When a wire carries a current of `1.20 A`, the drift velocity is `1.20xx10^-4 m//s`. What is the velocity when the current is `6.00 A`?

Explanation

To solve the problem, we need to understand the relationship between current (I) and drift velocity (Vd) in a wire. The formula that relates these quantities is:

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

Where:
- \( I \) is the current,
- \( n \) is the number density of charge carriers,
- \( e \) is the charge of an electron,
- \( A \) is the cross-sectional area of the wire,
- \( V_d \) is the drift velocity.

From this equation, we can see that the current is directly proportional to the drift velocity, assuming that \( n \), \( e \), and \( A \) remain constant.

### Step-by-step Solution:

1. **Identify the relationship between current and drift velocity**:
Since \( I \propto V_d \), we can express this relationship as:
\[ \frac{I_1}{I_2} = \frac{V_{d1}}{V_{d2}} \]
where \( I_1 \) and \( V_{d1} \) are the current and drift velocity for the first case, and \( I_2 \) and \( V_{d2} \) are for the second case.

2. **Substitute the known values**:
We know:
- \( I_1 = 1.20 \, A \)
- \( V_{d1} = 1.20 \times 10^{-4} \, m/s \)
- \( I_2 = 6.00 \, A \)

We need to find \( V_{d2} \):
\[ \frac{1.20}{6.00} = \frac{1.20 \times 10^{-4}}{V_{d2}} \]

3. **Cross-multiply to solve for \( V_{d2} \)**:
\[ 1.20 \cdot V_{d2} = 6.00 \cdot (1.20 \times 10^{-4}) \]
\[ V_{d2} = \frac{6.00 \cdot (1.20 \times 10^{-4})}{1.20} \]

4. **Simplify the equation**:
The \( 1.20 \) in the numerator and denominator cancels out:
\[ V_{d2} = 6.00 \times 10^{-4} \, m/s \]

5. **Final Answer**:
The drift velocity when the current is \( 6.00 A \) is:
\[ V_{d2} = 6.00 \times 10^{-4} \, m/s \]