Answer
(a) $11.7A$
(b) current would be cut to a fourth
Work Step by Step
(a) We can find the required field as follows:
$B=\frac{\phi}{A}$
$B=\frac{1.28\times 10^{-4}}{(0.085)^2}$
$B=5.64\times 10^{-3}T$
We know that
$B=\mu_{\circ}nI$
This can be rearranged as:
$I=\frac{B}{\mu_{\circ}n}$
$I=\frac{5.64\times 10^{-3}}{(4\times 10^{-3})(385)}$
$I=11.7A$
(b) We know that current is inversely proportional to the square of the diameter of the solenoid and hence the current would be cut to a fourth by doubling the diameter.
