\[ \text{Induced emf} \, (\mathcal{E}) = R \cdot I = 10 \times 2 = 20 \, \text{V} \]
Also, emf is related to change in flux by:
\[ \mathcal{E} = \frac{\Delta \Phi}{\Delta t} \Rightarrow \Delta \Phi = \mathcal{E} \cdot \Delta t = 20 \cdot 1 \times 10^{-3} = 20 \times 10^{-3} = 2 \times 10^{-2} \, \text{Wb} \]
The change in magnetic flux is \( {2 \times 10^{-2} \, \text{Wb}} \), so the correct answer is (B).
A battery of \( 6 \, \text{V} \) is connected to the circuit as shown below. The current \( I \) drawn from the battery is: