To determine the correct prediction based on the given interactions between the ends of the metallic bars, we need to analyze the patterns of attraction and repulsion in the interactions provided.
### Analyzing the Interactions
1. Interaction 1: A1 attracts B1
- This indicates that A1 and B1 are opposite poles; one is a north pole and the other is a south pole.
2. Interaction 2: A2 repels C1
- This indicates that A2 and C1 are like poles; both are either north poles or south poles.
3. Interaction 3: B2 attracts A1
- This further indicates that B2 and A1 are opposite poles, consistent with Interaction 1.
### Deductions:
- Since B2 attracts A1 and A1 attracts B1:
- A1 must be opposite B1 and B2. This implies that B1 and B2 are similar in polarity since they both interact with A1 oppositely.
- Since A2 repels C1:
- This indicates A2 and C1 are similar poles.
### Check Predictions:
- Option 1: A2 repels B1
- If A2 repels B1, they are similar poles.
- Option 2: C2 attracts B2
- If C2 attracts B2, they are opposite poles.
- Option 3: B1 repels C1
- If B1 repels C1, they are similar poles.
- Option 4: A1 attracts C2
- If A1 attracts C2, they are opposite poles.
Given the deductions:
- A1 attracts B1 => Opposite poles.
- A2 repels C1 => Similar poles.
- B2 attracts A1 => Opposite poles.
Since A1 attracts B1 and B2 attracts A1, then B1 and B2 are similar poles (both must be opposite to A1). Knowing A2 repels C1 implies A2 and C1 are the same, which means:
- A1 (opposite of A2) would attract C2 (opposite of C1).
Thus, the prediction A1 attracts C2 is consistent with all the given observations and deductions.
### Conclusion:
Therefore, Barton should expect:
[tex]\[ \boxed{A1 \, \text{attracts} \, C2} \][/tex]
Hence, the correct prediction is:
[tex]\[ \boxed{4} \][/tex]
