Answer
(a) $66\mu s$
(b) $43\mu J$
Work Step by Step
We know that
$I=(\frac{\epsilon}{R})(1-e^{\frac{-tR}{L}})$
This can be rearranged as:
$t=-(\frac{L}{R})(1-\frac{IR}{\epsilon})$
We plug in the known values to obtain:
$t=-(\frac{29}{110})(1-\frac{12\times 10^{-3}(110)}{6.0})$
$t=66\mu s$
(b) We know that
$U=(\frac{1}{2})LI^2$
$U=(\frac{1}{2})L(\frac{\epsilon}{R})^2$
We plug in the known values to obtain:
$U=(\frac{1}{2})(29\times 10^{-3})(\frac{6.0}{110})^2$
$U=43\mu J$
