Question

At the instant when the current in an inductor is increasing at a rate of $0.0640A/s$, the magnitude of the self-induced emf is $0.0160V$. If the inductor is a solenoid with $400$ turns, what is the average magnetic flux through each turn when the current is $0.720A$?

Answer 1

$Inductance,L=\frac{N\varphi }{i}$
$\therefore Flux,\varphi =\frac{Li}{N}=\frac{\left(0.25\right)\left(0.72\right)}{400}=4.5\times {10}^{-4}Wb$