The current in a wire varies with time according to the relation i = `2.0A + (0.6A//s^(2))t^(2)`. How many coulomb of charge pass a cross-section of the wire in the time interval between t = 0 and t = 10s?

Options

A. 200C
B. 110C
C. 220C
D. 250C

Answer

To find the total charge that passes through a cross-section of the wire in the time interval from \( t = 0 \) to \( t = 10 \) seconds, we need to integrate the current over that time interval. The current \( i(t) \) is given by the equation:

\[ i(t) = 2.0 \, \text{A} + (0.6 \, \text{A/s}^2) t^2 \]

### Step 1: Set up the integral for charge
The charge \( Q \) that passes through the wire can be calculated by integrating the current over time:

\[ Q = \int_{0}^{10} i(t) \, dt \]

### Step 2: Substitute the expression for current
Substituting the expression for \( i(t) \):

\[ Q = \int_{0}^{10} \left( 2.0 + 0.6 t^2 \right) dt \]

### Step 3: Integrate the function
Now, we will integrate the function:

\[ Q = \int_{0}^{10} 2.0 \, dt + \int_{0}^{10} 0.6 t^2 \, dt \]

Calculating each integral separately:

1. For the first integral:
\[ \int 2.0 \, dt = 2.0 t \Big|_{0}^{10} = 2.0 \times 10 - 2.0 \times 0 = 20.0 \, \text{C} \]

2. For the second integral:
\[ \int 0.6 t^2 \, dt = 0.6 \cdot \frac{t^3}{3} \Big|_{0}^{10} = 0.6 \cdot \frac{10^3}{3} - 0.6 \cdot \frac{0^3}{3} = 0.6 \cdot \frac{1000}{3} = 200.0 \, \text{C} \]

### Step 4: Add the results of the integrals
Now, we add the results of the two integrals to find the total charge:

\[ Q = 20.0 \, \text{C} + 200.0 \, \text{C} = 220.0 \, \text{C} \]

### Final Answer
Thus, the total charge that passes through the cross-section of the wire in the time interval from \( t = 0 \) to \( t = 10 \) seconds is:

\[ \boxed{220.0 \, \text{C}} \]