A

Step 1: When the motor is disconnected, the flywheel continues to spin due to its inertia. As it spins through the magnetic field B, it induces an electromotive force (emf) in accordance with Faraday's law of electromagnetic induction.

Step 2: The induced emf (\epsilon) can be calculated using the formula:
\[ \epsilon = -B \cdot A \cdot \frac{d\theta}{dt} \]
where \( A \) is the area of the flywheel, and \( \frac{d\theta}{dt} \) is the angular velocity (which is initially \( \omega_0 \)).

Step 3: Since the flywheel is essentially generating a back emf as it spins, the initial current (I) through the resistance R can be expressed as:
\[ I = \frac{\epsilon}{R} = \frac{B \cdot A \cdot \omega_0}{R} \]

Step 4: As the flywheel slows down due to the load, the angular velocity will decrease, causing the induced emf and thus the current to also decrease. The current reduces exponentially due to the inductive nature of the circuit.

Step 5: The time (t) it takes for the current to reduce to half its value can be expressed using the time constant for an RL circuit:\( \tau = \frac{L}{R} \), where L is the inductance of the system. The relationship for the current can be expressed as:
\[ I(t) = I_0 e^{-\frac{t}{\tau}} \]

We set \( I(t) = \frac{I_0}{2} \):
\[ \frac{I_0}{2} = I_0 e^{-\frac{t}{\tau}} \]

Step 6: Solving yields:
\[ \frac{1}{2} = e^{-\frac{t}{\tau}} \]

Taking the natural logarithm of both sides results in:
\[ \ln{\frac{1}{2}} = -\frac{t}{\tau} \]

Hence,
\[ t = -\tau \ln{\frac{1}{2}} \]

Since \( \ln{\frac{1}{2}} = -\ln{2} \), we can express the time as:
\[ t = \tau \ln{2} \]

This indicates how long it takes for the current to reduce to half its initial value. Therefore, the initial current is \( I = \frac{B \cdot A \cdot \omega_0}{R} \), and the time to halve is given by the relation derived above.