

(a) What magnitude Bof magnetic field is required to achieve resonance?

(b) At that field magnitude, what is the kinetic energy of a proton emerging from the cyclotron? Suppose, instead, that B = 1.57T.

(c) What oscillator frequency is required to achieve resonance now?

(d) At that frequency, what is the kinetic energy of an emerging proton?

In Fig. 28-55, an electron moves at speed v=100m/salong an xaxis through uniform electric and magnetic fields. The magnetic field is directed into the page and has magnitude5.00T. In unit-vector notation, what is the electric field?

The bent wire shown in Figure lies in a uniform magnetic field. Each straight section is 2.0 m long and makes an angle of θ=60owith the xaxis, and the wire carries a current of 2.0A. (a) What is the net magnetic force on the wire in unit vector notation if the magnetic field is given by 4.0k^ T? (b) What is the net magnetic force on the wire in unit vector notation if the magnetic field is given by 4.0i^T?

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.