A conducting wire of cross-sectional area `1 cm^(2)` has `3 xx 10^(23)` charge carriers per m3. If wire carries a current of `24 mA`, then drift velocity of carriers is

Options

A. `5xx10^(-2)ms^(-1)`
B. `0.5ms^(-1)`
C. `5xx10^(-3)ms^(-1)`
D. `5xx10^(-6)ms^(-1)`

Answer

To find the drift velocity of charge carriers in a conducting wire, we can use the formula:

\[ I = n \cdot e \cdot A \cdot V_d \]

Where:
- \( I \) = current (in Amperes)
- \( n \) = number of charge carriers per cubic meter (in \( m^{-3} \))
- \( e \) = charge of each carrier (in Coulombs)
- \( A \) = cross-sectional area of the wire (in \( m^2 \))
- \( V_d \) = drift velocity (in \( m/s \))

### Step-by-step Solution:

1. **Convert the given values to appropriate units:**
- Cross-sectional area \( A = 1 \, cm^2 = 1 \times 10^{-4} \, m^2 \)
- Current \( I = 24 \, mA = 24 \times 10^{-3} \, A \)
- Number of charge carriers \( n = 3 \times 10^{23} \, m^{-3} \)
- Charge of each carrier \( e = 1.6 \times 10^{-19} \, C \)

2. **Rearrange the formula to solve for drift velocity \( V_d \):**
\[ V_d = \frac{I}{n \cdot e \cdot A} \]

3. **Substitute the known values into the equation:**
\[ V_d = \frac{24 \times 10^{-3}}{(3 \times 10^{23}) \cdot (1.6 \times 10^{-19}) \cdot (1 \times 10^{-4})} \]

4. **Calculate the denominator:**
- First, calculate \( n \cdot e \cdot A \):
\[ n \cdot e \cdot A = (3 \times 10^{23}) \cdot (1.6 \times 10^{-19}) \cdot (1 \times 10^{-4}) \]
- Calculate \( 3 \times 1.6 = 4.8 \)
- Then, \( 4.8 \times 10^{23 - 19 - 4} = 4.8 \times 10^{0} = 4.8 \)

5. **Now substitute back into the formula for \( V_d \):**
\[ V_d = \frac{24 \times 10^{-3}}{4.8} \]

6. **Perform the division:**
\[ V_d = 5 \times 10^{-3} \, m/s \]

### Final Answer:
The drift velocity \( V_d \) of the charge carriers is \( 5 \times 10^{-3} \, m/s \).

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