Answer
a) $F=1.6N$
b) $F=1.54N$
Work Step by Step
(a) We know that
$F=ILBsin\theta$
We plug in the known values to obtain:
$F=(110)(250)(5.9\times 10^{-5})sin90^{\circ}$
$F=1.6N$
(b) As $F=ILBsin\theta$
We plug in the known values to obtain:
$F=(110)(250)(5.9\times 10^{-5})sin72{\circ}$
$F=1.54N$