Question

The electric current through a wire varies with time as \(I = I_0 + \beta t\) where \( I_0 = 20 \, \text{A} \) and \( \beta = 3 \, \text{A/s} \). The amount of electric charge that crosses through a section of the wire in \( 20 \, \text{s} \) is:

• A. 80C
• B. 1000C
• C. 800C
• D. 1600C

Answer 1

The Correct Option is B

Step 1: Express Current as a Function of Time

\[ I = I_0 + \beta t = 20 + 3t \]

Step 2: Calculate Charge \(q\) Over Time

The current \(I = \frac{dq}{dt}\), so we can write:

\[ dq = (20 + 3t) dt \]

Step 3: Integrate to Find Total Charge \(q\) from \(t = 0\) to \(t = 20\)

\[ q = \int_{0}^{20} (20 + 3t) dt \]

Split the integral:

\[ q = \int_{0}^{20} 20 dt + \int_{0}^{20} 3t dt \]

Step 4: Evaluate Each Integral

\[ q = \left[20t\right]_{0}^{20} + \left[\frac{3t^2}{2}\right]_{0}^{20} \]

\[ = (20 \times 20) + \frac{3 \times 20^2}{2} \]

\[ = 400 + \frac{3 \times 400}{2} \]

\[ = 400 + 600 = 1000 \, C \]

So, the correct answer is: 1000 C