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Chapter 27: Problem 5

An electron is moving in the \(x y\) -plane. If at time \(t\) a magnetic field \(B=0.200 \mathrm{~T}\) in the \(+z\) -direction exerts a force on the electron equal to \(F=5.50 \times 10^{-18} \mathrm{~N}\) in the \(-y\) -direction, what is the velocity (magnitude and direction) of the electron at this instant?

Short Answer

Expert verified
The velocity of the electron in this case will be the result from step 2 in the +x direction.

Step by step solution

01

Identify given values

The given values are Force \(F=5.50 \times 10^{-18} \mathrm{~N}\), magnetic field \(B=0.200 \mathrm{~T}\), and the charge of an electron \(q = -1.602 \times 10^{-19} \mathrm{~C}\). Note that the charge of an electron is negative, indicating that it's negatively charged.
02

Apply the magnetic force formula to find the velocity magnitude

We can rearrange the formula \(F = qvBsin(\Theta)\) to solve for velocity v. Thus, \(v = \frac{F}{q*B*sin(\Theta)}\). Substituting the given values, we get \(v = \frac{5.50 \times 10^{-18} \mathrm{~N}}{-1.602 \times 10^{-19} \mathrm{~C} * 0.200 \mathrm{~T} * sin(90 degrees)}\). Solving this equation will give us the velocity of the electron.
03

Determine the direction of the velocity

Knowing that the force is in the -y direction, and the force and velocity are always perpendicular to each other, we can deduce that the velocity of the electron is in the +x direction. This is in accordance with Fleming's left-hand rule.
04

Combine the magnitude and direction of the velocity

Finally, combine the calculated velocity magnitude from step 2 and the direction from step 3 to form the final answer for the velocity of the electron.

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Key Concepts

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

Lorentz Force
The Lorentz force is the force exerted on a charged particle moving through an electric and magnetic field. It is crucial in understanding the behavior of particles like electrons in fields.

The Lorentz force is given by the equation: \( F = q(E + v \times B) \), where \( F \) is the force on the particle, \( q \) is the charge of the particle, \( E \) is the electric field, \( v \) is the velocity of the particle, and \( B \) is the magnetic field. In the scenario where only a magnetic field is present, the equation simplifies to \( F = q(v \times B) \).

This force is always perpendicular to the direction of velocity and the magnetic field, thereby causing the charged particle to move in a curved path. In our exercise, the electron encounters only a magnetic field which results in the force exerted on it in the -y direction.
Fleming's Left-Hand Rule
Fleming's Left-Hand Rule is a simple mnemonic to determine the direction of force experienced by a charged particle moving in a magnetic field. To use this rule, extend the thumb, index finger, and middle finger of your left hand so that they are mutually perpendicular to each other.

If the index finger points in the direction of the magnetic field (\(+z\) direction, as per our exercise), and the middle finger points in the direction of the current (opposite to the velocity of the electron since electrons are negatively charged), then the thumb will point in the direction of the force acting on the electron. This rule is an excellent tool to visualize the interaction described by the Lorentz force formula and is particularly helpful in the situation described in our exercise.
Velocity of Charged Particle in Magnetic Field
The velocity of a charged particle in a magnetic field can be calculated using the rearranged form of the Lorentz force equation. Since the force is always perpendicular to the velocity, if we know the magnitude of the force, the charge of the particle, and the strength of the magnetic field, we can find the magnitude of the velocity.

The equation used is \( v = \frac{F}{|q|B} \), assuming the angle \(\theta\) between the velocity and the magnetic field is 90 degrees, which is typical when dealing with simple scenarios of perpendicular motion. In the exercise, we have all the necessary values to input into this formula, allowing us to calculate the electron's velocity. It's essential to consider the direction of velocity as well, which can be determined using Fleming's Left-Hand Rule.
Electron Charge
The charge of an electron is a fundamental constant in physics and is denoted by \( q \) or \( e \). Its value is approximately \( -1.602 \times 10^{-19} \) coulombs.

The negative sign indicates that the electron has a negative electrical charge. Understanding the electron charge is vital because it influences how electrons interact with electric and magnetic fields, determining the strength of the force experienced by the electron in these fields. Knowing the charge also aids in calculating the velocity of an electron moving through a magnetic field, as shown in our exercise.

The charge is integral to the equations that describe electromagnetic interactions, particularly in the Lorentz force equation, which is central to many principles of electromagnetism and electronics.

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Most popular questions from this chapter

A particle of mass \(0.195 \mathrm{~g}\) carries a charge of \(-2.50 \times 10^{-8} \mathrm{C}\). The particle is given an initial horizontal velocity that is due north and has magnitude \(4.00 \times 10^{4} \mathrm{~m} / \mathrm{s}\). What are the magnitude and direction of the minimum magnetic field that will keep the particle moving in the earth's gravitational field in the same horizontal, northward direction?

