Essential University Physics (3rd Edition)
3rd Edition
ISBN: 9780134202709
Author: Richard Wolfson
Publisher: PEARSON
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Chapter 38, Problem 44P
To determine
The strength of its unperturbed magnetic field.
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Chapter 38 Solutions
Essential University Physics (3rd Edition)
Ch. 38.1 - Prob. 38.1GICh. 38.2 - Prob. 38.2GICh. 38.3 - Prob. 38.3GICh. 38.4 - Prob. 38.4GICh. 38.5 - Prob. 38.5GICh. 38 - Prob. 1FTDCh. 38 - Prob. 2FTDCh. 38 - Prob. 3FTDCh. 38 - Prob. 4FTDCh. 38 - Prob. 5FTD
Ch. 38 - Why are iodine-131 and strontium-90 particularly...Ch. 38 - Prob. 7FTDCh. 38 - Prob. 8FTDCh. 38 - Prob. 9FTDCh. 38 - Prob. 10FTDCh. 38 - Prob. 11FTDCh. 38 - Prob. 12FTDCh. 38 - Prob. 13FTDCh. 38 - Prob. 14FTDCh. 38 - Explain the different approaches to the Lawson...Ch. 38 - Prob. 16FTDCh. 38 - Three radon isotopes have 125, 134, and 136...Ch. 38 - Prob. 18ECh. 38 - Prob. 19ECh. 38 - Prob. 20ECh. 38 - Prob. 21ECh. 38 - How many half-lives will it take for the activity...Ch. 38 - Prob. 23ECh. 38 - Prob. 24ECh. 38 - Prob. 25ECh. 38 - Prob. 26ECh. 38 - Prob. 27ECh. 38 - Prob. 28ECh. 38 - Use Fig. 38.9 to estimate the mass defect in...Ch. 38 - Find the total binding energy of oxygen-16, given...Ch. 38 - Determine the nuclear mass of nickel-60, given...Ch. 38 - Prob. 32ECh. 38 - Prob. 33ECh. 38 - Prob. 34ECh. 38 - Prob. 35ECh. 38 - Prob. 36ECh. 38 - Prob. 37ECh. 38 - Prob. 38ECh. 38 - Prob. 39ECh. 38 - Prob. 40ECh. 38 - Prob. 41ECh. 38 - Prob. 42PCh. 38 - Prob. 43PCh. 38 - Prob. 44PCh. 38 - Iron-56, with nuclear mass 55.9206 u, is among the...Ch. 38 - Prob. 46PCh. 38 - As a geologist, youre assessing the feasibility of...Ch. 38 - Prob. 48PCh. 38 - Prob. 49PCh. 38 - Nitrogen-13 is a 9.97-min-half-lifc isotope used...Ch. 38 - Prob. 51PCh. 38 - Prob. 52PCh. 38 - Prob. 53PCh. 38 - Prob. 54PCh. 38 - The table below lists reported levels of...Ch. 38 - Prob. 56PCh. 38 - Analysis of a Moon rock shows that 82% of its...Ch. 38 - Prob. 58PCh. 38 - Prob. 59PCh. 38 - Today, uranium-235 comprises only 0.72% of natural...Ch. 38 - Prob. 61PCh. 38 - Prob. 62PCh. 38 - Prob. 63PCh. 38 - Prob. 64PCh. 38 - Prob. 65PCh. 38 - Prob. 66PCh. 38 - Prob. 67PCh. 38 - Prob. 68PCh. 38 - Prob. 69PCh. 38 - Prob. 70PCh. 38 - Prob. 71PCh. 38 - Prob. 72PCh. 38 - Prob. 73PCh. 38 - Prob. 74PCh. 38 - Bismuth-209 and chromium-54 combine to form a...Ch. 38 - Prob. 76PCh. 38 - Prob. 77PCh. 38 - Prob. 78PCh. 38 - Prob. 79PCh. 38 - Prob. 80PCh. 38 - Prob. 81PCh. 38 - Prob. 82PCh. 38 - Prob. 83PCh. 38 - Prob. 84PCh. 38 - Prob. 85PCh. 38 - Prob. 86PPCh. 38 - Prob. 87PPCh. 38 - Prob. 88PPCh. 38 - Prob. 89PP
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- A deuteron is accelerated from rest through a 5-kV potential difference and then moves perpendicularly to a uniform magnetic field with B = 1.2 T. What is the radius of the resulting circular path? (deuteron: m = 3.3 x 10-27kg, q = 1.6 x 10-19C)arrow_forwardF.j = (-e)(-v,k) × (B,i + B.k) ev Bzj. Solving for B gives F. Br 8.50x10 16 N = 1.13 T. Therefore B = 1.13 Ti – 0.772 Tk. The magnitude o A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.70 km/s in the +x-direction experiences a force of 2.10x10-16 N in the +y- direction, and an electron moving at 4.70 km/s in the eve (1.60x10 19 C)(4700 m/s) is B = VB? + B = (1.13 T)² + (-0.772 T)² = 1.37 T. -z-direction experiences a force of 8.50x10-16 N in the +y-direction. Part B For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Magnetic force on a proton. What is the direction of the magnetic field? (in the xz-plane) Express your answer in degrees. 0 = 60.59 ° from the -z-direction to the +x-direction Submit Previous Answers Request Answer X Incorrect; Try Again; 9 attempts remainingarrow_forwardThe magnetism of permanent magnets arises because the inherent magnetic moment of electrons causes them to act like little compass needles. Protons also have an inherent magnetic moment, and this is the basis for magnetic resonance imaging (MRI) in medicine.Although a compass needle would prefer to align with a magnetic field, the needle can point in any direction. This isn’t the case for the magnetic moment of a proton. Quantum physics tells us that the proton’s energy must be quantized. There are only two possible energy levels—and thus two possible orientations—for protons in a magnetic field: E1 = -μB magnetic moment aligned with the field E2 = +μB magnetic moment aligned opposite the fieldwhere μ = 1.41 x 10-26 J/T is the known value of the proton’s magnetic moment. FIGURE 28.24 shows the two possible energy states. The magnetic moment, like a compass needle, “wants” to align with the field, so that is the lower-energy state.…arrow_forward
