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Chapter 28: Problem 21

An electron of kinetic energy \(1.20 \mathrm{keV}\) circles in a plane perpendicular to a uniform magnetic field. The orbit radius is \(25.0 \mathrm{~cm}\). Find (a) the electron's speed, (b) the magnetic field magnitude, (c) the circling frequency, and (d) the period of the motion.

Short Answer

Expert verified
(a) Speed = \(2.05 \times 10^7 \text{ m/s}\), (b) Field = \(5.01 \times 10^{-4} \text{ T}\), (c) Frequency = \(1.31 \times 10^8 \text{ Hz}\), (d) Period = \(7.63 \times 10^{-9} \text{ s}\)

Step by step solution

01

Convert Kinetic Energy to Joules

First, convert the electron's kinetic energy from kiloelectronvolts to joules. The conversion is needed because we generally work in SI units.\[1.20 \text{ keV} = 1.20 \times 10^3 \text{ eV} = 1.20 \times 10^3 \times 1.602 \times 10^{-19} \text{ J}\]Calculating gives:\[1.20 \text{ keV} = 1.92 \times 10^{-16} \text{ J}\]
02

Calculate Electron's Speed

Use the kinetic energy formula to find the speed. The kinetic energy \(K\) of the electron is given by \[K = \frac{1}{2} mv^2\]where \(m\) is the electron's mass \((9.11 \times 10^{-31} \text{ kg})\) and \(v\) is the speed. Solving for \(v\), we get:\[v = \sqrt{\frac{2K}{m}}\]Substitute the values:\[v = \sqrt{\frac{2 \times 1.92 \times 10^{-16}}{9.11 \times 10^{-31}}} = 2.05 \times 10^7 \text{ m/s}\]
03

Calculate Magnetic Field Magnitude

For an electron moving in a magnetic field in a circular path, the centripetal force is provided by the magnetic Lorentz force:\[\frac{mv^2}{r} = qvB\]Where:- \(r\) is the orbit radius, \(25.0 \text{ cm} = 0.25 \text{ m}\)- \(q\) is the electron's charge, \(-1.602 \times 10^{-19} \text{ C}\)- \(B\) is the magnetic field magnitude.Rearranging for \(B\), we find:\[B = \frac{mv}{qr}\]Substitute the values:\[B = \frac{9.11 \times 10^{-31} \times 2.05 \times 10^7}{1.602 \times 10^{-19} \times 0.25} = 5.01 \times 10^{-4} \text{ T}\]
04

Calculate Circling Frequency

The frequency \(f\) is the number of orbits per second. It is related to the speed and the radius of the circle:\[f = \frac{v}{2\pi r}\]Substitute the values:\[f = \frac{2.05 \times 10^7}{2\pi \times 0.25} = 1.31 \times 10^8 \text{ Hz}\]
05

Calculate the Period of the Motion

The period of the motion \(T\) is the reciprocal of the frequency.\[T = \frac{1}{f}\]Substitute the calculated frequency:\[T = \frac{1}{1.31 \times 10^8} = 7.63 \times 10^{-9} \text{ s}\]

