A circular loop of radius $20\text{ cm}$ and resistance $2\ \Omega$ is placed in a time varying magnetic field $\vec{B} = (2t^2 + 2t + 3)\text{ T}$. At $t = 0$, for the plane of the loop being perpendicular to the magnetic field and, the induced current in the loop at $t = 3\text{ s}$ is $\dfrac{\alpha}{50}\text{ A}$. The value of $\alpha$ is _______.
(Take $\pi = 22/7$)

The magnetic flux linked with the circular loop is given by $\Phi = BA \cos \theta$. Since the plane of the loop is perpendicular to the magnetic field, the angle between the area vector and the magnetic field is $\theta = 0^{\circ}$.

$\Phi = B \cdot \pi r^2$

Given $r = 20\text{ cm} = 0.2\text{ m}$, the area of the loop is:

$A = \pi (0.2)^2 = 0.04\pi\text{ m}^2$

The induced emf in the loop is given by Faraday's law of induction:

$|E| = \dfrac{d\Phi}{dt} = A \dfrac{dB}{dt}$

Given the magnetic field $B = 2t^2 + 2t + 3$, we differentiate it with respect to time $t$:

$\dfrac{dB}{dt} = \dfrac{d}{dt}(2t^2 + 2t + 3) = 4t + 2$

At $t = 3\text{ s}$, the rate of change of the magnetic field is:

$\dfrac{dB}{dt} = 4(3) + 2 = 14\text{ T/s}$

Now, substituting the values into the emf equation:

$|E| = 0.04\pi \times 14 = 0.56\pi\text{ V}$

The induced current $I$ in the loop is:

$I = \dfrac{|E|}{R} = \dfrac{0.56\pi}{2} = 0.28\pi\text{ A}$

Using $\pi = \dfrac{22}{7}$, we get:

$I = 0.28 \times \dfrac{22}{7} = 0.04 \times 22 = 0.88\text{ A}$

We are given that the induced current is $\dfrac{\alpha}{50}\text{ A}$. Equating the two values:

$\dfrac{\alpha}{50} = 0.88$

$\alpha = 0.88 \times 50 = 44$

Answer: $44$