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 - General Problems - Page 588: 80

Answer

$3.2 \times10^{-4}m\cdot N$

Work Step by Step

See Figure 20-34b. Use equation 20–10 to find the torque. $$\tau=NIABsin\theta$$ Use Ohm’s law to equate the current to the voltage divided by the resistance. $$\tau=N\frac{V}{R}ABsin\theta $$ The torque is maximum when the angle between the perpendicular to the loop, and the magnetic field, is 90 degrees. The sine is 1. $$\tau_{max}= N\frac{V}{R}AB $$ $$\tau_{max}= 20\frac{9.0V}{28\Omega}(0.050m)^2(0.020T) $$ $$ =3.2 \times10^{-4}m\cdot N$$
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