Answer
$2.4A$
Work Step by Step
We know that
$F=mg=ILBsin90^{\circ}=ILB$
This can be rearranged as:
$I=\frac{mg}{LB}$
We plug in the known values to obtain:
$I=\frac{(0.75)(9.81)}{3.6(0.84)}$
$I=2.4A$
Physics Technology Update (4th Edition)
by Walker, James S.
Published by
Pearson
ISBN 10:
0-32190-308-0
ISBN 13:
978-0-32190-308-2
Chapter 22 - Magnetism - Problems and Conceptual Exercises - Page 793: 35
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