College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 20 - Problems - Page 794: 13

Answer

(a) $\Phi = 0.090~Wb$ (b) $\Phi = 0.16~Wb$ (c) The current through the top side of the loop flows in the negative z direction.

Work Step by Step

(a) We can find the magnetic flux: $\Phi = B~A~cos~\theta$ $\Phi = (0.32~T)~(0.75~m)(0.75~m)~cos~60^{\circ}$ $\Phi = 0.090~Wb$ (b) We can find the magnetic flux: $\Phi = B~A~cos~\theta$ $\Phi = (0.32~T)~(0.75~m)(0.75~m)~cos~30^{\circ}$ $\Phi = 0.16~Wb$ (c) The magnetic flux in the positive x direction is increasing, thus, the induced current will produce a magnetic field with a component in the negative x direction. By the right hand rule, this current flows around the square loop such that the current through the top side of the loop flows in the negative z direction.
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