Physics for Scientists and Engineers, Vol. 1
6th Edition
ISBN: 9781429201322
Author: Paul A. Tipler, Gene Mosca
Publisher: Macmillan Higher Education
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Chapter 26, Problem 11P
To determine
The explanation for force on electron.
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Chapter 26 Solutions
Physics for Scientists and Engineers, Vol. 1
Ch. 26 - Prob. 1PCh. 26 - Prob. 2PCh. 26 - Prob. 3PCh. 26 - Prob. 4PCh. 26 - Prob. 5PCh. 26 - Prob. 6PCh. 26 - Prob. 7PCh. 26 - Prob. 8PCh. 26 - Prob. 9PCh. 26 - Prob. 10P
Ch. 26 - Prob. 11PCh. 26 - Prob. 12PCh. 26 - Prob. 13PCh. 26 - Prob. 14PCh. 26 - Prob. 15PCh. 26 - Prob. 16PCh. 26 - Prob. 17PCh. 26 - Prob. 18PCh. 26 - Prob. 19PCh. 26 - Prob. 20PCh. 26 - Prob. 21PCh. 26 - Prob. 22PCh. 26 - Prob. 23PCh. 26 - Prob. 24PCh. 26 - Prob. 25PCh. 26 - Prob. 26PCh. 26 - Prob. 27PCh. 26 - Prob. 28PCh. 26 - Prob. 29PCh. 26 - Prob. 30PCh. 26 - Prob. 31PCh. 26 - Prob. 32PCh. 26 - Prob. 33PCh. 26 - Prob. 34PCh. 26 - Prob. 35PCh. 26 - Prob. 36PCh. 26 - Prob. 37PCh. 26 - Prob. 38PCh. 26 - Prob. 39PCh. 26 - Prob. 40PCh. 26 - Prob. 41PCh. 26 - Prob. 42PCh. 26 - Prob. 43PCh. 26 - Prob. 44PCh. 26 - Prob. 45PCh. 26 - Prob. 46PCh. 26 - Prob. 47PCh. 26 - Prob. 48PCh. 26 - Prob. 49PCh. 26 - Prob. 50PCh. 26 - Prob. 51PCh. 26 - Prob. 52PCh. 26 - Prob. 53PCh. 26 - Prob. 54PCh. 26 - Prob. 55PCh. 26 - Prob. 56PCh. 26 - Prob. 57PCh. 26 - Prob. 58PCh. 26 - Prob. 59PCh. 26 - Prob. 60PCh. 26 - Prob. 61PCh. 26 - Prob. 62PCh. 26 - Prob. 63PCh. 26 - Prob. 64PCh. 26 - Prob. 65PCh. 26 - Prob. 66PCh. 26 - Prob. 67PCh. 26 - Prob. 68PCh. 26 - Prob. 69PCh. 26 - Prob. 70PCh. 26 - Prob. 71PCh. 26 - Prob. 72PCh. 26 - Prob. 73PCh. 26 - Prob. 74PCh. 26 - Prob. 75PCh. 26 - Prob. 76PCh. 26 - Prob. 77PCh. 26 - Prob. 78PCh. 26 - Prob. 79PCh. 26 - Prob. 80PCh. 26 - Prob. 81P
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- Repeat the previous problem, but with the loop lying flat on the ground with its current circulating counterclockwise (when viewed from above) in a location where Earth’s field is north, but at an angle 45.0° below the horizontal and with a strength of 6.0105T.arrow_forwardAn electron of kinetic energy 2000 eV passes between parallel plates that are 1.0 an apart and kept at a potential difference of 300 V. What is the strength of the uniform magnetic field B that will allow the electron to travel undeflected through the plates? Assume E and B are perpendicular.arrow_forwardDescribe the following physical occurrences as events, that is, in the form (x, y, z, t): (a) A postman rings a doorbell of a house precisely at noon. (b) At the same lime as the doorbell is lung, a slice of bread pops out of a toaster that is located 10 1T1 from the door in the east direction from the door. (c) Tell seconds later, an airplane arrives at the airport, which is 10 km from the door in the east direction and 2 km to the south.arrow_forward
- An electron moving with a velocity v=(4.0i+3.0j+2.0k)106m/s enters a region where there is a uniform electric field and a uniform magnetic field. The magnetic field is given by v=(1.0i2.0j+4.0k)102T. If the electron travels through a region without being deflected, what is the electric field?arrow_forwardCheck your Understanding Repeat the calculations from the previous example for I0=0.040A .arrow_forward(a) An oxygen16 ion with a mass at 2.661026kg travels at 5.00106m/s perpendicular to a 1.20T magnetic field, which makes it move in a circular arc with a 0.231-m radius. What positive charge is on the ion? (b) What is the radio of this charge to the charge of an electron? (c) Discuss why the radio found in (b) should be an integer.arrow_forward
- CASE STUDY In Example 34.6 (page 1111), we imagined equipping 1950DA, an asteroid on a collision course with the Earth, with a solar sail in hopes of ejecting it from the solar system. We found that the enormous size required for the solar sail makes the plan impossible at this time. Of course, there is no need to eject such an object from the solar system: we only need to change the orbit. A much more pressing problem is Apophis, a 300-m asteroid that may be on a collision course with the Earth and is due to come by on April 13, 2029. It is unlikely to hit the Earth on that pass, but it will return again in 2036. If Apophis passes through a 600-m keyhole on its 2029 pass, it is expected to hit the Earth in 2036. causing great damage. There are plans to deflect Apophis when it comes by in 2029. For example, we could hit it with a 10- to 150-kg impactor accelerated by a solar sail. The impactor is launched from the Earth to start orbiting the Sun in the same direction as the Earth and Apophis. The idea is to use a solar sail to accelerate the impactor so that it reverses direction and collides head-on with Apophis at 8090 km/s and thereby keeps Apophis out of the keyhole. Consider the momentum in the impactors orbit (Fig. P34.75) when the solar sail makes an angle of = 60 with the tangent to its orbit. Current solar sails may be about 40 m on a side, but the hope is to construct some that are about 160 m on a side. Estimate the impactors tangential acceleration when it is about 1 AU from the Sun. Keep in mind that the sail is neither a perfect absorber nor a perfect reflector, and a heavier impactor would presumably be equipped with a larger sail. Dont be surprised by what may seem like a very small acceleration. FIGURE P34.75arrow_forwardA long, straight, horizontal wire carries a left-to-right current of 20 A. If the wire is placed in a uniform magnetic field of magnitude 4.0105 T that is directed vertically downward, what is tire resultant magnitude of the magnetic field 20 cm above the wire? 20 cm below the wire?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
- The magnetic field between the poles of a horseshoe electromagnet is uniform and has a cylindrical symmetry about an axis from the middle of the South Pole to the middle of the North Pole. The magnitude of the magnetic field changes as a rate of dB/dt due to the changing current through the electromagnet, Determine the electric field at a distance r from the center.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_forwardCheck 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_forward
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