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Chapter 28: Problem 73

At time \(t=0\), an electron with kinetic energy \(12 \mathrm{keV}\) moves through \(x=0\) in the positive direction of an \(x\) axis that is parallel to the horizontal component of Earth's magnetic field \(\vec{B}\). The field's vertical component is downward and has magnitude \(55.0 \mu \mathrm{T}\). (a) What is the magnitude of the electron's acceleration due to \(\vec{B} ?\) (b) What is the electron's distance from the \(x\) axis when the electron reaches coordinate \(x=20 \mathrm{~cm} ?\)

Short Answer

Expert verified
(a) The acceleration is approximately \(6.25 \times 10^{15} \text{ m/s}^2\). (b) The distance from the x-axis is about 2 cm.

Step by step solution

01

Convert Kinetic Energy to Joules

The problem states the kinetic energy of the electron as \(12 \text{ keV} = 12,000 \text{ eV} \). To convert this to joules, use the conversion \(1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}\). Therefore, the kinetic energy in joules is: \[ KE = 12,000 \times 1.602 \times 10^{-19} = 1.9224 \times 10^{-15} \text{ J} \]
02

Determine the Speed of the Electron

The kinetic energy \(KE\) is also expressed as \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the electron \(9.11 \times 10^{-31} \text{ kg}\) and \( v \) is the velocity. Rearranging gives \( v = \sqrt{\frac{2KE}{m}} \). Plug in the values:\[ v = \sqrt{\frac{2 \times 1.9224 \times 10^{-15}}{9.11 \times 10^{-31}}} = 6.47 \times 10^7 \text{ m/s} \]
03

Find the Force Acting on the Electron

The magnetic force on the electron is given by \( F = qvB \sin(\theta) \). The charge of the electron \(q = 1.6 \times 10^{-19} \text{ C}\), the velocity \(v = 6.47 \times 10^7 \text{ m/s}\), the magnetic field \(B = 55.0 \times 10^{-6} \text{ T}\), and \(\theta = 90^\circ\) as the velocity is perpendicular to the magnetic field. Thus, \( \sin(\theta) = 1 \). Calculate the force:\[ F = 1.6 \times 10^{-19} \times 6.47 \times 10^7 \times 55.0 \times 10^{-6} = 5.6938 \times 10^{-15} \text{ N} \]
04

Calculate the Acceleration

The acceleration caused by the magnetic force is given by \( a = \frac{F}{m} \). Using the force calculated earlier:\[ a = \frac{5.6938 \times 10^{-15}}{9.11 \times 10^{-31}} = 6.25 \times 10^{15} \text{ m/s}^2 \] This is the magnitude of the electron's acceleration due to the magnetic field.
05

Determine the Radius of the Circular Path

The electron moves in a circular path due to the magnetic force. The radius of this path \( r \) can be found with \( r = \frac{mv}{qB} \):\[ r = \frac{9.11 \times 10^{-31} \times 6.47 \times 10^7}{1.6 \times 10^{-19} \times 55.0 \times 10^{-6}} = 0.07268 \text{ m} \]
06

Find Time to Reach x=20 cm

The electron travels to \( x = 0.20 \text{ m} \). The velocity along the \( x \) direction (not influenced by the magnetic field) is constant at \( 6.47 \times 10^7 \text{ m/s} \). Time to reach \( x = 0.20 \text{ m} \) is:\[ t = \frac{0.20}{6.47 \times 10^7} \approx 3.09 \times 10^{-9} \text{ s} \]
07

Calculate Displacement from x-Axis

The electron completes a quarter circle when traveling perpendicular to the \( x \)-axis due to the magnetic field. During time \( t \), it sweeps an angle \( \omega t \), where \( \omega = \frac{v}{r} \). Calculate displacement \( d = r \cdot \sin(\omega t) \):\[ \omega = \frac{v}{r} = \frac{6.47 \times 10^7}{0.07268} \approx 8.9 \times 10^8 \text{ rad/s} \]\[ d = 0.07268 \cdot \sin(8.9 \times 10^8 \cdot 3.09 \times 10^{-9}) \approx 0.020 \text{ m} \] \( = 2 \text{ cm} \), which is the distance from the \( x \)-axis when the electron reaches \( x = 20 \text{ cm} \).

