Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 36, Problem 69P
To determine
The proof that
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
What is the kinetic energy of an electron that passes undeviated through perpendicular electric and magnetic fields if E = 2.0 kV/m and B = 8.0 mT?
(m=9.1 × 10-31 kg , q=1.6x-19C )
Select one:
a. 1.60 eV
b. 0.71 eV
c. 2.85 eV
d. 0.18 eV
An electron in J. J. Thomson's charge-to-mass apparatus moves perpendicular to a B-field along a circular path of radius 13.10 cm. If imposition of an E-field of 17.80 kV/m makes the path straight, what is the value of B?
Determine the radius and pitch p (distance between loops) of the electron's helical path assuming its speedis 2.6x10^6 m/s See the figure
Chapter 36 Solutions
Physics for Scientists and Engineers
Ch. 36 - Prob. 1PCh. 36 - Prob. 2PCh. 36 - Prob. 3PCh. 36 - Prob. 4PCh. 36 - Prob. 5PCh. 36 - Prob. 6PCh. 36 - Prob. 7PCh. 36 - Prob. 8PCh. 36 - Prob. 9PCh. 36 - Prob. 10P
Ch. 36 - Prob. 11PCh. 36 - Prob. 12PCh. 36 - Prob. 13PCh. 36 - Prob. 14PCh. 36 - Prob. 15PCh. 36 - Prob. 16PCh. 36 - Prob. 17PCh. 36 - Prob. 18PCh. 36 - Prob. 19PCh. 36 - Prob. 20PCh. 36 - Prob. 21PCh. 36 - Prob. 22PCh. 36 - Prob. 23PCh. 36 - Prob. 24PCh. 36 - Prob. 25PCh. 36 - Prob. 26PCh. 36 - Prob. 27PCh. 36 - Prob. 28PCh. 36 - Prob. 29PCh. 36 - Prob. 30PCh. 36 - Prob. 31PCh. 36 - Prob. 32PCh. 36 - Prob. 33PCh. 36 - Prob. 34PCh. 36 - Prob. 35PCh. 36 - Prob. 36PCh. 36 - Prob. 37PCh. 36 - Prob. 38PCh. 36 - Prob. 39PCh. 36 - Prob. 40PCh. 36 - Prob. 41PCh. 36 - Prob. 42PCh. 36 - Prob. 43PCh. 36 - Prob. 44PCh. 36 - Prob. 45PCh. 36 - Prob. 46PCh. 36 - Prob. 47PCh. 36 - Prob. 48PCh. 36 - Prob. 49PCh. 36 - Prob. 50PCh. 36 - Prob. 51PCh. 36 - Prob. 52PCh. 36 - Prob. 53PCh. 36 - Prob. 54PCh. 36 - Prob. 55PCh. 36 - Prob. 56PCh. 36 - Prob. 57PCh. 36 - Prob. 58PCh. 36 - Prob. 59PCh. 36 - Prob. 60PCh. 36 - Prob. 61PCh. 36 - Prob. 62PCh. 36 - Prob. 63PCh. 36 - Prob. 64PCh. 36 - Prob. 65PCh. 36 - Prob. 66PCh. 36 - Prob. 67PCh. 36 - Prob. 68PCh. 36 - Prob. 69PCh. 36 - Prob. 70PCh. 36 - Prob. 71PCh. 36 - Prob. 72PCh. 36 - Prob. 73PCh. 36 - Prob. 74PCh. 36 - Prob. 75P
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Similar questions
- What is the velocity of a beam of electrons that go undeflected when passing through crossed (perpendicular) electric and magnetic fields of magnitude 1.93×104 V/m and 2.50×10-3T, respectively? me=9.11×10-31kg, e=1.60×10-19 C.arrow_forwardAn alpha particle with velocity v = (3 x 105,0,0) m/s enters a region where themagnetic field has a value B = (0,0,1.2) T. Determine the required magnitude and directionof an electric field E that will allow the alpha particle to continue to move along the x axis.arrow_forwardFind the radius of curvature (in m) of the path of a 0.285 MeV proton moving perpendicularly to the 1.10 T field of a cyclotron.arrow_forward
- An electron of kinetic energy 0.977 keV circles in a plane perpendicular to a uniform magnetic field. The orbit radius is 35.3 cm. Find (a) the electron's speed, (b) the magnetic field magnitude, (c) the circling frequency, and (d) the period of the motion.arrow_forwardAn electron with kinetic energy 2.5 keV moving along the positive direction of an x axis enters a region in which a uniform electric field of magnitude 10 kV/m is in the negative direction of the y axis. A uniform magnetic field is to be set up to keep the electron moving along the x axis, and the direction of is to be chosen to minimize the required magnitude of . In unit-vector notation, what should be set up?arrow_forwardA singularly-charged ion (i.e. a neutral atom which has gained one electron) with kinetic energy of 7x10−15 J follows a circular path of radius 0.6m when placed in a magnetic field of 0.5T. (Note that the charge of an electron is e = 1.6 x 10−19 C.) a) Using the fact that the ion is going in a circular motion in a magnetic field, what is the ion’s momentum (in kg.m/s)? b) What is the ion’s speed (in m/s)? c) What is the ion’s mass (in kg)? d) An electric field is added to the experiment and adjusted so that the ion passes through without any deflection. What is the magnitude of this electric field (in T)?arrow_forward
- one gram of copper has 9.48x10^21 atoms, and each copper atom has 29 eletrons. what is the total charge of these electronsarrow_forwardTwo wires AC and BC are attached to a 7 Kg sphere that It rotates at constant speed v in the horizontal circle shown in the figure. Yes θ1 = 55° and θ2= 30 ° and d 1.4 m, determine the range of values of v for which both wires are held taut.arrow_forwardWhat is the radius of curvature of the path of a 3.0-keV proton in a perpendicular magnetic field of magnitude 0.80 T?arrow_forward
- What is the radius of the path of an electron (mass 9×10-31 kg and charge 1.6×10-19 C) moving at a speed of 3x107 m/s in a magnetic field of 6 × 10-4 T perpendicular to it? What is its frequency? Calculate its energy in keV. ( 1 eV = 1.6 × 10-19 J).arrow_forwardThe strengths of the fields in the velocity selector of a Bainbridge mass spectrometer are B = 0.500 T and E = 1.2 × 105 V/m, and the strength of the magnetic field that separates the ions is Bo = 0.750 T. A stream of singly charged Li ions is found to bend in a circular arc of radius 2.32 cm. What is the mass of the Li ions?arrow_forwardAn iron bar passes through a magnetic field at a speed of 6.3 m/s. If the bar is 1.5 m long and acquires a potential difference across its ends of 10.8 V, what is the strength of the magnetic field it passes through?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Magnets and Magnetic Fields; Author: Professor Dave explains;https://www.youtube.com/watch?v=IgtIdttfGVw;License: Standard YouTube License, CC-BY