/*! 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 3 An electron accelerated through ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An electron accelerated through a potential difference \(V\) passes through a uniform transverse magnetic field and experiences a force \(F\). If the accelerating potential is increased to \(2 V\), the electron in the same magnetic field will experience a force a \(F\) h \(F / 2\) c. \(\sqrt{2} F\) d \(2 F\)

Short Answer

Expert verified
The force experienced will be \(\sqrt{2} F\).

Step by step solution

01

Understand the Relationship

The force experienced by a charged particle, such as an electron, moving through a magnetic field is given by the equation \( F = qvB \), where \( q \) is the charge of the electron, \( v \) is its velocity, and \( B \) is the magnetic field strength.
02

Relate Kinetic Energy to Potential Difference

The kinetic energy gained by the electron when accelerated through a potential difference \( V \) is \( eV \), where \( e \) is the charge of the electron. The relationship between kinetic energy and velocity is given by \( \frac{1}{2}mv^2 = eV \).
03

Solve for Velocity

Rearrange the kinetic energy equation to solve for velocity: \( v = \sqrt{\frac{2eV}{m}} \). When the potential is increased to \( 2V \), \( v_2 = \sqrt{\frac{2e(2V)}{m}} = \sqrt{2} \cdot v \).
04

Determine the New Force

Substitute \( v_2 = \sqrt{2} \cdot v \) into the force equation: \( F_2 = qv_2B = q(\sqrt{2}v)B = \sqrt{2}(qvB) \). Therefore, \( F_2 = \sqrt{2}F \).

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electron Acceleration
When discussing electron acceleration, it's essential to understand how an electron is influenced by an electric field. Electrons, being negatively charged particles, experience a force when placed in an electric field. This force causes them to accelerate. The source of this acceleration in our context is a potential difference, often observed as a voltage across a pair of electrodes.The potential difference essentially does work on the electron, providing it with additional energy. This work is directly proportional to the charge of the electron and the voltage applied, represented by the equation: \[ W = eV \]where:
  • \( W \) is the work done on the electron.
  • \( e \) is the charge of the electron.
  • \( V \) is the potential difference.
This work results in an increase in the electron's kinetic energy, leading to a rise in velocity as it moves through the field. The greater the potential difference, the faster the electron accelerates.
Potential Difference
The potential difference, often simply called voltage, is crucial in understanding energy transfer to electrons. It's the difference in electric potential between two points and accomplishes the work of moving a charge between these points. For an electron, this potential difference translates directly into kinetic energy. When an electron is accelerated through a potential difference \( V \), it gains kinetic energy equivalent to \( eV \). This relationship sets the foundation for predicting the electron's behavior in an electric field. The connection between potential difference and kinetic energy can be expressed through this simple energy equivalence:\[ KE = eV \]Increasing the potential difference means giving the electron more energy, enhancing its velocity as it travels through the field. This idea is central to many technologies, such as cathode ray tubes and electron microscopes, where manipulating electron behavior with potential differences is critical.
Kinetic Energy and Velocity
Kinetic energy is the energy a particle possesses due to its motion. When a charged particle like an electron moves faster, its kinetic energy increases. The formula that relates kinetic energy \( KE \) to an electron's velocity \( v \) is:\[ KE = \frac{1}{2}mv^2 \]where \( m \) is the mass of the electron.In scenarios where electrons are accelerated by a potential difference, we find their velocity from their kinetic energy. By equating the gain in kinetic energy to \( eV \), the expression for velocity becomes:\[ v = \sqrt{\frac{2eV}{m}} \]When the accelerating potential is increased, say from \( V \) to \( 2V \), the velocity of the electron changes accordingly:\[ v_2 = \sqrt{2} \cdot v \]Here, the velocity is multiplied by \( \sqrt{2} \), indicating how even a simple change in potential difference can noticeably affect the speed of the electron. This change, in turn, influences the magnetic force experienced by the electron as it moves through a magnetic field, since force is directly related to velocity.

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

The value of the electric field strength in vacuum if the energy density is same as that due to a magnetic field of induction \(1 \mathrm{~T}\) in vacuum is a. \(3 \times 10^{8} \mathrm{NC}^{-1}\) b \(1.5 \times 10^{8} \mathrm{NC}^{-1}\) c. \(2.0 \times 10^{8} \mathrm{NC}^{-1}\) d. \(1.0 \times 10^{8} \mathrm{NC}^{-1}\)

Five very long, straight insulated wires are closely bound together to form a small cable. Currents carried by the wires are: \(I_{1}=20 \mathrm{~A}, I_{2}=-6 \mathrm{~A}_{1} I_{3}=12 \mathrm{~A}, I_{4}=-7 \mathrm{~A}, I_{5}=18 \mathrm{~A}\). [Negative currents are opposite in direction to the positive.] The magnetic field induction at a distance of \(10 \mathrm{~cm}\) from the cable is a. \(5 \mu \mathrm{T}\) b. \(15 \mu \mathrm{T}\) c. \(74 \mu \mathrm{T}\) d. \(128 \mathrm{uT}\)

A particle of charge per unit mass \(\alpha\) is released from origin with velocity \(\vec{v}=v_{0} \hat{i}\) in a magnetic field $$ \bar{B}=-B_{0} \hat{k} \quad \text { for } x \leq \frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} \alpha} $$ and \(\bar{B}=0\) $$ \text { for } x>\frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} \alpha} $$ The \(x\) -coordinate of the particle at time \(t\left(>\frac{\pi}{3 B_{0} \alpha}\right)\) would be a. \(\frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} \alpha}+\frac{\sqrt{3}}{2} v_{0}\left(t-\frac{\pi}{B_{0} \alpha}\right)\) b \(\frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} \alpha}+v_{0}\left(t-\frac{\pi}{3 B_{0} \alpha}\right)\) c. \(\frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} \alpha}+\frac{v_{0}}{2}\left(t-\frac{\pi}{3 B_{0} \alpha}\right)\) d. \(\frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} \alpha}+\frac{v_{0} t}{2}\)

A current of \(1 /(4 \pi)\) ampere is flowing in a long straight conductor. The line integral of magnetic induction around \(\mathrm{a}\) closed path enclosing the current carrying conductor is a. \(10^{-7} \mathrm{~Wb} \mathrm{~m}^{-1}\) b \(4 \pi \times 10^{-7} \mathrm{~Wb} \mathrm{~m}^{-1}\) c. \(16 \pi^{2} \times 10^{-1} \mathrm{~Wb} \mathrm{~m}^{-1}\) d zero

A long cylindrical wire of radius ' \(a\) ' carrles a current \(l\) distributed uniformly over its cross section. If the magnetic fields at distances ' \(r\) and \(R\) from the axis have equal magnitude, then a \(a=\frac{R+r}{2}\) b \(a=\sqrt{R r}\) c. \(a=\operatorname{Rr} / R+r\) d \(a=R^{2} / r\)

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.