Answer
a. The inductance is $L= 0.250H$.
b. $\Phi_B =4.50\times10^{-4}Wb$.
Work Step by Step
a. The inductance is $L=\frac{\epsilon}{|di/dt|}$.
$$ L=\frac{0.0160V}{0.0640A}=0.250H$$
b. We know that $L=\frac{N\Phi_B}{i}$.
The average flux through each turn is:
$$\Phi_B =\frac{L i}{N}=\frac{(0.250H)(0.720A)}{400}=4.50\times10^{-4}Wb$$
