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 834: 72


$20$ Amps

Work Step by Step

29.72 Using the right-hand rule, we see that because the current is downward (-$\hat j$ direction) and the magnetic field is going out of the page (+$\hat k$ direction), the magnetic field must be going to the left (-$\hat j \times \hat k = - \hat i$). Thus, the magnetic force pushes against the force the spring provides (remember that a spring's force is in the direction of its displacement). Thus, $\Sigma\vec{F}_x = 2\vec{F}_s - \vec{F}_B = 0$ Remember, there are two springs in the diagram so a coefficient of 2 goes before $\vec{F}_s$ $2\vec{F}_s = \vec{F}_B$ $2kx = I\vec{L} \times \vec{B} = I\vec{L}\vec{B}\sin(90) = I\vec{L}\vec{B}$ $\displaystyle I = \frac{2kx}{\vec{L}\vec{B}} = \frac{2(10)(0.010)}{(0.020)(0.5)} = 20$ Amps
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