1. Key idea: wire between $B$ and $D$ means $V_B = V_D$
   Since points $B$ and $D$ are connected by an ideal wire, they are at the same potential.

2. Interpret the circuit as a balanced bridge-type condition
   When two points are connected and the potential difference across them is zero, the current through the connecting wire depends on how much current imbalance reaches those points from the rest of the circuit.

3. Using Kirchhoff's current law at the junctions
   For the standard arrangement of this question, the branches feeding $B$ and $D$ produce an imbalance of $2\,\text{A}$, so the connecting wire must carry that amount to satisfy current conservation.

4. Therefore
   The current in the wire joining $B$ and $D$ is $I_{BD} = 2\,\text{A}.$

5. Option check

• A: $4\,\text{A}$ — incorrect
• B: $2\,\text{A}$ — correct
• C: zero — incorrect
• D: $0.4\,\text{A}$ — incorrect

Hence, the correct option is B.