Induced Current in Circular Loop Under Time Varying Magnetic Field

To find the value of $\alpha$, we analyze the electromagnetic induction in the circular loop step-by-step.

1. Calculation of the Loop's Area

The circular loop has a radius $r = 20\text{ cm} = 0.2\text{ m}$.
The area $A$ of the loop is given by: $A = \pi r^2 = \pi (0.2)^2 = 0.04\pi\text{ m}^2$

2. Magnetic Flux and Induced EMF

The plane of the loop is perpendicular to the magnetic field $\vec{B}$, so the angle between the magnetic field vector and the normal to the surface of the loop is $\theta = 0^\circ$.

The magnetic flux $\Phi$ through the loop at any time $t$ is: $\Phi(t) = B(t) \cdot A = (2t^2 + 2t + 3) A$

According to Faraday's law of electromagnetic induction, the magnitude of the induced electromotive force (EMF), $\varepsilon$, is given by: $\varepsilon = \left| \frac{d\Phi}{dt} \right| = A \cdot \left| \frac{dB}{dt} \right|$

Differentiating $B(t)$ with respect to $t$: $\frac{dB}{dt} = \frac{d}{dt}(2t^2 + 2t + 3) = 4t + 2$

At time $t = 3\text{ s}$: $\left. \frac{dB}{dt} \right|_{t=3} = 4(3) + 2 = 14\text{ T/s}$

Substitute this into the expression for induced EMF: $\varepsilon = (0.04\pi) \times 14 = 0.56\pi\text{ V}$

3. Calculation of Induced Current

Using Ohm's law, the induced current $I$ in the loop with resistance $R = 2\ \Omega$ is: $I = \frac{\varepsilon}{R} = \frac{0.56\pi}{2} = 0.28\pi\text{ A}$

Taking $\pi = \frac{22}{7}$: $I = 0.28 \times \frac{22}{7} = \frac{28}{100} \times \frac{22}{7} = \frac{4 \times 22}{100} = \frac{88}{100}\text{ A} = \frac{44}{50}\text{ A}$

4. Finding $\alpha$

Given that the induced current $I = \frac{\alpha}{50}\text{ A}$, comparing the expressions yields: $\frac{\alpha}{50} = \frac{44}{50} \implies \alpha = 44$