(N/A) The magnetic field produced by a long solenoid is given by $B = \mu_0 n I$.
The magnetic flux $\phi$ linked with the smaller coil is $\phi = N A B$,where $A$ is the area of the smaller coil.
$\phi = N (\pi (b/2)^2) (\mu_0 n I) = \frac{\mu_0 N n \pi b^2 I}{4}$.
The induced emf $\varepsilon$ in the smaller coil is given by Faraday's law: $\varepsilon = -\frac{d\phi}{dt}$.
$\varepsilon = -\frac{d}{dt} \left( \frac{\mu_0 N n \pi b^2 I}{4} \right) = -\frac{\mu_0 N n \pi b^2}{4} \frac{dI}{dt}$.
Given the current varies as $I(t) = mt^2 + C$,we have $\frac{dI}{dt} = 2mt$.
Substituting this into the emf equation:
$\varepsilon = -\frac{\mu_0 N n \pi b^2}{4} (2mt) = -\left( \frac{\mu_0 N n \pi b^2 m}{2} \right) t$.
The magnitude of the induced emf is $|\varepsilon| = \left( \frac{\mu_0 N n \pi b^2 m}{2} \right) t$.
This represents a linear relationship between $|\varepsilon|$ and $t$,which is a straight line passing through the origin.