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:
$I=\frac{(0.75)(9.81)}{3.6(0.84)}$
$I=2.4A$
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