A current of 4 A is flowing in a cylindrical conductor. The number of free electrons passing per second through the cross-section of conductor is

Options

A. `2 * 10^(19)`
B. `3 * 10^(20)`
C. `2.5 * 10^(19)`
D. `4 * 10^(18)`

Answer

To find the number of free electrons passing per second through the cross-section of a cylindrical conductor with a current of 4 A, we can follow these steps:

### Step 1: Understand the relationship between current, charge, and electrons
The current (I) in a conductor is defined as the rate of flow of charge. Mathematically, it can be expressed as:
\[ I = \frac{Q}{t} \]
where \( Q \) is the total charge flowing through the conductor in time \( t \).

### Step 2: Relate charge to the number of electrons
The charge of a single electron (\( e \)) is approximately:
\[ e = 1.6 \times 10^{-19} \, \text{C} \]
If \( n \) is the number of electrons passing through the cross-section per second, then the total charge \( Q \) can also be expressed as:
\[ Q = n \cdot e \]

### Step 3: Substitute into the current equation
Substituting the expression for \( Q \) into the current equation gives:
\[ I = \frac{n \cdot e}{t} \]
For \( t = 1 \) second, this simplifies to:
\[ I = n \cdot e \]

### Step 4: Solve for \( n \)
Rearranging the equation to solve for \( n \):
\[ n = \frac{I}{e} \]

### Step 5: Plug in the values
Now, substitute the known values into the equation:
- Current \( I = 4 \, \text{A} \)
- Charge of an electron \( e = 1.6 \times 10^{-19} \, \text{C} \)

So,
\[ n = \frac{4}{1.6 \times 10^{-19}} \]

### Step 6: Calculate \( n \)
Calculating \( n \):
\[ n = \frac{4}{1.6 \times 10^{-19}} = \frac{4 \times 10^{19}}{1.6} = 2.5 \times 10^{20} \]

### Conclusion
The number of free electrons passing per second through the cross-section of the conductor is:
\[ n = 2.5 \times 10^{20} \]