A current , 32 A, is made to pass through a conductor where the free electrons density is `4 xx 10^(28) m^(-3) ` and its area of cross section is `10^(-6) m^(2)`. Find out the value of the drift velocity ( in mm`//` s ) of free electrons.

Answer

To find the drift velocity of free electrons in the given conductor, we can use the formula relating current (I), electron density (N), charge of an electron (E), area of cross-section (A), and drift velocity (Vd):

\[ I = N \cdot E \cdot A \cdot V_d \]

### Step 1: Identify the given values
- Current, \( I = 32 \, A \)
- Free electron density, \( N = 4 \times 10^{28} \, m^{-3} \)
- Charge of an electron, \( E = 1.6 \times 10^{-19} \, C \)
- Area of cross-section, \( A = 10^{-6} \, m^{2} \)

### Step 2: Rearrange the formula to solve for drift velocity (Vd)
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, substituting the known values into the rearranged formula:

\[ V_d = \frac{32}{(4 \times 10^{28}) \cdot (1.6 \times 10^{-19}) \cdot (10^{-6})} \]

### Step 4: Calculate the denominator
First, calculate the product of \( N \), \( E \), and \( A \):

\[ N \cdot E \cdot A = (4 \times 10^{28}) \cdot (1.6 \times 10^{-19}) \cdot (10^{-6}) \]

Calculating \( 4 \cdot 1.6 = 6.4 \):

\[ N \cdot E \cdot A = 6.4 \times 10^{28} \cdot 10^{-19} \cdot 10^{-6} = 6.4 \times 10^{28 - 19 - 6} = 6.4 \times 10^{3} \]

### Step 5: Substitute back to find Vd
Now substitute back into the equation for \( V_d \):

\[ V_d = \frac{32}{6.4 \times 10^{3}} = \frac{32}{6400} = 0.005 \, m/s \]

### Step 6: Convert to mm/s
To convert from meters per second to millimeters per second, we multiply by 1000:

\[ V_d = 0.005 \, m/s \times 1000 = 5 \, mm/s \]

### Final Answer
The drift velocity of free electrons is \( 5 \, mm/s \).

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