A circular coil of 50 turns and cross-sectional area 2 m^2 is...

The magnetic flux linked with the coil is given by $\Phi = NBA\cos\theta$.

Since the plane of the coil is perpendicular to the magnetic field, the angle between the area vector and the magnetic field is $\theta = 0^{\circ}$. Thus, $\cos(0^{\circ}) = 1$.

Substituting the given values:

$\Phi(t) = 50 \times 2 \times (0.2t^2 - 0.05t)$

$\Phi(t) = 100(0.2t^2 - 0.05t) = 20t^2 - 5t$

According to Faraday's law, the magnitude of the induced emf is $|e| = \dfrac{d\Phi}{dt}$.

Differentiating the flux equation with respect to time:

$|e| = \dfrac{d}{dt}(20t^2 - 5t) = 40t - 5$

At $t = 2\text{ s}$, the magnitude of the induced emf is:

$|e| = 40(2) - 5 = 80 - 5 = 75\text{ V}$

Answer: $75\text{ V}$