A uniform copper wire of length `1m` and cross section area `5 xx 10^(-7)m^(2)` carries a current of `1A`. Assuming that are `8 xx 10^(28)` free electron per `m^(3)` in copper, how long will an electron take to drift from one end of the wire an electron the other. Charge on an electron `= 1.6 xx 10^(-19)C`

Options

A. `0.8xx10^3s`
B. `1.6xx10^3s`
C. `3.2xx10^3s`
D. `6.4xx10^3s`

Answer

To solve the problem, we need to determine how long it takes for an electron to drift from one end of a copper wire to the other. We will follow these steps:

### Step 1: Understand the relationship between current, drift velocity, and charge density
The current \( I \) in a conductor is given by the equation:
\[ I = n \cdot A \cdot V_d \]
where:
- \( I \) is the current (in Amperes),
- \( n \) is the number of free electrons per unit volume (in \( m^{-3} \)),
- \( A \) is the cross-sectional area of the wire (in \( m^2 \)),
- \( V_d \) is the drift velocity of the electrons (in \( m/s \)).

### Step 2: Rearranging the formula to find drift velocity
From the equation above, we can solve for the drift velocity \( V_d \):
\[ V_d = \frac{I}{n \cdot A} \]

### Step 3: Substitute the known values
We know:
- \( I = 1 \, A \)
- \( n = 8 \times 10^{28} \, m^{-3} \)
- \( A = 5 \times 10^{-7} \, m^2 \)

Substituting these values into the drift velocity equation:
\[ V_d = \frac{1}{8 \times 10^{28} \cdot 5 \times 10^{-7}} \]

### Step 4: Calculate the drift velocity
Calculating the denominator:
\[ 8 \times 10^{28} \cdot 5 \times 10^{-7} = 40 \times 10^{21} = 4 \times 10^{22} \]
Now substituting back:
\[ V_d = \frac{1}{4 \times 10^{22}} = 2.5 \times 10^{-23} \, m/s \]

### Step 5: Calculate the time taken for an electron to drift the length of the wire
The time \( T \) taken for an electron to drift from one end of the wire to the other is given by:
\[ T = \frac{L}{V_d} \]
where \( L \) is the length of the wire (1 m).

Substituting the values:
\[ T = \frac{1}{2.5 \times 10^{-23}} = 4 \times 10^{22} \, seconds \]

### Step 6: Final calculation
To express this in a more manageable form:
\[ T = 4 \times 10^{22} \, seconds \]

### Conclusion
Thus, the time taken for an electron to drift from one end of the wire to the other is approximately \( 4 \times 10^{22} \, seconds \).

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