Answer
Please see the work below.
Work Step by Step
(a) The required inductance can be determined as follows:
$ L=\frac{\epsilon}{\frac{\Delta I}{\Delta t}}$
We plug in the known values to obtain:
$ L==\frac{75\times10^{-3}V}{2A/s}$
$ L=37.5mH $
(b) We know that the induced emf is directly proportional to the inductance. The inductance is inversely proportional to the length of the solenoid, so the induced emf is inversely proportional to the length. Thus, if the length is doubled then the induced emf will be less than $75mV $.
(c) We know that
$\frac{\epsilon_2}{\epsilon_1}=[\frac{L_2}{L_1}]\epsilon_1$
$\frac{\epsilon_2}{\epsilon_1}=[\frac{\frac{1}{2}L_1}{L_1}]\epsilon_1$
We plug in the known values to obtain:
$\frac{\epsilon_2}{\epsilon_1}=\frac{1}{2}(75mV)=38mV $