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. For an electron moving in a magnetic field, it plays a key role in determining the speed of the electron. The formula for kinetic energy is:\[ K = \frac{1}{2}mv^2 \]Here, \( m \) represents the mass of the electron, and \( v \) is its speed. This formula can be rearranged to solve for speed if the kinetic energy is known, as shown in the step-by-step solution. Understanding how to convert and use kinetic energy in different units, such as from kiloelectronvolts to joules, is essential since physicists typically use SI units for calculations.
Magnetic Field
A magnetic field is a vector field that influences the motion of charged particles like electrons. In this exercise, the magnetic field is uniform, meaning its strength and direction are constant in space. The motion of the electron is directly affected by the magnetic field, which causes it to travel in a circular path. The strength of the magnetic field can be determined using the equation:\[ B = \frac{mv}{qr} \]where \( q \) is the charge of the electron, \( r \) is the radius of the circular path, and \( v \) is its velocity. This relationship reveals how magnetic fields can provide the necessary centripetal force for circular motion.
Centripetal Force
Centripetal force is the force that keeps an object moving in a circular path. For an electron in a magnetic field, this force is provided by the magnetic Lorentz force. The balance of forces ensures that the electron maintains its circular trajectory:\[ \frac{mv^2}{r} = qvB \]This equation shows the equality between the magnetic force and the centripetal force. Each of these variables is interconnected, meaning that changes in the magnetic field or the electron's speed impact the overall movement and trajectory of the electron.
Frequency and Period
The frequency of a circling electron tells us how many times per second the electron completes a full circle around its path. It is calculated as:\[ f = \frac{v}{2\pi r} \]where \( v \) is the velocity and \( r \) is the radius of the circle. Its reciprocal, the period \( T \), represents the time it takes to complete one full circle:\[ T = \frac{1}{f} \]These concepts are crucial for understanding oscillatory and rotational motions in physics. For example, in this case, the high frequency results in a very short period, indicating rapid motion of the electron.
Electromagnetic Theory
Electromagnetic theory encompasses the study of electric fields, magnetic fields, and their interactions with matter. The motion of electrons in a magnetic field is governed by these principles. Utilizing equations derived from electromagnetic theory, such as those for calculating kinetic energy, magnetic field strength, and centripetal force, allows us to predict the behavior of charged particles. - The electron's trajectory in the magnetic field exemplifies how magnetic fields bend and guide charged particles. - Applications of these principles can be seen in technologies such as cyclotrons and magnetic resonance imaging (MRI). Understanding electromagnetic theory provides the foundation needed to explore the larger realm of electromagnetism in physics.

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Most popular questions from this chapter

A circular coil of 160 turns has a radius of \(1.90 \mathrm{~cm} .\) (a) Calculate the current that results in a magnetic dipole moment of magnitude \(2.30 \mathrm{~A} \cdot \mathrm{m}^{2}\). (b) Find the maximum magnitude of the torque that the coil, carrying this current, can experience in a uniform \(35.0 \mathrm{mT}\) magnetic field.

An electron follows a helical path in a uniform magnetic field given by \(\vec{B}=(20 \hat{\mathrm{i}}-50 \hat{\mathrm{j}}-30 \hat{\mathrm{k}}) \mathrm{mT} .\) At time \(t=0,\) the elec- tron's velocity is given by \(\vec{v}=(20 \hat{\mathrm{i}}-30 \mathrm{j}+50 \hat{\mathrm{k}}) \mathrm{m} / \mathrm{s} .\) (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?

A long, rigid conductor, lying along an \(x\) axis, carries a current of \(5.0 \mathrm{~A}\) in the negative \(x\) direction. A magnetic field \(\vec{B}\) is present, given by \(\vec{B}=3.0 \hat{\mathrm{i}}+8.0 x^{2} \mathrm{j},\) with \(x\) in meters and \(\vec{B}\) in milliteslas. Find, in unit-vector notation, the force on the \(2.0 \mathrm{~m}\) segment of the conductor that lies between \(x=1.0 \mathrm{~m}\) and \(x=3.0 \mathrm{~m}\).

A proton circulates in a cyclotron, beginning approximately at rest at the center. Whenever it passes through the gap between dees, the electric potential difference between the dees is \(200 \mathrm{~V}\). (a) By how much does its kinetic energy increase with each passage through the gap? (b) What is its kinetic energy as it completes 100 passes through the gap? Let \(r_{100}\) be the radius of the proton's circular path as it completes those 100 passes and enters a dee, and let \(r_{101}\) be its next radius, as it enters a dee the next time. (c) By what percentage does the radius increase when it changes from \(r_{100}\) to \(r_{101} ?\) That is, what is $$ \text { percentage increase }=\frac{r_{101}-r_{100}}{r_{100}} 100 \% ? $$

A current loop, carrying a current of \(5.0 \mathrm{~A},\) is in the shape of a right triangle with sides \(30,40,\) and \(50 \mathrm{~cm} .\) The loop is in a uniform magnetic field of magnitude \(80 \mathrm{mT}\) whose direction is parallel to the current in the \(50 \mathrm{~cm}\) side of the loop. Find the magnitude of (a) the magnetic dipole moment of the loop and (b) the torque on the loop.
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