(N/A) From Faraday's law,the magnitude of the induced electromotive force (emf) is given by:
$|\varepsilon| = \frac{\Delta \Phi_{B}}{\Delta t} \quad \dots(1)$
We also know that the induced current $I$ is related to the induced emf by Ohm's law,$|\varepsilon| = I r$,where $r$ is the resistance of the coil. Since $I = \frac{\Delta Q}{\Delta t}$,we can write:
$|\varepsilon| = \frac{\Delta Q}{\Delta t} r \quad \dots(2)$
Equating $(1)$ and $(2)$:
$\frac{\Delta \Phi_{B}}{\Delta t} = \frac{\Delta Q}{\Delta t} r$
Canceling $\Delta t$ from both sides,we get:
$\Delta \Phi_{B} = \Delta Q \cdot r$
Therefore,the induced charge is:
$\Delta Q = \frac{\Delta \Phi_{B}}{r}$
This shows that the induced charge depends only on the total change in magnetic flux and the resistance of the circuit,and is independent of the time taken or the rate of change of flux.