/*! 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} Q3P An electron that has velocity ν... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An electron that has velocity ν→=(2.0×106m/s)i^+ (3.0×106m/s)moves through the uniform magnetic field B→= (0.030T)i^-(0.15T)j^.(a)Find the force on the electron due to the magnetic field. (b)Repeat your calculation for a proton having the same velocity.

Short Answer

Expert verified

j^a. Magnetic force experienced by the electron isFB=0.624×10-13Nk^~6.2×10-14Nk^

b. Magnetic force experienced by the proton isFB=-0.624×10-13Nk¯~6.2×10-14Nk^

Step by step solution

01

Given 

v→=2.0×106i¯+3.0×106j^B→=0.030i^-0.15j^

02

Determining the concept 

Find the magnetic force on the electron using the formula for magnetic force in terms of charge, magnetic field strength, and velocity of electron.

Formulae are as follow:

FB→=ev→×B→

Where, FB is magnetic force, v is velocity, B is magnetic field, e is charge of particle.

03

(a) Determining the magnetic force experienced by the electron 

The magnetic force experienced by the electron is,

FB→=ev→×B→

FB=-1.6×10-192.0×106i^+3.0×106j^×0.030i^-0.15j^

v→×B→=i^j^k^2.0×1063.0×10600.030-0.150v→×B→=2.0×106-0.15-3.0×1060.030k^v→×B→=-0.39×106k^msT

Hence,

FB=-1.6×10-19-0.39×106k^FB=0.624×10-13Nk^~6.2×10-14Nk^

Hence, the magnetic force experienced by the electron islocalid="1663047307015" FB=-0.624×10-13Nk^~6.2×10-14Nk^

04

(b) Determining the magnetic force experienced by the proton

Since, proton will experience the same magnetic field, and its velocity is the same but charge on it is positive.

Therefore, the magnetic force experienced by proton is FB=-0.624×10-13Nk^~-6.2×10-14Nk^

Hence, the magnetic force experienced by the proton is FB=-0.624×10-13Nk^~-6.2×10-14Nk^

Therefore, the magnetic force on the electron and proton can be found using the formula for magnetic force in terms of charge, magnetic field strength, and velocity of electron.

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

A proton traveling at23.0°with respect to the direction of a magnetic field of strength2.60mT experiences a magnetic force of6.50×10-17N.

(a) Calculate the proton’s speed

. (b)Find its kinetic energy in electron-volts.

Figure 28-27 shows the path of an electron that passes through two regions containing uniform magnetic fields of magnitudesB1and.B2

Its path in each region is a half-circle.

(a) Which field is stronger?

(b) What is the direction of each field?

(c) Is the time spent by the electron in theB1→region greater than,

less than, or the same as the time spent in theB2→region?

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 distanceΔxbetween the points where the ions strike the detector?

Figure shows a rectangular 20-turn coil of wire, of dimensions 10cmby 5.0cm. It carries a current of 010Aand is hinged along one long side. It is mounted in the x-yplane, at angle θ=300to the direction of a uniform magnetic field of magnitude 0.50T. In unit-vector notation, what is the torque acting on the coil about the hinge line?

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 B⃗=(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?

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.