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Chapter 28: Problem 39

(a) What uniform magnetic field, applied perpendicular to a beam of electrons moving at \(1.30 \times 10^{6} \mathrm{~m} / \mathrm{s}\), is required to make the electrons travel in a circular arc of radius \(0.500 \mathrm{~m} ?(\mathrm{~b})\) What is the period of the motion?

Short Answer

Expert verified
(a) The magnetic field is approximately 0.0148 T. (b) The period is approximately 2.42 碌s.

Step by step solution

01

Understanding the Problem

We're tasked with finding (a) the uniform magnetic field required for electrons to move in a circular path of a given radius and (b) the period of this motion.
02

Relate Magnetic Force to Centripetal Force

For an electron moving in a circle under a magnetic field, the magnetic force provides the necessary centripetal force to keep the electron in its circular path. The magnetic force is given by the equation \( F_B = qvB \) where \( q \) is the charge of the electron, \( v \) is its velocity, and \( B \) is the magnetic field. The centripetal force is given by \( F_c = \frac{mv^2}{r} \) where \( m \) is the mass of the electron and \( r \) is the radius of the path.
03

Set Magnetic Force Equal to Centripetal Force

Set \( qvB = \frac{mv^2}{r} \). We'll solve for \( B \). Simplifying gives \( B = \frac{mv}{qr} \). We will substitute for \( m = 9.11 \times 10^{-31} \text{ kg} \), \( v = 1.30 \times 10^6 \text{ m/s} \), \( q = 1.60 \times 10^{-19} \text{ C} \), and \( r = 0.500 \text{ m} \).
04

Calculate the Magnetic Field

Plugging in the values, we get:\[ B = \frac{(9.11 \times 10^{-31} \text{ kg})(1.30 \times 10^6 \text{ m/s})}{(1.60 \times 10^{-19} \text{ C})(0.500 \text{ m})} \]After calculation, \( B \approx 0.0148 \text{ T} \).
05

Calculate the period of the Electron's Circular Motion

The period \( T \) is the time it takes for one complete revolution. It can be found using \( T = \frac{2\pi r}{v} \). Substitute \( r = 0.500 \text{ m} \) and \( v = 1.30 \times 10^6 \text{ m/s} \).
06

Calculate the Period

Plugging in the values, we get:\[ T = \frac{2\pi (0.500 \text{ m})}{1.30 \times 10^6 \text{ m/s}} \]After calculation, \( T \approx 2.42 \times 10^{-6} \text{ s} \).

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Key Concepts

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

Centripetal Force
When any object moves in a circular path, it's constantly changing direction. This change in direction requires a force, called the centripetal force, which is always directed toward the center of the circle. It's important to understand that even if an object moves at a constant speed around a circle, its velocity is not constant due to the changing direction.

For an electron traveling in a circular path due to a magnetic field, this centripetal force is provided by the magnetic force. The relationship between these forces is crucial in understanding electron motion in magnetic fields:
  • The centripetal force is calculated as \( F_c = \frac{mv^2}{r} \), where \( m \) is the mass, \( v \) is the velocity, and \( r \) is the radius of the path.
  • The magnetic force which acts as the centripetal force is \( F_B = qvB \), where \( q \) is the charge and \( B \) is the magnetic field.
Understanding this relationship helps in calculating unknown variables like the magnetic field when other parameters are given.
Electron Motion
Electrons are subatomic particles with a small mass and a negative charge. When they move through a magnetic field, they experience a magnetic force because of their charge. This force influences their motion and can cause them to travel in circular paths.

In the case where a uniform magnetic field is applied perpendicular to the path of electron motion:
  • The electron enters the field and begins to move in a circular orbit.
  • The motion is dictated by the balance between the magnetic force and the centripetal force required for circular motion.
  • The radius of the circular path can be adjusted by changing the strength of the magnetic field or the speed of the electron.
This motion doesn't change the speed of the electron but redirects its path, making it essential to relate the input conditions to the resulting curve or path.
Magnetic Field Calculation
To calculate the magnetic field needed to keep an electron moving in a circle, you must balance the forces acting on the electron. This requires knowing the mass, velocity, charge of the electron, and radius of the circle.

Use the formula for magnetic force, \( F_B = qvB \), and set it equal to the centripetal force, \( F_c = \frac{mv^2}{r} \), to solve for the magnetic field \( B \). Simplifying gives:
  • \( B = \frac{mv}{qr} \)
  • By substituting known values for the mass \( m = 9.11 \times 10^{-31} \) kg, velocity \( v = 1.30 \times 10^6 \) m/s, charge \( q = 1.60 \times 10^{-19} \) C, and radius \( r = 0.500 \) m, you can find the magnetic field \( B \).
Solving this gives a magnetic field of approximately 0.0148 T. This calculation is crucial in understanding how different forces affect electron paths and can be applied in various magnetic applications.

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Most popular questions from this chapter

An electron is accelerated from rest by a potential difference of \(380 \mathrm{~V}\). It then enters a uniform magnetic field of magnitude \(200 \mathrm{mT}\) with its velocity perpendicular to the field. Calculate (a) the speed of the electron and (b) the radius of its path in the magnetic field.

An electron moves through a uniform magnetic field given by \(\vec{B}=B_{x} \hat{\mathrm{i}}+\left(-3.0 B_{x}\right) \hat{\mathrm{j}}\). At a particular instant, the electron has velocity \(\vec{v}=(2.0 \hat{\mathrm{i}}+4.0 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}\) and the magnetic force acting on it is \(\left(6.4 \times 10^{-19} \mathrm{~N}\right) \hat{\mathrm{k}}\). Find \(B_{x^{-}}\)

In a certain cyclotron a proton moves in a circle of radius \(0.500 \mathrm{~m}\). The magnitude of the magnetic field is \(1.00 \mathrm{~T}\). (a) What is the oscillator frequency? (b) What is the kinetic energy of the proton, in electron-volts?

An electron that has an instantaneous velocity of $$ \vec{v}=\left(-5.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\right) \hat{\mathrm{i}}+\left(3.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\right) \hat{\mathrm{j}} $$ is moving through the uniform magnetic field \(\vec{B}=(0.030 \mathrm{~T}) \hat{\mathrm{i}}-\) \((0.15 \mathrm{~T}) \hat{j}\). (a) Find the force on the electron due to the magnetic field. (b) Repeat your calculation for a proton having the same velocity.

An electron follows a helical path in a uniform magnetic field of magnitude \(1.30 \mathrm{~T}\). The pitch of the path is \(6.00 \mu \mathrm{m}\), and the magnitude of the magnetic force on the electron is \(2.00 \times 10^{-14} \mathrm{~N}\). What is the electron's speed?
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