COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Question
Chapter 19, Problem 50QAP
To determine
The direction of the magnetic field due to the current at the points O, P, Q and R
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COLLEGE PHYSICS
Ch. 19 - Prob. 1QAPCh. 19 - Prob. 2QAPCh. 19 - Prob. 3QAPCh. 19 - Prob. 4QAPCh. 19 - Prob. 5QAPCh. 19 - Prob. 6QAPCh. 19 - Prob. 7QAPCh. 19 - Prob. 8QAPCh. 19 - Prob. 9QAPCh. 19 - Prob. 10QAP
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- A proton enters a region with a uniform electric field E=5.0kV/m and a uniform magnetic field B=5.0104kT. The proton has initial velocity v0=2.5105m/s. How far along the z axis does the proton travel after it undergoes three complete revolutions?arrow_forwardA proton of speed v=6105m/s enters a region of uniform magnetic field of B = 0.5 T at an angle of q = 30° to the magnetic field. In the region of magnetic field proton describes a helical path with radius R and pitch p (distance between loops.) Find R and p.arrow_forwardTo see why an MRI utilizes iron to increase the magnetic field created by a coil, calculate the current needed in a 400-looppermeter circular coil 0.660 m in radius to create a 1.20T field (typical of an MRI instrument) at its center with no iron present. The magnetic field of a proton is approximately like that of a circular current loop 0.6501015m in radius carrying 1.05104A. What is the field at the center of such a loop?arrow_forward
- Four different proton velocities are given. For each case, determine the magnetic force on the proton in terms of e, v0and B0.arrow_forwardConstruct Your Own Problem Consider a mass separator that applies a magnetic field perpendicular to the velocity of ions and separates the ions based on the radius of curvature of their paths in the field. Construct a problem in which you calculate the magnetic field strength needed to separate two ions that differ in mass, but not charge, and have the same initial velocity. Among the things to consider are the types of ions, the velocities they can be given before entering the magnetic field, and a reasonable value for the radius of curvature of the paths they follow. In addition, calculate the separation distance between the ions at the point where they are detected.arrow_forwardUnreasonable Results A charged particle having mass 6.641027kg (that of a helium atom) moving at 8.70105m/s perpendicular to a 1.50T magnetic field travels in a circular path of radius 16.0 mm. (a) What is the charge of the particle? (b) What is unreasonable about this result? (c) Which assumptions are responsible?arrow_forward
- A particle’s path is bent when it passes through a region of non-zero magnetic field although its speed remains unchanged. This is very useful for “beam steering’’ in particle accelerators. Consider a proton of speed 4106m/s entering a region of uniform magnetic field 0.2 T over a 5-cm-wide region. Magnetic field is perpendicular to the velocity of the particle. By how much angle will the path of the proton be bent? (Hint: the particle comes out tangent to a circle.arrow_forward(a) A physicist performing a sensitive measurement wants to limit the magnetic force on a moving charge in her equipment to less than 1.001012N. What is the greatest the charge can be if it moves at a maximum speed of 30.0 m/s in the Earth’s field? (b) Discuss whether it would be difficult to limit the charge to less than the value found in (a) by competing it with typical static electricity and noting that static is often absent.arrow_forwardA charged particle is traveling through a uniform magnetic field. Which of the following statements are true of the magnetic field? There may be more than one correct statement. (a) It exerts a force on the particle parallel to the field. (b) It exerts a force on the particle along the direction of its motion. (c) It increases the kinetic energy of the particle. (d) It exerts a force that is perpendicular to the direction of motion. (e) It does not change the magnitude of the momentum of the particle.arrow_forward
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Magnets and Magnetic Fields; Author: Professor Dave explains;https://www.youtube.com/watch?v=IgtIdttfGVw;License: Standard YouTube License, CC-BY