Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 20 - Magnetism - Problems - Page 583: 1

Answer

a. 5.8 N/m b. 3.3 N/m

Work Step by Step

a. Use equation 20–1 to calculate the force with an angle of 90 degrees, and a length of 1 meter. $$F=I\mathcal{l}Bsin\theta$$ $$\frac{F}{\mathcal{l}}=IBsin\theta=(6.40A)(0.90T)sin90^{\circ}=5.8N/m$$ b. Change the angle. $$\frac{F}{\mathcal{l}}=IBsin\theta=(6.40A)(0.90T)sin35^{\circ}=3.3N/m$$
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