A silver wire of radius `0.1 cm` carries a current of 2A. If the charge density in silver is `5.86xx10^(28)m^(-3)`, then the drif velocity is

Options

A. `0.2xx10^(-3)ms^(-1)`
B. `0.4xx10^(-4)ms^(-1)`
C. `0.68xx10^(-4)ms^(-1)`
D. `7xx10^(-4)ms^(-1)`

Answer

To find the drift velocity of the silver wire, we will use the formula:

\[ v_d = \frac{I}{n \cdot e \cdot A} \]

Where:
- \( v_d \) = drift velocity
- \( I \) = current (2 A)
- \( n \) = charge density (5.86 x \( 10^{28} \) m\(^{-3}\))
- \( e \) = charge of an electron (approximately \( 1.6 \times 10^{-19} \) C)
- \( A \) = cross-sectional area of the wire

### Step 1: Calculate the cross-sectional area \( A \)
The radius of the wire is given as \( 0.1 \) cm. First, we need to convert this to meters:

\[ r = 0.1 \, \text{cm} = 0.1 \times 10^{-2} \, \text{m} = 0.001 \, \text{m} \]

Now, we can calculate the area using the formula for the area of a circle:

\[ A = \pi r^2 \]

Substituting the radius:

\[ A = \pi (0.001)^2 = \pi \times 10^{-6} \, \text{m}^2 \]

### Step 2: Substitute the values into the drift velocity formula
Now we can substitute the values into the drift velocity formula:

\[ v_d = \frac{I}{n \cdot e \cdot A} \]

Substituting the known values:

\[ v_d = \frac{2}{(5.86 \times 10^{28}) \cdot (1.6 \times 10^{-19}) \cdot (\pi \times 10^{-6})} \]

### Step 3: Calculate the denominator
First, calculate \( n \cdot e \cdot A \):

\[ n \cdot e = 5.86 \times 10^{28} \cdot 1.6 \times 10^{-19} = 9.376 \times 10^{9} \]

Now, multiply by the area:

\[ n \cdot e \cdot A = 9.376 \times 10^{9} \cdot (\pi \times 10^{-6}) \approx 9.376 \times 10^{9} \cdot 3.14 \times 10^{-6} \approx 2.94 \times 10^{4} \]

### Step 4: Calculate the drift velocity
Now substitute this back into the drift velocity equation:

\[ v_d = \frac{2}{2.94 \times 10^{4}} \approx 0.68 \times 10^{-4} \, \text{m/s} \]

### Final Answer
The drift velocity \( v_d \) is approximately:

\[ v_d \approx 0.68 \times 10^{-4} \, \text{m/s} \]

---