/*! 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 An \(\alpha\) -particle has a ch... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An \(\alpha\) -particle has a charge of \(+2 e\) and a mass of \(6.64 \times 10^{-27}\) kg. It is accelerated from rest through a potential difference that has a value of \(1.20 \times\) \(10^{6} \mathrm{V}\) and then enters a uniform magnetic field whose magnitude is \(2.20 \mathrm{T}\) The \(\alpha\) -particle moves perpendicular to the magnetic field at all times. What is (a) the speed of the \(\alpha\) -particle, (b) the magnitude of the magnetic force on it, and (c) the radius of its circular path?

Short Answer

Expert verified
(a) Speed: \(1.54 \times 10^7\) m/s, (b) Magnetic force: \(1.08 \times 10^{-12}\) N, (c) Radius: 4.63 cm.

Step by step solution

01

Calculate the Speed of the Alpha Particle

First, use the energy conservation principle. The kinetic energy gained by the alpha particle equals the potential energy provided by the electric field. This is given by the equation\[K.E. = qV = \frac{1}{2}mv^2\]where \(q = +2e\) is the charge of the alpha particle, \(V = 1.20 \times 10^6 \text{ V}\) is the potential difference, and \(m = 6.64 \times 10^{-27} \text{ kg}\) is its mass.Plugging these values into the equation, \[(2 \times 1.60 \times 10^{-19} \text{ C} \times 1.20 \times 10^6 \text{ V}) = \frac{1}{2}(6.64 \times 10^{-27} \text{ kg})v^2\]Solving for \(v\), we find\[v = \sqrt{\frac{2 \times 2 \times 1.60 \times 10^{-19} \times 1.20 \times 10^6}{6.64 \times 10^{-27}}}\]Thus, the speed \(v\) is approximately \(1.54 \times 10^7 \text{ m/s}\).
02

Determine the Magnetic Force on the Alpha Particle

The magnetic force \(F_m\) on a charge moving in a magnetic field is given by the equation:\[F_m = qvB \sin(\theta)\]where \(q = 2e\) is the charge, \(v\) is the speed found in Step 1, \(B = 2.20 \text{ T}\) is the magnetic field strength, and \(\theta = 90^\circ\) since the particle moves perpendicular to the magnetic field.Substituting in the values, \(\sin(90^\circ) = 1\), so the force is:\[F_m = (2 \times 1.60 \times 10^{-19} \text{ C}) \cdot (1.54 \times 10^7 \text{ m/s}) \cdot (2.20 \text{ T})\]This results in \(F_m \approx 1.08 \times 10^{-12} \text{ N}\).
03

Find the Radius of the Circular Path

Since the force is centripetal, equate the magnetic force to the centripetal force required to keep the particle moving in a circle. The expression for centripetal force \(F_c\) is:\[F_c = \frac{mv^2}{r}\]where \(r\) is the radius of the circular path.Setting \(F_m = F_c\),\[qvB = \frac{mv^2}{r}\]Solve for \(r\):\[r = \frac{mv}{qB}\]Substitute in the values:\[r = \frac{6.64 \times 10^{-27} \cdot 1.54 \times 10^7}{2 \times 1.60 \times 10^{-19} \cdot 2.20}\]The radius \(r\) is approximately \(4.63 \times 10^{-2} \text{ m}\) or 4.63 cm.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Magnetic Field
When an alpha particle, a type of charged particle, enters a magnetic field, its motion is affected by the field. Imagine pushing something into a stream of water; the flow will change its path. Similarly, a magnetic field exerts a force on charged particles, causing them to move in a circular path. This is because the magnetic field applies a force that is always perpendicular to the particle's velocity, pulling it into a curved motion.

An alpha particle, with a charge of +2e, can be significantly influenced by a magnetic field. The strength of the effect depends on several factors:
  • The magnitude of the magnetic field, which in our example is 2.20 Tesla (T).
  • The charge on the particle, which is the fundamental charge multiplied by 2 for alpha particles.
  • The velocity of the particle as it enters the field.
In problems involving a magnetic field, it's crucial to remember that these factors work together to determine the particle's behavior and how much it will curve as it moves through the field.
Potential Difference
The potential difference, also known as voltage, is fundamental in propelling an alpha particle forward. When the particle is accelerated from rest, it gains energy from the electric field caused by the potential difference. Think of it like a hill that the particle rolls down, gaining speed as it goes.

