MESSAGE MY INSTRUCTOR FULL SCREENRINTER VERSIONB Chapter 28, Problem 2 Your answ

Question

MESSAGE MY INSTRUCTOR FULL SCREENRINTER VERSIONB Chapter 28, Problem 2 Your answer is partially correct. Try again. A particle of mass 11.9 g and charge 89.4 HC moves through a uniform magnetic field, in a region where the free-fall acceleration is -9.8 J m/s? without falling. The velocity of the particle is a constant 23.6 i km/s, which is perpendicular to the magnetic field. What, then, is the magnetic field? Number l O k Units J/T Question Attempts: 4 of 10 used SAVE FOR LATER SUBMIT ANSWER SUBMIT ANSWER Copyright 2000-2016 by John Wiley& Sons, Inc. or related companies. All rights reserved velicy I 2000-20162ohn liley &Sons;,Inc. All Rights Reserved. A Division of 2ohnwileyasons.lne Version 4.17.3.3

Explanation / Answer

F = q (v cross B)

F = qvB

B = F/qv = mg/qv = 11.9*10^-3*9.8/ 89.4*10^-6*23.6*10^3 =

K component is

B = 0.0552 T

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