Answer
The maximum speed is $~~3.02~mm/s$
Work Step by Step
When we compare the two systems, the velocity $v$ corresponds to the current $i$
We can find the maximum current:
$U_B = \frac{i^2~L}{2} = E$
$i^2 = \frac{2E}{L}$
$i = \sqrt{\frac{2E}{L}}$
$i = \sqrt{\frac{(2)(5.70\times 10^{-6}~J)}{1.25~H}}$
$i = 3.02~mA$
Since the maximum current is $3.02~mA$, then the maximum speed is $~~v = 3.02~mm/s$
