/*! 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} Problem 79 A positive charge is passing thr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 79

A positive charge is passing through an electromagnetic field in which \(\vec{E}\) and \(\vec{B}\) are directed towards \(y\)-axis and \(z\)-axis, respectively. If a charged particle passes through the region undeviated, then its velocity is/are represented by (here \(a, b\), and \(c\) are constant) (A) \(\vec{v}=\frac{E}{B} \hat{i}+a \hat{j}\) (B) \(\vec{v}=\frac{E}{B} \hat{i}+b \hat{k}\) (C) \(\vec{v}=\frac{E}{B} \hat{i}+c \hat{i}\) (D) \(\vec{v}=\frac{E}{B} \hat{i}\)

Short Answer

Expert verified
The correct answer is (B) \( \vec{v} = \frac{E}{B} \hat{i} + b \hat{k} \), where \(b = v_z\), as it matches the derived velocity vector form.

Step by step solution

01

\( \vec{F} = q(\vec{E} + \vec{v} \times \vec{B}) \) where \( \vec{F} \) is the Lorentz force, \(q\) is the charge of the particle, \( \vec{E} \) is the electric field, \( \vec{v} \) is the velocity of the particle, and \( \vec{B} \) is the magnetic field. If the particle is not deflected, the net force experienced must be zero. #Step 2: Substitute given values for \( \vec{E} \) and \( \vec{B} \) into the Lorentz force equation# We are given that the electric field and magnetic field are directed along the \(y\)-axis and \(z\)-axis, respectively. Therefore, we can write the electric and magnetic fields as:

\( \vec{E} = E \hat{j} \) and \( \vec{B} = B \hat{k} \) Substituting these values into the Lorentz force equation, we get:
02

\( \vec{F} = q\left( E \hat{j} + \vec{v} \times B \hat{k} \right) \) #Step 3: Set the Lorentz force equation to zero and analyze the particle's velocity components# Since the charged particle is not deflected, there should be no net force acting on it:

\( 0 = q\left( E \hat{j} + \vec{v} \times B \hat{k} \right) \) Now we need to consider the velocity components of the particle. The particle will have three components of velocity, represented by \( \vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k} \). #Step 4: Calculate cross product of \( \vec{v} \) and \( \vec{B} \) # To move forward, let's determine the cross product term in the Lorentz force equation:
03

\( \vec{v} \times \vec{B} = (v_x \hat{i} + v_y \hat{j} + v_z \hat{k}) \times (B \hat{k}) \) After performing the cross product, we obtain:

\( \vec{v} \times \vec{B} = -v_x B \hat{j} + v_y B \hat{i} \) #Step 5: Substitute cross product result into the Lorentz force equation and analyze the components# Now we can substitute the cross product result into the Lorentz force equation:
04

\( 0 = q\left( E \hat{j} - v_x B \hat{j} + v_y B \hat{i} \right) \) To ensure the net force is zero, the coefficients of \( \hat{i} \) and \( \hat{j} \) must be separately equal to zero. So, we have:

\(q v_y B = 0\) and \(q (E - v_x B) = 0\). Since \(q\) and \(B\) are not zero, we can deduce the following:
05

\(v_y = 0\) and \(v_x = \frac{E}{B}\). The particle's velocity vector can then be expressed as:

\( \vec{v} = \frac{E}{B} \hat{i} + 0 \hat{j} + v_z \hat{k} \) #Step 6: Compare the resulting velocity vector with given choices# Looking at the velocity vector, we derive that it can be expressed as \( \vec{v} = \frac{E}{B} \hat{i} + v_z \hat{k} \). Comparing this with the given choices, the correct answer is:
06

(B)

\( \vec{v} = \frac{E}{B} \hat{i} + b \hat{k} \), where \(b = v_z\), as it matches the derived velocity vector form.

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

Two coils of self-inductance \(4 \mathrm{H}\) and \(16 \mathrm{H}\) are wound on the same iron core. The coefficient of mutual inductance for them will be (A) \(8 \mathrm{H}\) (B) \(10 \mathrm{H}\) (C) \(20 \mathrm{H}\) (D) \(64 \mathrm{H}\)

A horizontal telegraph wire \(0.5 \mathrm{~km}\) long running east and west is a part of a circuit whose resistance is \(2.5 \Omega\). The wire falls to the ground from a height of \(5 \mathrm{~m}\). If \(g=10.0 \mathrm{~m} / \mathrm{s}^{2}\) and horizontal component of earth's magnetic field is \(2 \times 10^{-5}\) weber \(/ \mathrm{m}^{2}\), then the current induced in the circuit just before the wire hits the ground will be (A) \(0.7 \mathrm{~A}\) (B) \(0.04 \mathrm{~A}\) (C) \(0.02 \mathrm{~A}\) (D) \(0.01 \mathrm{~A}\)

A magnetized wire of moment \(M\) is bent into an arc of a circle subtending an angle of \(60^{\circ}\) at the centre, the new magnetic moment is (A) \(\frac{2 M}{\pi}\) (B) \(\frac{M}{\pi}\) (C) \(\frac{3 \sqrt{3} M}{\pi}\) (D) \(\frac{3 M}{\pi}\)

Two particles each of mass \(m\) and charge \(q\) are attached to the two ends of a light rigid rod of length \(2 \ell\). The rod is rotated at a constant angular speed about a perpendicular axis passing through its centre. The ratio of the magnitudes of the magnetic moment of the system and its angular momentum about the centre of the rod is (A) \(\frac{q}{2 m} \quad\) (B) \(\frac{q}{m}\) (C) \(\frac{2 q}{m}\) (D) \(\frac{q}{\pi m}\)

A circular loop of radius \(0.3 \mathrm{~cm}\) lies parallel to a much bigger circular loop of radius \(20 \mathrm{~cm}\). The centre of the small loop is on the axis of the bigger loop. The distance between their centres is \(15 \mathrm{~cm}\). If a current of \(2.0 \mathrm{~A}\) flows through the smaller loop, then the flux linked with bigger loop is (A) \(6 \times 10^{-11}\) weber (B) \(3.3 \times 10^{-11}\) weber (C) \(6.6 \times 10^{-9}\) weber (D) \(9.1 \times 10^{-11}\) weber
See all solutions

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

Read Explanation

Linear Momentum

Read Explanation

Radiation

Read Explanation

Classical Mechanics

Read Explanation

Particle Model of Matter

Read Explanation

Quantum Physics

Read Explanation
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.