/*! 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 16 \(\bullet\) A beam of protons tr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(\bullet\) A beam of protons traveling at 1.20 \(\mathrm{km} / \mathrm{s}\) enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field in a direction perpendicular to its original direction (Fig. \(20.60 ) .\) The beam travels a distance of 1.18 cm while in the field. What is the magnitude of the magnetic field?

Short Answer

Expert verified
The magnitude of the magnetic field is approximately 0.0226 T.

Step by step solution

01

Understanding the Problem

We need to find the magnitude of the magnetic field that causes the proton beam to change direction from its original direction to a perpendicular one. To achieve this, we will use the fact that protons in a magnetic field follow a circular path, with the beam's path length corresponding to a quarter of the circle since it exits perpendicular to the original direction.
02

Identify Known Variables

The speed of the protons, \( v = 1.20 \) km/s, must be converted into meters per second. Therefore, \( v = 1200 \) m/s. The path or arc length \( L = 1.18 \) cm is converted into meters, \( L = 0.0118 \) m. The charge of a proton is \( q = 1.6 \times 10^{-19} \) C, and the mass of a proton is \( m = 1.67 \times 10^{-27} \) kg.
03

Connecting Concepts with Circular Motion

Since the beam’s path is one quarter of a circular orbit, the relationship between the angle subtended by the path (90 degrees, or \( \frac{\pi}{2} \) radians) and the radius of the circle \( r \) is given by \( L = r \cdot \frac{\pi}{2} \). Therefore, \( r = \frac{2 \cdot L}{\pi} \).
04

Apply Centripetal Motion Formula

The centripetal force needed to keep the protons in circular motion is given by \( F = \frac{mv^2}{r} \). Since this force is provided by the magnetic force \( F = qvB \) (where \( B \) is the magnetic field strength), we equate: \[ \frac{mv^2}{r} = qvB \]. Solving for \( B \), we get \( B = \frac{mv}{qr} \).
05

Calculate the Radius

Substitute \( L = 0.0118 \) m into the equation for radius: \[ r = \frac{2 \cdot 0.0118}{\pi} \approx 0.00751 \text{ m}\].
06

Solve for Magnetic Field Magnitude

Using the formula for \( B \), substitute known values: \[ B = \frac{(1.67 \times 10^{-27} \text{ kg})(1200 \text{ m/s})}{(1.6 \times 10^{-19} \text{ C})(0.00751 \text{ m})} \approx 0.0226 \text{ T}\].
07

Final Result

After calculating the aforementioned expression, the magnitude of the magnetic field is approximately 0.0226 Tesla.

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.

Protons
Protons are subatomic particles with a positive electric charge. They are located in the nucleus of an atom, along with neutrons. Protons carry a charge of about \( +1.6 \times 10^{-19} \ \text{C} \) and possess a mass of approximately \( 1.67 \times 10^{-27} \ \text{kg} \). These small particles play a crucial role in forming matter as the number of protons in the nucleus determines the atomic number, classifying the element.
In experiments, managing beams of protons allows scientists to explore the behavior of elementary particles under various forces, such as magnetic fields. This quality makes them ideal to study electromagnetic properties because their motion is easily influenced by their positive charge. Protons have applications in medical imaging, particle accelerators, and in probing the fundamental properties of matter.
Circular Motion
When objects, such as a beam of protons, move in circular paths, they exhibit circular motion. This motion occurs when there is a constant force acting perpendicular to the velocity of the protons, deflecting them into a curved path. In the case of this exercise, we see how magnetic force causes the protons to travel in a circular arc.
A key feature of circular motion is that even if the speed is constant, the object is continuously accelerating towards the center of the circle due to the change in direction. This inward acceleration is what causes the circular motion.
Understanding this type of motion is essential in many fields, such as engineering and physics. It helps us in designing safe structures like amusement rides or understanding planetary motions around stars.
Centripetal Force
Centripetal force is the "center-seeking" force required to keep an object moving along a circular path. In the case of protons entering a magnetic field, the magnetic force acts as the centripetal force. This force doesn't exist on its own but emerges from interactions—such as friction, tension, or magnetism—that happen to point towards the circle's center.
Mathematically, centripetal force \( F_c \) is given by the formula:
  • \( F_c = \frac{mv^2}{r} \)
Where:
  • \( m \) is the mass of the object
  • \( v \) is the velocity
  • \( r \) is the radius of the path
By understanding centripetal force, students can grasp how objects like planets stay in orbit and why roller coasters are designed the way they are.
Magnetic Force
Magnetic force is a vital concept in the realm of electromagnetism. When charged particles, such as protons, move through a magnetic field, they experience a force perpendicular to both the field and their direction of motion. This movement in the presence of a magnetic field causes them to follow a circular path.
The magnitude of magnetic force \( F_m \) influencing a charged particle can be calculated using the equation:
  • \( F_m = qvB \)
Which involves:
  • \( q \) - the electric charge of the particle
  • \( v \) - the velocity of the particle
  • \( B \) - the strength of the magnetic field
Magnetic force is essential in devices like motors, generators, and magnetic storage media. By examining how proton beams are deflected in magnetic fields, scientists can deduce important properties of the field and the particle's characteristics.

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

\(\bullet\) A 150 \(\mathrm{V}\) battery is connected across two parallel metal plates of area 28.5 \(\mathrm{cm}^{2}\) and separation 8.20 \(\mathrm{mm} .\) A beam of alpha particles (charge \(+2 e,\) mass \(6.64 \times 10^{-27} \mathrm{kg} )\) is accelerated from rest through a potential difference of 1.75 \(\mathrm{kV}\) and enters the region between the plates perpendicular to the electric field. What magnitude and direction of magnetic field are needed so that the alpha particles emerge undeflected from between the plates?

\(\bullet\) A 3.25 g bullet picks up an electric charge of 1.65\(\mu C\) as it travels down the barrel of a rifle. It leaves the barrel at a speed of 425 \(\mathrm{m} / \mathrm{s}\) , traveling perpendicular to the earth's magnetic field, which has a magnitude of \(5.50 \times 10^{-4} \mathrm{T} .\) Calculate (a) the magnitude of the magnetic force on the bullet and (b) the magnitude of the bullet's acceleration due to the magnetic force at the instant it leaves the rifle barrel.

\(\bullet\) In a 1.25 T magnetic field directed vertically upward, a particle having a charge of magnitude 8.50\(\mu \mathrm{C}\) and initially moving northward at 4.75 \(\mathrm{km} / \mathrm{s}\) is deflected toward the east. (a) What is the sign of the charge of this particle? Make a sketch to illustrate how you found your answer. (b) Find the magnetic force on the particle.

(a) What is the speed of a beam of electrons when the simultaneous influence of an electric field of \(1.56 \times 10^{4} \mathrm{V} / \mathrm{m}\) and a magnetic field of \(4.62 \times 10^{-3} \mathrm{T}\) , with both fields normal to the beam and to each other, produces no deflection of the electrons? (b) In a diagram, show the relative orientation of the vectors \(\vec{\boldsymbol{v}}\) , \(\vec{\boldsymbol{E}}\) and \(\vec{\boldsymbol{B}}\) . (c) When the electric field is removed, what is the radius of the electron orbit? What is the period of the orbit?

An ion having charge \(+6 e\) is traveling horizontally to the left at 8.50 \(\mathrm{km} / \mathrm{s}\) when it enters a magnetic field that is perpendicular to its velocity and deflects it downward with an initial magnetic force of \(6.94 \times 10^{-15} \mathrm{N} .\) What are the direction and magnitude of this field? Illustrate your method of solving this problem with a diagram.

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.