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Chapter 4: Problem 102

A magnetic field forces an electron to move in a circle with radial acceleration \(3.0 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2} .\) (a) What is the speed of the electron if the radius of its circular path is \(15 \mathrm{~cm} ?\) (b) What is the period of the motion?

Short Answer

Expert verified
(a) The speed of the electron is \(6.7 \times 10^6 \mathrm{~m/s}\). (b) The period of the motion is \(1.41 \times 10^{-7}\) seconds.

Step by step solution

01

Understanding Radial Acceleration

Radial acceleration in a circular motion is given by the formula \(a = \frac{v^2}{r}\), where \(a\) is radial acceleration, \(v\) is the speed, and \(r\) is the radius. We need to find the speed \(v\) using the given acceleration and radius, \(a = 3.0 \times 10^{14} \mathrm{~m/s^2}\) and \(r = 0.15 \mathrm{~m}\).
02

Solving for the Speed of the Electron

Using the radial acceleration formula, \(v^2 = a \cdot r\). Substitute \(a = 3.0 \times 10^{14}\) and \(r = 0.15\) into the equation: \(v^2 = 3.0 \times 10^{14} \times 0.15\), which gives \(v^2 = 4.5 \times 10^{13}\). Solving for \(v\), we get \(v = \sqrt{4.5 \times 10^{13}} = 6.7 \times 10^6 \mathrm{~m/s}\).
03

Understanding the Period of Motion

The period \(T\) of circular motion is the time taken for one complete revolution. It is calculated as \(T = \frac{2\pi r}{v}\), where \(r\) is the radius, and \(v\) is the speed.
04

Solving for the Period of the Motion

Using \(r = 0.15\) m and \(v = 6.7 \times 10^6 \mathrm{~m/s}\), substitute into the period formula: \(T = \frac{2\pi \times 0.15}{6.7 \times 10^6}\). Calculating, \(T \approx 1.41 \times 10^{-7}\) seconds.

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Key Concepts

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

Magnetic Field
Magnetic fields are invisible forces that significantly influence charged particles, like electrons. When a charged particle enters a magnetic field, it experiences a force perpendicular to both its velocity and the magnetic field itself. This phenomenon is key to understanding circular motion under magnetic influence.
In practical terms, the magnetic field causes an electron to follow a curved path instead of moving in a straight line. This effect is utilized in many technologies, such as in CRT monitors and particle accelerators. In these cases, the magnetic field alters the path of electrons, guiding them as needed.
To summarize, magnetic fields can change the motion of electrons, forcing them to move in circles and if the field and speed remain constant, they will continue to move in a stable circular path.
Electron Dynamics
Electron dynamics refers to how electrons behave when subjected to different forces and fields. When electrons are in a magnetic field, they experience a force known as the Lorentz force, which is perpendicular to both the magnetic field and the direction of the electron's velocity.
This causes them to move in circular or helical paths, dictated by their speed and the strength of the magnetic field. The radius of this path is determined by the balance of this force with the centripetal force needed to maintain circular motion.
Thus, understanding electron dynamics is crucial when analyzing their paths in magnetic fields, as it tells us how various factors like velocity and field strength affect their motion. It also forms the basis for complex phenomena in electromagnetic theory and applications.
Radial Acceleration
Radial acceleration is the acceleration experienced by an object moving in a circle, which points towards the center of the circle. In electron dynamics, radial acceleration (\(a\)) can be expressed as \(a = \frac{v^2}{r}\), where \(v\) is the speed of the electron and \(r\) is the path's radius.
This acceleration is responsible for changing the direction of the electron's velocity as it moves through the magnetic field. Even though the speed may remain constant, without radial acceleration, the electron would not stay on its circular path.
Solving for the electron's speed involves rearranging the radial acceleration formula: \(v^2 = a \cdot r\). By plugging in the known values, one can calculate the electron鈥檚 speed on its circular path. Radial acceleration is fundamental in understanding circular motion dynamics.
Period of Motion
The period of motion in circular dynamics refers to the time it takes for an object to complete one full cycle around its path. For an electron in a magnetic field, this is an important measure of how it travels in a circle. The period (\(T\)) is determined by the equation \(T = \frac{2\pi r}{v}\), where \(r\) is the radius and \(v\) is the electron鈥檚 speed.
This formula shows that the period is directly proportional to the radius of the circular path and inversely proportional to the speed. A longer path or slower speed results in a longer period.
  • Radius affects the path length, thus affecting the period.
  • Speed, being inversely related, means faster electrons complete cycles more quickly.
Understanding the period of motion is crucial for predictions about electron behavior in fields and for calculating timing in circuits and devices.

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Most popular questions from this chapter

A graphing surprise. At time \(t=0,\) a burrito is launched from level ground, with an initial speed of \(16.0 \mathrm{~m} / \mathrm{s}\) and launch angle \(\theta_{0}\). Imagine a position vector \(\vec{r}\) continuously directed from the launching point to the burrito during the flight. Graph the magnitude \(r\) of the position vector for (a) \(\theta_{0}=40.0^{\circ}\) and (b) \(\theta_{0}=80.0^{\circ} .\) For \(\theta_{0}=40.0^{\circ},\) (c) when does \(r\) reach its maximum value, (d) what is that value, and how far (e) horizontally and (f) vertically is the burrito from the launch point? For \(\theta_{0}=80.0^{\circ},(\mathrm{g})\) when does \(r\) reach its maximum value, (h) what is that value, and how far (i) horizontally and (j) vertically is the burrito from the launch point?

A rugby player runs with the ball directly toward his opponent's goal, along the positive direction of an \(x\) axis. He can legally pass the ball to a teammate as long as the ball's velocity relative to the field does not have a positive \(x\) component. Suppose the player runs at speed \(4.0 \mathrm{~m} / \mathrm{s}\) relative to the field while he passes the ball with velocity \(\vec{v}_{B P}\) relative to himself. If \(\vec{v}_{B P}\) has magnitude \(6.0 \mathrm{~m} / \mathrm{s},\) what is the smallest angle it can have for the pass to be legal?

The pilot of an aircraft flies due east relative to the ground in a wind blowing \(20.0 \mathrm{~km} / \mathrm{h}\) toward the south. If the speed of the aircraft in the absence of wind is \(70.0 \mathrm{~km} / \mathrm{h},\) what is the speed of the aircraft relative to the ground?

A batter hits a pitched ball when the center of the ball is \(1.22 \mathrm{~m}\) above the ground. The ball leaves the bat at an angle of \(45^{\circ}\) with the ground. With that launch, the ball should have a horizontal range (returning to the launch level) of \(107 \mathrm{~m}\). (a) Does the ball clear a 7.32 -m-high fence that is \(97.5 \mathrm{~m}\) horizontally from the launch point? (b) At the fence, what is the distance between the fence top and the ball center?

A purse at radius \(2.00 \mathrm{~m}\) and a wallet at radius \(3.00 \mathrm{~m}\) travel in uniform circular motion on the floor of a merry-go-round as the ride turns. They are on the same radial line. At one instant, the acceleration of the purse is \(\left(2.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(4.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}} .\) At that instant and in unit-vector notation, what is the acceleration of the wallet?
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