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 584: 22

Answer

a. 0.53 T. b. 0.61 T.

Work Step by Step

A perpendicular field of $B_1$ makes a Hall emf of 12 mV. On page 569, it is stated that the Hall emf is proportional to the magnetic field strength. Use this proportionality to find the unknown field, designated “2”. a. The unknown magnetic field is oriented at $90^{\circ}$ to the surface, so all of the magnetic field is effective at creating the Hall emf. $$\frac{V_{H,2}}{ V_{H,1}}=\frac{B_2}{B_1}$$ $$ B_2= B_1\frac{V_{H,2}}{ V_{H,1}}=(0.10T)\frac{63mV }{12mV }=0.53T$$ b. The unknown magnetic field is oriented at $60^{\circ}$ to the surface, so only the component of the magnetic field that is perpendicular to the slab is effective at creating the Hall emf. $$\frac{V_{H,2}}{ V_{H,1}}=\frac{B_2sin60^{\circ}}{B_1}$$ $$ B_2= \frac{B_1}{ sin60^{\circ}}\frac{V_{H,2}}{ V_{H,1}}= \frac{0.10T}{ sin60^{\circ}}\frac{63mV }{12mV }=0.61T$$
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