A

Step 1: Given the current through the wire is \( i(t) = i_0 + \alpha \sin(\pi t) \) with \( i_0 = 10 \, \text{A} \) and \( \alpha = \frac{1}{2} \, \text{A} \). Therefore, \( i(t) = 10 + \frac{1}{2} \sin(\pi t) \).

Step 2: To find the charge \( Q \) that crosses through a section of the wire in 3 seconds, we use the formula: \( Q = \int_{0}^{3} i(t) \, dt \).
Thus, \( Q = \int_{0}^{3} \left( 10 + \frac{1}{2} \sin(\pi t) \right) dt \).

Step 3: Calculating the integral:
\( Q = \int_{0}^{3} 10 \, dt + \int_{0}^{3} \frac{1}{2} \sin(\pi t) \, dt \).
\( Q = 10t \big|_{0}^{3} + \frac{1}{2}\left(-\frac{1}{\pi}\cos(\pi t)\right)\big|_{0}^{3} = 30 + \frac{1}{4}(\cos(3\pi) - \cos(0)) \).
Since \( \cos(3\pi) = -1 \) and \( \cos(0) = 1 \), we obtain:
\( Q = 30 + \frac{1}{4}(-1 - 1) = 30 - \frac{1}{2} = 29.5 \, \text{C}.

Step 4: To find the average current \( I_{avg} \), use the formula: \( I_{avg} = \frac{Q}{\Delta t} = \frac{29.5}{3} \approx 9.83 \, \text{A}.

Therefore, the charge crossed through a section of the wire in 3 seconds is 29.5 C, and the average current for that interval is approximately 9.83 A.