Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 29 - The Magnetic Field - Exercises and Problems - Page 830: 1

Answer

$B_2 = 40~mT$ $B_3 = 0$ $B_4 = 40~mT$

Work Step by Step

Near point 2, there is twice the current flowing in the same direction compared with point 1. Therefore, the magnetic field is $B_2 = 40~mT$ Near point 3, the current is flowing one direction in one part of the wire, and in the opposite direction in the other part of the wire. By the right hand rule, the magnetic fields produced by each part of the wire will cancel out. Therefore, the magnetic field is $B_3 = 0$ Near point 4, the current is flowing in one direction in one part of the wire, and in the opposite direction in the other part of the wire. However, the currents in the two sections of the wire are on the opposite sides of point 4. By the right hand rule, the magnetic fields produced by each part of the wire will add together. Therefore, the magnetic field is $B_4 = 40~mT$
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