A current of 5 ampere is passing through a metallic wire of cross-sectional area `4xx10^(-6) m^(2)`. If the density of the charge-carriers in the wire is `5xx10^(26) m^(-3)`, find the drift speed of the electrons.

Options

A. `1xx10^(2)ms^(-1)`
B. `1.56xx10^(-2)ms^(-1)`
C. `1.56xx10^(-3)ms^(-1)`
D. `1xx10^(-2)ms^(-1)`

Answer

To find the drift speed of the electrons in the metallic wire, we can use the formula that relates current (I), charge carrier density (n), charge of an electron (e), cross-sectional area (A), and drift speed (v_d):

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

Where:
- \( I \) = current in amperes (A)
- \( n \) = density of charge carriers in \( m^{-3} \)
- \( e \) = charge of an electron (approximately \( 1.6 \times 10^{-19} \) coulombs)
- \( A \) = cross-sectional area in \( m^2 \)
- \( v_d \) = drift speed in \( m/s \)

### Step 1: Identify the given values
- Current, \( I = 5 \, A \)
- Cross-sectional area, \( A = 4 \times 10^{-6} \, m^2 \)
- Density of charge carriers, \( n = 5 \times 10^{26} \, m^{-3} \)
- Charge of an electron, \( e = 1.6 \times 10^{-19} \, C \)

### Step 2: Rearrange the formula to solve for drift speed \( v_d \)
We can rearrange the formula to isolate \( v_d \):

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

### Step 3: Substitute the known values into the equation
Now, we substitute the known values into the rearranged formula:

\[ v_d = \frac{5}{(5 \times 10^{26}) \cdot (1.6 \times 10^{-19}) \cdot (4 \times 10^{-6})} \]

### Step 4: Calculate the denominator
First, calculate the product in the denominator:

1. Calculate \( n \cdot e \):
\[ n \cdot e = (5 \times 10^{26}) \cdot (1.6 \times 10^{-19}) = 8 \times 10^{7} \]

2. Now, calculate \( n \cdot e \cdot A \):
\[ n \cdot e \cdot A = (8 \times 10^{7}) \cdot (4 \times 10^{-6}) = 3.2 \times 10^{2} \]

### Step 5: Substitute back into the drift speed equation
Now substitute back into the drift speed equation:

\[ v_d = \frac{5}{3.2 \times 10^{2}} \]

### Step 6: Perform the division
Now, perform the division:

\[ v_d = \frac{5}{320} = 0.015625 \, m/s \]

### Step 7: Convert to scientific notation
Convert \( 0.015625 \, m/s \) to scientific notation:

\[ v_d = 1.5625 \times 10^{-2} \, m/s \]

### Step 8: Round to two significant figures
Rounding to two significant figures gives:

\[ v_d \approx 1.6 \times 10^{-2} \, m/s \]

### Final Answer
The drift speed of the electrons is approximately \( 1.6 \times 10^{-2} \, m/s \).

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