Answer:
please see the answers below
Explanation:
a) the magnetic field is given by
[tex]B=\frac{\mu_0NI(t)}{l}[/tex]
N is the number of turns, l is the length of the solenoid, mu_0 is the magnetic permeability of vacuum and I(t) is the current.
b)
[tex]\Phi_B=BS=\frac{\pi r^2 \mu_0 N I(t)}{l}[/tex]
for a single turn:
[tex]\Phi_B=BS=\frac{\pi r^2 \mu_0I(t)}{l}[/tex]
c)
[tex]emf=-\frac{d\Phi_B}{dt}=-\frac{\pi r^2 \mu_0 N}{l}\frac{dI(t)}{dt}[/tex]
d)
[tex]emf=-L\frac{dI(t)}{dt}\\\\L=-\frac{emf}{\frac{dI(t)}{dt}}=-\frac{-\frac{\pi r^2\mu_0N}{l}\frac{dI(t)}{dt}}{\frac{dI(t)}{dt}}=\frac{\pi r^2 \mu_0N}{l}[/tex]
hope this helps!!