Charge passing through a conductor of cross-section area `A=0.3 m^(2)` is given by `q=3t^(2)+5t+2` in coluomb, where t is second. What is the value of drift velocity at `t = 2 s` ? (Given, `n=2xx10^(25)//m^(3)`)

Options

A. `0.77 xx 10^(-5) m//s`
B. `1.77 xx 10^(-5) m//s`
C. `2.08 xx 10^(-5) m//s`
D. `0.57 xx 10^(-5) m//s`

Answer

To find the drift velocity at \( t = 2 \, \text{s} \), we will follow these steps:

### Step 1: Determine the Current \( I \)
The charge \( q \) passing through the conductor is given by the equation:
\[ q(t) = 3t^2 + 5t + 2 \]
To find the current \( I \), we need to calculate the derivative of \( q \) with respect to time \( t \):
\[ I = \frac{dq}{dt} \]
Calculating the derivative:
\[ I = \frac{d}{dt}(3t^2 + 5t + 2) = 6t + 5 \]

### Step 2: Calculate the Current at \( t = 2 \, \text{s} \)
Now, we substitute \( t = 2 \) into the current equation:
\[ I(2) = 6(2) + 5 = 12 + 5 = 17 \, \text{A} \]

### Step 3: Use the Formula for Drift Velocity
The relationship between current \( I \), number density \( n \), charge of an electron \( e \), area \( A \), and drift velocity \( v_d \) is given by:
\[ I = n e A v_d \]
We can rearrange this formula to solve for drift velocity \( v_d \):
\[ v_d = \frac{I}{n e A} \]

### Step 4: Substitute the Known Values
We know:
- \( I = 17 \, \text{A} \)
- \( n = 2 \times 10^{25} \, \text{m}^{-3} \)
- \( e = 1.6 \times 10^{-19} \, \text{C} \) (charge of an electron)
- \( A = 0.3 \, \text{m}^2 \)

Substituting these values into the drift velocity formula:
\[ v_d = \frac{17}{(2 \times 10^{25}) \times (1.6 \times 10^{-19}) \times (0.3)} \]

### Step 5: Calculate the Drift Velocity
Calculating the denominator:
\[ (2 \times 10^{25}) \times (1.6 \times 10^{-19}) \times (0.3) = 9.6 \times 10^{6} \]
Now, substituting back to find \( v_d \):
\[ v_d = \frac{17}{9.6 \times 10^{6}} \approx 1.77 \times 10^{-6} \, \text{m/s} \]

### Final Answer
The drift velocity at \( t = 2 \, \text{s} \) is approximately:
\[ v_d \approx 1.77 \times 10^{-6} \, \text{m/s} \]

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