Ch 29: The Magnetic Field

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 29: The Magnetic Field

Problem 61

Chapter 29, Problem 61

An antiproton (same properties as a proton except that q = -e) is moving in the combined electric and magnetic fields of FIGURE P29.61. What are the magnitude and direction of the antiproton's acceleration at this instant?

Verified step by step guidance

Step 1: Identify the forces acting on the antiproton. The antiproton experiences two forces: the electric force due to the electric field $ E $ and the magnetic force due to the magnetic field $ B $. The electric force is given by $ F_E = qE $, and the magnetic force is given by $ F_B = qvB \sin \theta $, where $ \theta $ is the angle between the velocity $ v $ and the magnetic field $ B $.

Step 2: Calculate the electric force. The charge of the antiproton is $ q = -e $, where $ e $ is the elementary charge $ e = 1.6 \times 10^{-19} \, \text{C} $. The electric field $ E $ is given as $ 1000 \, \text{V/m} $. Use the formula $ F_E = qE $ to find the magnitude of the electric force.

Step 3: Calculate the magnetic force. The velocity of the antiproton is $ v = 500 \, \text{m/s} $, and the magnetic field $ B $ is $ 2.5 \, \text{T} $. Since the velocity is perpendicular to the magnetic field ($ \theta = 90^\circ $), $ \sin \theta = 1 $. Use the formula $ F_B = qvB \sin \theta $ to find the magnitude of the magnetic force.

Step 4: Determine the net force acting on the antiproton. The electric force and magnetic force act in perpendicular directions. Use vector addition to find the net force. The magnitude of the net force can be calculated using the Pythagorean theorem: $ F_{\text{net}} = \sqrt{F_E^2 + F_B^2} $.

Step 5: Calculate the acceleration of the antiproton. Use Newton's second law $ F = ma $, where $ m $ is the mass of the antiproton (same as the proton, $ m = 1.67 \times 10^{-27} \, \text{kg} $). The direction of the acceleration will be the same as the direction of the net force.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Lorentz Force

The Lorentz force describes the force experienced by a charged particle moving through electric and magnetic fields. It is given by the equation F = q(E + v × B), where F is the force, q is the charge, E is the electric field, v is the velocity of the particle, and B is the magnetic field. This concept is crucial for determining the acceleration of the antiproton in the given fields.

Acceleration

Acceleration is the rate of change of velocity of an object and is defined as a vector quantity. According to Newton's second law, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass (a = F/m). In this scenario, the net force on the antiproton, derived from the Lorentz force, will determine its acceleration.

Direction of Forces

The direction of forces acting on a charged particle in electric and magnetic fields is determined by the right-hand rule. For the electric field, the force direction is along the field lines for positive charges and opposite for negative charges. For the magnetic field, the force direction is perpendicular to both the velocity of the particle and the magnetic field direction, which is essential for understanding how the antiproton will move in the combined fields.

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Related Practice

Textbook Question

FIGURE P29.64 shows a mass spectrometer, an analytical instrument used to identify the various molecules in a sample by measuring their charge-to-mass ratio q/m. The sample is ionized, the positive ions are accelerated (starting from rest) through a potential difference ∆V, and they then enter a region of uniform magnetic field. The field bends the ions into circular trajectories, but after just half a circle they either strike the wall or pass through a small opening to a detector. As the accelerating voltage is slowly increased, different ions reach the detector and are measured. Consider a mass spectrometer with a 200.00 mT magnetic field and an 8.0000 cm spacing between the entrance and exit holes. To five significant figures, what accelerating potential differences ∆V are required to detect the ions (a) O₂⁺ (b) N₂⁺ and (c) CO⁺? See Exercise 29 for atomic masses; the mass of the missing electron is less than 0.001 u and is not relevant at this level of precision. Although N₂⁺ and CO⁺ both have a nominal molecular mass of 28, they are easily distinguished by virtue of their slightly different accelerating voltages. Use the following constants: 1 u = 1.6605 x 10⁻²⁷ kg, e = 1.6022 x 10⁻¹⁹ C.

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Textbook Question

A long, hollow wire has inner radius R₁ and outer radius R₂. The wire carries current I uniformly distributed across the area of the wire. Use Ampère's law to find an expression for the magnetic field strength in the three regions 0 < r < R₁, R₁ < r < R₂, and R₂ < r.

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Textbook Question

The uniform 30 mT magnetic field in FIGURE P29.65 points in the positive z-direction. An electron enters the region of magnetic field with a speed of 5.0 x 10⁶ m/s and at an angle of 30° above the xy-plane. Find the radius r and the pitch p of the electron's spiral trajectory.

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Textbook Question

A 65-cm-diameter cyclotron uses a 500 V oscillating potential difference between the dees. What is the maximum kinetic energy of a proton if the magnetic field strength is 0.75 T?

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Textbook Question

A proton moving in a uniform magnetic field with ${\vec{v}}_{1} = (1.00 \times 10^{6} \hat{i}) m/s$ experiences force ${\vec{F}}_{1} = (1.20 \times 10^{- 16} \hat{k}) N$ . A second proton with ${\vec{v}}_{2} = (2.0 \times 10^{6} \hat{j}) m/s$ experiences ${\vec{F}}_{2} = (- 4.16 \times 10^{- 16} \hat{k}) N$ in the same field. What is $\vec{B}$ ? Give your answer as a magnitude and an angle measured counter-clockwise from the $+ x$ -axis.

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Textbook Question

An electron in a cathode-ray tube is accelerated through a potential difference of 10 kV, then passes through the 2.0-cm-wide region of uniform magnetic field in FIGURE P29.60. What field strength will deflect the electron by 10°?

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