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 63a

Chapter 29, Problem 63a

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?

Verified step by step guidance

Convert the diameter of the cyclotron to radius in meters. The radius is half the diameter, so $ r = \frac{65}{2} $ cm, and then convert to meters: $ r = \frac{65}{2} \times 0.01 $ m.

Determine the maximum speed of the proton using the relationship between the magnetic force and centripetal force in a cyclotron: $ qvB = \frac{mv^2}{r} $, where $ q $ is the charge of the proton, $ v $ is the velocity, $ B $ is the magnetic field strength, $ m $ is the mass of the proton, and $ r $ is the radius. Solve for $ v $: $ v = \frac{qBr}{m} $.

Calculate the maximum kinetic energy of the proton using the formula $ KE = \frac{1}{2}mv^2 $. Substitute the expression for $ v $ from the previous step into this formula: $ KE = \frac{1}{2}m\left(\frac{qBr}{m}\right)^2 $. Simplify to $ KE = \frac{q^2B^2r^2}{2m} $.

Substitute the known values into the formula for $ KE $: $ q = 1.6 \times 10^{-19} $ C (charge of a proton), $ B = 0.75 $ T, $ r $ (calculated in step 1), and $ m = 1.67 \times 10^{-27} $ kg (mass of a proton).

Ensure the result is consistent with the energy provided by the oscillating potential difference. The maximum kinetic energy of the proton is limited by the energy gained per cycle, which is $ q \times V $, where $ V = 500 $ V. Compare this value with the calculated $ KE $ to confirm consistency.

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Key Concepts

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

Cyclotron

A cyclotron is a type of particle accelerator that uses a magnetic field and an oscillating electric field to accelerate charged particles, such as protons. The particles move in a circular path due to the magnetic field, and the oscillating electric field provides energy to the particles at each half-cycle, increasing their speed and kinetic energy.

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion, calculated using the formula KE = 0.5 * m * v^2, where m is mass and v is velocity. In the context of a cyclotron, the maximum kinetic energy of a proton can be determined by the potential difference it experiences as it accelerates through the electric field.

Magnetic Field Strength

Magnetic field strength, measured in teslas (T), influences the radius of the circular path that charged particles take in a cyclotron. A stronger magnetic field allows for greater acceleration of the particles, which can lead to higher kinetic energy. The relationship between the magnetic field, particle charge, and velocity is crucial for understanding the dynamics of particle motion in a cyclotron.

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Related Practice

Textbook Question

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?

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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

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 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

Particle accelerators, such as the Large Hadron Collider, use magnetic fields to steer charged particles around a ring. Consider a proton ring with 36 identical bending magnets connected by straight segments. The protons move along a 1.0-m-long circular arc as they pass through each magnet. What magnetic field strength is needed in each magnet to steer protons around the ring with a speed of 2.5 x 10⁷ m/s? Assume that the field is uniform inside the magnet, zero outside.

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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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