In a closed circuit, the current `I` (in ampere) at an instant of time t (in second) is given by `I = 4-0.08t`. The number of electrons flowing in 50 s through the cross-section of the conductor is

Options

A. `1.25xx10^(19)`
B. `6.25xx10^(20)`
C. `5.25xx10^(19)`
D. `2.25xx10^(20)`

Answer

To solve the problem, we need to find the total charge flowing through the conductor in 50 seconds and then calculate the number of electrons corresponding to that charge. Here’s the step-by-step solution:

### Step 1: Understand the relationship between current and charge
The current \( I \) is defined as the rate of flow of charge \( Q \) with respect to time \( t \):
\[ I = \frac{dQ}{dt} \]
This can be rearranged to express the charge in terms of current:
\[ dQ = I \, dt \]

### Step 2: Substitute the expression for current
We are given the current as a function of time:
\[ I = 4 - 0.08t \]
Substituting this into the equation for charge gives:
\[ dQ = (4 - 0.08t) \, dt \]

### Step 3: Integrate to find total charge over the interval
To find the total charge \( Q \) that flows through the conductor from \( t = 0 \) to \( t = 50 \) seconds, we integrate:
\[ Q = \int_0^{50} (4 - 0.08t) \, dt \]

### Step 4: Perform the integration
Calculating the integral:
\[ Q = \int_0^{50} 4 \, dt - \int_0^{50} 0.08t \, dt \]
Calculating each part:
1. \(\int_0^{50} 4 \, dt = 4t \Big|_0^{50} = 4 \times 50 - 4 \times 0 = 200\)
2. \(\int_0^{50} 0.08t \, dt = 0.08 \cdot \frac{t^2}{2} \Big|_0^{50} = 0.08 \cdot \frac{50^2}{2} = 0.08 \cdot 1250 = 100\)

Thus, we have:
\[ Q = 200 - 100 = 100 \, \text{Coulombs} \]

### Step 5: Calculate the number of electrons
The charge of a single electron \( e \) is approximately \( 1.6 \times 10^{-19} \) Coulombs. The number of electrons \( n \) can be calculated using:
\[ Q = n \cdot e \]
Rearranging gives:
\[ n = \frac{Q}{e} = \frac{100}{1.6 \times 10^{-19}} \]

### Step 6: Perform the calculation
Calculating \( n \):
\[ n = \frac{100}{1.6 \times 10^{-19}} = 62.5 \times 10^{20} = 6.25 \times 10^{20} \]

### Conclusion
The number of electrons flowing through the cross-section of the conductor in 50 seconds is:
\[ \boxed{6.25 \times 10^{20}} \]