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: \(44\)