Answer
(a) $1.8N$
(b) $0.73A$
Work Step by Step
(a) We can find the required force as follows:
$F=\frac{B^2vl^2}{R}$
We plug in the known values to obtain:
$F=\frac{(2.0)^2(3.49)(1.25)}{12}$
$F=1.8N$
(b) We can find the required current as
$I=\frac{Bvl}{R}$
$I=\frac{(2.0)(3.49)(1.25)}{12}$
$I=0.73A$