- When you lie in an MRI machine, you lie in a strong magnetic field and the protons in your body align in the z-direction to give a net magnetic moment. We can now flip all these magnetic moment by sending in a magnetic pulse from another direction. The pulse causes the arrow that represents the magnetic moment to lie flat in the xy plane. After the pulse is gone, the magnetic moment will slowly recover back to the z-direction because this is the lowest energy configuration (remember that the magnetic field in the z-direction remains present during this whole process).arrow_forwardAt the heart of a magnetic resonance imaging scanner is a large electromagnet capable of generating strong uniform magnetic fields of around one Tesla or more. a) Describe the effect of this large magnetic field on the protons of the hydrogen nuclei within the molecules that make up tissue. A pulse of radio waves is used to excite these protons to a higher energy state. b) Explain why a precise frequency of radiation is required and how this affects the motion of the proton spins.arrow_forwardAn alpha particle is whizzing by at a velocity determined by its KE of 5 Mev. a ) What is the velocity of the alpha particle? b) What is the magnetic field at the position of a nucleus 2nm away from the alpha particle at angle of 900 to the velocity of the alpha. Alpha particle has charge of 2e but mass of 4 protons.arrow_forward
- Using magnetic confinement, one particle is restrained to levitate over another. What is this separation when the particles are two electrons in m?arrow_forwardProtons are placed in a 2.30 T magnetic field that points in the positive z-direction. (a) What is the energy difference between states with the z-component of proton spin angular momentum parallel and antiparallel to the field? (b) A proton can make a transition from one of these states to the other by emitting or absorbing a photon with the appropriate energy. Find the frequency and wavelength of such a photon.arrow_forwardIn tokamaks, strong B fields are necessary for confinement. If a tokamak has a minor radius of 0.6m, and the Deuterium ions are at 200 keV, find out if confinement can be achieved with a 5T field.arrow_forward
- A beam of alpha particles is incident on a target of lead. A particular alpha particle comes in “head-on” to a particular lead nu- cleus and stops 6.50 * 10-14 m away from the center of the nucleus. (This point is well outside the nucleus.) Assume that the lead nucleus, which has 82 protons, remains at rest. The mass of the alpha particle is 6.64 * 10-27 kg. (a) Calculate the electrostatic potential energy at the instant that the alpha particle stops. Express your result in joules and in MeV. (b) What initial kinetic energy (in joules and in MeV) did the alpha particle have? (c) What was the initial speed of the alpha particle?arrow_forwardSuppose you want to identify a bunch of particles using mass spectrometer. First, they are ionized with charge q= +1.6e-19 C. Then the particles pass through the velocity selector with a speed v= 6.7e6 m/s and enters a magnetic field B= 1.4 T. If they hit the detector 0.4 m away from the entry opening, what kind of particle are they? Options are: A. Deuterium (m= 3.34e-27 kg) B. Proton (m= 1.67e-27 kg) C. Helium 4 (m= 6.65e-27 kg) D. Beryllium 8 (m= 13.3e-27 kg)arrow_forwardSuppose a cyclotron is operated at an oscillator frequency of 12.4 MHz and has a dee radius 45.3 cm. Estimate the total path length traveled by a deuteron in the cyclotron during the (entire) acceleration process. Assume that the accelerating potential between the dees is 84.3 kV. The deuteron mass is m - 3.34 x 10-27 kg.arrow_forward
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