To find the current in the wire, we use the formula:

$I = \frac{Q}{t}$

where:

• $I$ is the current,
• $Q$ is the charge flowing through the wire,
• $t$ is the time.

Given:

• $Q = 93.7 \, \text{mC} = 93.7 \times 10^{-3} \, \text{C} = 0.0937 \, \text{C}$,
• $t = 33.5 \, \text{ms} = 33.5 \times 10^{-3} \, \text{s} = 0.0335 \, \text{s}$.

Substitute these values into the formula:

$I = \frac{0.0937}{0.0335} \approx 2.8 \, \text{A}$

Answer:

The current in the wire is approximately 2.8 A.

Would you like further details on the calculation steps?

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Here are some additional questions for further exploration:

1. What happens to the current if the time decreases to 20 ms?
2. How would the current change if the charge was increased to 150 mC?
3. What units are used for current, and why is it measured in those units?
4. How would this calculation change if we measured the charge in microcoulombs (μC)?
5. Can we calculate the power dissipated in the wire with this information alone?

Tip: Always convert units to the SI system (e.g., Coulombs, seconds) for consistency in calculations.