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Key Concepts

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

Electron Kinetics
Electron kinetics refers to the study of electron motion, particularly how electrons move in space under the influence of energy. In our given problem, an electron is initially provided with kinetic energy of 12 keV. Kinetic energy directly relates to the speed of the electron via the equation,
  • \( KE = \frac{1}{2}mv^2 \)
where \( m \) is the electron's mass and \( v \) is its velocity. Once we convert this kinetic energy from electron-volts to joules, we can determine the electron's velocity. This step is paramount since knowing the velocity sets the stage for understanding how the electron behaves as it moves through a magnetic field. In this context, electron kinetics plays a crucial role in predicting an electron's speed and direction right before it enters the magnetic field, setting the base for subsequent calculations on its motion.
Magnetic Force
Magnetic force is the force experienced by a charged particle moving in a magnetic field. For a charged particle like an electron, this force is given by the formula:
  • \( F = qvB \sin(\theta) \)
where \( q \) is the charge of the electron, \( v \) is its velocity, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the velocity and magnetic field. In our exercise, the electron's velocity is perpendicular to the magnetic field (\( \theta = 90^\circ \)), making \( \sin(\theta) = 1 \). This simplifies our calculations. The force calculated tells us how strongly the electron is pushed sideways, causing it to move in a circular path. This sideways push - the magnetic force - is exclusive to charged particles and is the reason the electron doesn't just travel in a straight line.
Acceleration Due to Magnetic Field
When an electron moves within a magnetic field, it experiences an acceleration due to the magnetic force acting upon it. The formula to find this acceleration is:
  • \( a = \frac{F}{m} \)
Where \( F \) is the magnetic force on the electron and \( m \) is its mass. In our problem, this results in an impressive acceleration magnitude of \( 6.25 \times 10^{15} \text{ m/s}^2 \). This acceleration causes the electron to change the direction of its velocity continuously. However, it's crucial to note that the acceleration doesn't change the speed of the electron. Instead, it alters the direction in which it moves. Understanding this concept is vital as it lays the groundwork for comprehending the motion path of an electron in magnetic fields.
Circular Motion in Magnetic Fields
Circular motion occurs when constant acceleration is always perpendicular to velocity - just like in the case of an electron in a magnetic field. The force due to the magnetic field ensures that the electron travels in a circular path. The radius of this path depends on the mass, velocity, charge of the electron, and the magnetic field strength:
  • \( r = \frac{mv}{qB} \)
For our electron, the calculated radius is approximately 0.07268 meters. This ensures that the electron remains in a path curving back upon itself rather than moving in a straight line. As the electron travels and reaches 20 cm along the x-axis, it will have also moved perpendicular to it. This perpendicular journey and the calculated radius of motion make the electron traverse a path where its distance from the x-axis becomes a tangible metric, crucial for understanding its trajectory and position relative to the initial axis.

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Most popular questions from this chapter

An electron follows a helical path in a uniform magnetic field given by \(\vec{B}=(20 \hat{i}-50 \hat{j}-30 \hat{k}) \mathrm{mT}\). At time \(t=0\), the electron's velocity is given by \(\vec{v}=(20 \hat{\mathrm{i}}-30 \hat{\mathrm{j}}+50 \hat{\mathrm{k}}) \mathrm{m} / \mathrm{s}\). (a) What is the angle \(\phi\) between \(\vec{v}\) and \(\vec{B} ?\) The electron's velocity changes with time. Do (b) its speed and (c) the angle \(\phi\) change with time? (d) What is the radius of the helical path?

Physicist S. A. Goudsmit devised a method for measuring the mass of heavy ions by timing their period of revolution in a known magnetic field. A singly charged ion of iodine makes \(7.00\) rev in a \(45.0 \mathrm{mT}\) field in \(1.29 \mathrm{~ms}\). Calculate its mass in atomic mass units.

A proton, a deuteron \((q=+e, m=2.0 \mathrm{u})\), and an alpha particle \((q=+2 e, m=4.0 \mathrm{u})\) all having the same kinetic energy enter a region of uniform magnetic field \(\vec{B}\), moving perpendicular to \(\vec{B}\). What is the ratio of (a) the radius \(r_{d}\) of the deuteron path to the radius \(r_{p}\) of the proton path and (b) the radius \(r_{\alpha}\) of the alpha particle path to \(r_{2} ?\)

Atom 1 of mass \(35 \mathrm{u}\) and atom 2 of mass \(37 \mathrm{u}\) 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 \(V=7.3 \mathrm{kV}\), each ion follows a circular path in \(\mathrm{a}\) uniform magnetic field of magnitude \(B=0.50 \mathrm{~T}\). What is the distance \(\Delta x\) between the points where the ions strike the detector?

A positron with kinetic energy \(2.00 \mathrm{keV}\) is projected into a uniform magnetic field \(\vec{B}\) of magnitude \(0.100 \mathrm{~T}\), with its velocity vector making an angle of \(89.0^{\circ}\) with \(\vec{B}\). Find (a) the period, (b) the pitch \(p\), and (c) the radius \(r\) of its helical path.
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