/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 A beam of protons is accelerated... [FREE SOLUTION] | 91Ó°ÊÓ

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A beam of protons is accelerated through a potential difference of \(0.745 \mathrm{kV}\) and then enters a uniform magnetic field traveling perpendicular to the field. (a) What magnitude of field is needed to bend these protons in a circular are of diameter \(1.75 \mathrm{~m} ?\) (b) What magnetic field would be needed to produce a path with the same diameter if the particles were electrons having the same speed as the protons?

Short Answer

Expert verified
(a) B is calculated using proton's mass; (b) B_e uses electron's mass with the same speed.

Step by step solution

01

Calculate the Speed of Protons

First, we need to determine the speed of the protons after being accelerated through the potential difference. The kinetic energy (KE) gained by the protons is equal to the electric potential energy (qV). For protons, \( q = e \) (the elementary charge), so \( KE = eV \). The kinetic energy is also given by \( KE = \frac{1}{2}mv^2 \). Equating these expressions, we have:\[ eV = \frac{1}{2}mv^2 \]Solving for \( v \), we find:\[ v = \sqrt{\frac{2eV}{m}} \]For protons, \( e = 1.602 \times 10^{-19} \text{ C} \), \( m = 1.673 \times 10^{-27} \text{ kg} \), and \( V = 745 \text{ V} \). Plug these values into the equation and calculate \( v \).
02

Use Magnetic Force to Find Magnetic Field for Protons

Next, we use the fact that the magnetic force provides the centripetal force necessary to bend the proton's path into a circle. The magnetic force is given by \( F = qvB \), and the centripetal force is \( F = \frac{mv^2}{r} \). Set these two expressions equal:\[ qvB = \frac{mv^2}{r} \]Simplify to solve for the magnetic field:\[ B = \frac{mv}{qr} \]The radius \( r \) is half of the given diameter, so \( r = \frac{1.75}{2} = 0.875 \text{ m} \). Substitute \( m \), \( v \), \( q \), and \( r \) to find \( B \).
03

Express the Speed of Electrons

For the electrons, we maintain the same speed as the protons. Recall from Step 1 that the speed \( v \) calculated for protons should be used for electrons in this scenario.
04

Use Magnetic Force to Find Magnetic Field for Electrons

Similar to the proton case, apply the expression for the magnetic force equaling the centripetal force for electrons:\[ e v B_e = \frac{m_e v^2}{r} \]Solve for \( B_e \):\[ B_e = \frac{m_e v}{e r} \]Here, \( m_e = 9.109 \times 10^{-31} \text{ kg} \). Use the same \( v \) and \( r \) as in the proton case, and the same charge \( e \), to find \( B_e \).

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Key Concepts

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

Proton Acceleration
When protons are accelerated through a potential difference, they gain kinetic energy, which leads to an increase in their speed. Imagine you have a proton starting from rest. As it is accelerated by a potential difference, its potential energy is converted into kinetic energy. This process can be described by the equation:
  • Potential energy: \(qV\)
  • Kinetic energy: \(\frac{1}{2}mv^2\)
The charge of a proton is represented by \(e\), and its mass is a known constant. By equating these two forms of energy (i.e., \(eV = \frac{1}{2}mv^2\)), we can solve for the velocity \(v\) of the protons. This is the speed they will have after being accelerated by the potential difference. Once we calculate this velocity using the given potential difference and known constants, we gain insight into how fast the proton beam is moving as it enters the magnetic field. This step is crucial since the speed of the protons will greatly influence subsequent calculations related to magnetic field interactions.
Centripetal Force
Centripetal force is the force required to keep an object moving in a circular path. For charged particles like protons moving through a magnetic field, this force is provided by the magnetic force.
As the protons enter the magnetic field, they experience a force that is perpendicular to both their motion and the magnetic field. This magnetic force keeps them moving along a circular path, ensuring they don't drift away in a straight line.
To understand this interaction, recall the equation for magnetic force:
  • Magnetic force: \(qvB\)
Meanwhile, the formula for centripetal force is:
  • Centripetal force: \(\frac{mv^2}{r}\)
Setting these two forces equal allows us to find the magnetic field \(B\) needed to bend the proton's path into a circle of the desired diameter, where \(r\) is the radius of the path (half the diameter). This ensures that the proton beam follows a circular trajectory inside the magnetic field.
Electron Path
Electrons, much like protons, also follow a circular path when subjected to a magnetic field, but there are key differences due to their significantly smaller mass. To make electrons travel in a circular path of the same size as protons, we use the same speed calculated earlier for the protons.
The expression for magnetic force and centripetal force for electrons remains similar:
  • Magnetic force (electron): \(e v B_e\)
  • Centripetal force: \(\frac{m_e v^2}{r}\)
Here, \(m_e\) represents the mass of an electron, which is much smaller than that of a proton. To maintain the same radius of curvature, the magnetic field \(B_e\) for electrons will differ, targeting an appropriately different magnitude due to the electron's smaller mass. By calculating \(B_e\), we adjust for these differences, enabling the electrons to navigate the same path as the protons in the magnetic field.
Potential Difference
Potential difference, often termed as voltage, is a measure of electric potential energy per unit charge. It's like a push that accelerates charged particles, such as protons or electrons. When these particles move through a potential difference, they gain kinetic energy which manifests as an increase in their velocity.
The formula connecting potential difference \(V\) with energy is:
  • Energy gained: \(qV\)
Here, \(q\) represents the charge of the particle. In this context, when a proton or an electron moves through a potential difference, its speed changes based on this energy gain.
This is the driving force behind the initial speed of particles entering a magnetic field. Understanding potential difference helps clarify how energy is transformed as charged particles accelerate, paving their way into more complex interactions, such as those involving magnetic fields.

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Most popular questions from this chapter

As a new electrical technician, you are designing a large solenoid to produce a uniform 0.150 T magnetic field near its center. You have enough wire for 4000 circular turns, and the solenoid must be \(1.40 \mathrm{~m}\) long and \(2.00 \mathrm{~cm}\) in diameter. What current will you need to produce the necessary field?

The magnetic field around the human head has been measured to be approximately \(3.0 \times 10^{-8}\) gauss. Although the currents that cause this field are quite complicated, we can get a rough estimate of their size by modeling them as a single circular current loop \(16 \mathrm{~cm}\) (the width of a typical head) in diameter. What is the current needed to produce such a field at the center of the loop?

A long, straight, cylindrical wire of radius \(R\) carries a current uniformly distributed over its cross section. At what location is the magnetic field produced by this current equal to half of its largest value? Use Ampère's law and consider points inside and outside the wire.

A current-carrying wire of length \(0.15 \mathrm{~m}\) is in a perpendicular magnetic field. The magnetic force on the wire is measured as a function of the current through the wire. The resulting data are given in the table. $$\begin{array}{ll}\hline \text { Current (A) } & \text { Force (N) } \\\\\hline 0.5 & 3.0 \times 10^{-3} \\\0.8 & 9.6 \times 10^{-3} \\\1.4 & 1.7 \times 10^{-2} \\\1.9 & 2.3 \times 10^{-2} \\\\\hline\end{array}$$ Make a plot of the force as a function of the current. Using a "best fit" to the data, determine the magnitude of the magnetic field.

A solenoid is designed to produce a 0.0279 T magnetic field near its center. It has a radius of \(1.40 \mathrm{~cm}\) and a length of \(40.0 \mathrm{~cm},\) and the wire carries a current of \(12.0 \mathrm{~A}\). (a) How many turns must the solenoid have? (b) What total length of wire is required to make this solenoid?

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