In mathematical terms, the potential energy provided by this electric field is equal to the charge times the potential difference (qV). For an alpha particle, this equation looks like:
  • The charge \( q = +2e \)
  • The potential difference \( V = 1.20 \times 10^6 \, \text{V} \)
This potential energy is then converted entirely into kinetic energy as the particle starts moving. This concept explains why the alpha particle has a specific speed when entering the magnetic field, directly related to the voltage; higher the voltage, higher the speed.
Centripetal Force
Once the alpha particle is moving through the magnetic field, it travels in a circular path. This is due to the centripetal force. Imagine spinning a ball on a string; the string pulls the ball towards the center as it moves in a circle. Similarly, the magnetic force acts as the 'string', providing the centripetal force that keeps the particle moving in its curved path.

Mathematically, centripetal force is given by the expression:
  • \( F_c = \frac{mv^2}{r} \)
Where:
  • \( m \) is the mass of the particle
  • \( v \) is its velocity
  • \( r \) is the radius of the circle
In the presence of a magnetic field, this centripetal force is exactly equal to the magnetic force acting on the particle, which is calculated as \( qvB \). Understanding this balance helps us determine the radius of the path; knowing how these forces work together is key to predicting the motion of charged particles in magnetic fields.
Kinetic Energy
Kinetic energy is what an alpha particle gains when it is accelerated by a potential difference. This energy is the energy of motion, which means the faster the particle moves, the more kinetic energy it has.

The relationship between kinetic energy and speed can be defined using this equation:
  • \( K.E. = \frac{1}{2}mv^2 \)
Where:
  • \( m \) is the mass of the alpha particle
  • \( v \) is its speed
The entire energy imparted to the particle through the voltage (potential difference) is converted into kinetic energy. Because energy is conserved, the initial potential energy equals the final kinetic energy, allowing us to solve for the particle's speed. This conversion outlines how the alpha particle reaches its maximum speed before entering the magnetic field, showing us the direct effect of the potential difference on kinetic energy.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose that a uniform magnetic field is everywhere perpendicular to this page. The field points directly upward toward you. A circular path is drawn on the page. Use Ampère's law to show that there can be no net current passing through the circular surface.

An electron is moving through a magnetic field whose magnitude is \(8.70 \times 10^{-4} \mathrm{T}\). The electron experiences only a magnetic force and has an acceleration of magnitude \(3.50 \times 10^{14} \mathrm{m} / \mathrm{s}^{2} .\) At a certain instant, it has a speed of \(6.80 \times 10^{6} \mathrm{m} / \mathrm{s}\). Determine the angle \(\theta\) (less than \(90^{\circ}\) ) between the electron's velocity and the magnetic field.

The two conducting rails in the drawing are tilted upward so they each make an angle of \(30.0^{\circ}\) with respect to the ground. The vertical magnetic field has a magnitude of \(0.050 \mathrm{T}\). The \(0.20-\mathrm{kg}\) aluminum rod (length \(=\) \(1.6 \mathrm{m}\) ) slides without friction down the rails at a constant velocity. How much current flows through the rod?

Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The length of each rod is \(0.85 \mathrm{m},\) and the mass of each is \(0.073 \mathrm{kg} .\) One rod is held in place above the ground, while the other floats beneath it at a distance of \(8.2 \times 10^{-3} \mathrm{m} .\) Determine the current in the rods.

Suppose that an ion source in a mass spectrometer produces doubly ionized gold ions \(\left(\mathrm{Au}^{2+}\right),\) each with a mass of \(3.27 \times 10^{-25} \mathrm{kg} .\) The ions are accelerated from rest through a potential difference of \(1.00 \mathrm{kV} .\) Then, a 0.500-T magnetic field causes the ions to follow a circular path. Determine the radius of the path.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.