Answer
$emf_{max}=5.50kV$
Work Step by Step
We know that;
$emf_{max}=NAB\omega $
We plug in the known values to obtain:
$emf_{max}=100(0.50)(0.30)(3.5)(2\times 3.1416\times\frac{ 1000}{60})$
$emf_{max}=5.497\times 10^3=5.50\times 10^3=5.50kV$
Fundamentals of Physics Extended (10th Edition)
by Halliday, David; Resnick, Robert; Walker, Jearl
Published by
Wiley
ISBN 10:
1-11823-072-8
ISBN 13:
978-1-11823-072-5
Chapter 30 - Induction and Inductance - Problems - Page 897: 19
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