Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 26, Problem 42P
(a)
To determine
The speed of the ion enters the mass spectrometer.
(b)
To determine
The difference in the diameter of orbits.
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Before entering a mass spectrometer, ions pass through a velocity selector consisting of parallel plates separated by 2.1 mm and having a potential difference
200 V. The magnetic field between the plates is 0.42 T. The magnetic field in the mass spectrometer is 1.2 T.
(a) Find the speed of the ions entering the mass spectrometer.
m/s
(b) Find the difference in the diameters of the orbits of singly ionized 238U and 235U. (The mass of a 235U ion is 3.903 x
10-25
kg.)
mm
Chapter 26 Solutions
Physics for Scientists and Engineers
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- Check Your Understanding Repeat the previous problem with the magnetic field in the x-direction rather than in the z-direction. Check your answers with RHR-1.arrow_forwardCheck Your Understanding In what orientation would a magnetic dipole have to be to produce (a) a maximum torque in a magnetic field? (b) A maximum energy of the dipole?arrow_forwardA proton precesses with a frequency p in the presence of a magnetic field. If the intensity of the magnetic field is doubled, what happens to the precessional frequency?arrow_forward
- A mass spectrometer (Fig. 30.40, page 956) operates with a uniform magnetic field of 20.0 mT and an electric field of 4.00 103 V/m in the velocity selector. What is the radius of the semicircular path of a doubly ionized alpha particle (ma = 6.64 1027 kg)?arrow_forwardAn electron is accelerated through 2.40 103 V from rest and then enters a uniform 1.70-T magnetic field. What are (a) the maximum and (b) the minimum values of the magnetic force this particle experiences?arrow_forwardIs B constant in magnitude for points that lie on a magnetic field line?arrow_forward
- At a particular instant an electron is traveling west to east with a kinetic energy of 10 keV. Earth's magnetic field has a horizontal component of 1.8105 T north and a vertical component of 5.0105 T down. (a) What is the path of the election? (b) What is the radius of curvature of the path?arrow_forwardWhen the current through a circular loop is 6.0 A, the magnetic field at its center is 2.0104 T. What is the radius of the loop?arrow_forwardConsider the mass spectrometer shown schematically in Active Figure 22.12. The magnitude of the electric field between the plates of the velocity selector is 2.50 103 V/m, and the magnetic field in both the velocity selector and the deflection chamber has a magnitude of 0.035 0 T. Calculate the radius of the path for a singly charged ion having a mass m = 2.18 1026 kg.arrow_forward
- Check Your Understanding The wire loop forms a full circle of radius R and current I. What is the magnitude of the magnetic field at the center?arrow_forwardThe magnetic field in a cyclotron is 1.25 T, and the maximum orbital radius of the circulating protons is 0.40 m. (a) What is the kinetic energy of the protons when they are ejected from the cyclotron? (b) What Is this energy in MeV? (c) Through what potential difference would a proton have to be accelerated to acquire this kinetic energy? (d) What is the period of tire voltage source used to accelerate the piotons? (e) Repeat tire calculations for alpha-particles.arrow_forwardAn electron in a TV CRT moves with a speed of 6.0107 m/s, in a direction perpendicular to Earth's field, which has a strength of 5.0105 T. (a) What strength electric field must be applied perpendicular to the Earth’s field to make the election moves in a straight line? (b) If this is done between plates separated by 1.00 cm, what is the voltage applied? (Note that TVs are usually surrounded by a ferromagnetic material to shield against external magnetic fields and avoid the need for such a collection,)arrow_forward
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