Suppose the electric field between the plates in Fig. 27.22 is \(1.88 \times 10^{4} \mathrm{~V} / \mathrm{m}\) and the magnetic field in both regions is \(0.682 \mathrm{~T}\). If the source contains the three isotopes of krypton, \({ }^{82} \mathrm{Kr},{ }^{84} \mathrm{Kr},\) and \({ }^{86} \mathrm{Kr},\) and the ions are singly charged, find the distance between the lines formed by the three isotopes on the particle detector. Assume the atomic masses of the isotopes (in atomic mass units) are equal to their mass numbers, \(82,84,\) and \(86 .\) (One atomic mass unit \(=1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{~kg} .\) )

A dc motor with its rotor and field coils connected in series has an internal resistance of \(3.2 \Omega .\) When the motor is running at full load on a \(120 \mathrm{~V}\) line, the emf in the rotor is \(105 \mathrm{~V}\). (a) What is the current drawn by the motor from the line? (b) What is the power delivered to the motor? (c) What is the mechanical power developed by the motor?

A particle with charge \(q\) and mass \(m\) is dropped at time \(t=0\) from rest at its origin in a region of constant magnetic field \(\overrightarrow{\boldsymbol{B}}\) that points horizontally. What happens? To answer, construct a Cartesian coordinate system with the \(y\) -axis pointing downward and the \(z\) -axis pointing in the direction of the magnetic field. At time \(t \geq 0\) the particle has velocity \(\overrightarrow{\boldsymbol{v}}=v_{x} \hat{\imath}+v_{y} \hat{\jmath} .\) The net force \(\overrightarrow{\boldsymbol{F}}=F_{x} \hat{\imath}+F_{y} \hat{\jmath}\) on the particle is the vector sum of its weight and the magnetic force. (a) Using Newton's second law, write equations for \(a_{x}\) and \(a_{y},\) where \(\vec{a}=a_{x} \hat{\imath}+a_{y} \hat{\jmath}\) is the acceleration of the particle. (b) Differentiate the second of these equations with respect to time. Then substitute your expression for \(a_{x}=d v_{x} / d t\) to determine an equation for \(d v_{y}^{2} / d t^{2}\) in terms of \(v_{y}\) (c) This result shows that \(v_{y}\) is a simple harmonic oscillator. Use the initial conditions to determine \(v_{y}(t) .\) Write your answer in terms of the angular frequency \(\omega=q B / m\). Note that \(\left(d v_{y} / d t\right)_{0}=g .\) (d) Substitute your result for \(v_{y}(t)\) into your equation for \(d v_{x} / d t .\) Integrate using the initial conditions to determine \(v_{x}(t) .\) (e) Integrate your expressions for \(v_{x}(t)\) and \(v_{y}(t)\) to determine \(x(t)\) and \(y(t) .\) (f) If \(m=1.00 \mathrm{mg}, q=19.6 \mu \mathrm{C}\) and \(B=10.0 \mathrm{~T}\), what maximum vertical distance does the particle drop before returning upward?

Determine the magnetic moment \(\overrightarrow{\boldsymbol{\mu}}\) of a spherical shell with radius \(R\) and uniform charge \(Q\) rotating with angular speed \(\vec{\Omega}=\omega \hat{k} .\) Use the following steps: (a) Consider a coordinate system with the origin at the center of the sphere. Parameterize each latitude on the sphere with the angle \(\theta\) measured from the positive \(z\) -axis. There is a circular current loop at each value of \(\theta\) for \(0 \leq \theta \leq \pi .\) What is the radius of the loop at latitude \(\theta ?\) (b) The differential current carried by that loop is \(d I=\sigma v d W,\) where \(\sigma\) is the charge density of the sphere, \(v\) is the tangential speed of the loop, and \(d W=R d \theta\) is its differential width. Express \(d I\) in terms of \(\sigma, R, \omega, \theta,\) and \(d \theta .\) (c) The differential magnetic moment of the loop is \(d \mu=A d I,\) where \(A\) is the area enclosed by the loop. Express \(d \mu\) in terms of \(R, \omega, \theta,\) and \(d \theta\) (d) Integrate over the sphere to determine the magnetic moment. Express your result as a vector; use the total charge \(Q\) rather than the charge density \(\sigma\). (e) If the sphere is in a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}=(\sin \alpha \hat{\imath}+\cos \alpha \hat{\jmath}) B,\) what is the torque on the sphere?
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