The current density in a cylindrical wire of radius ` r = 4 ` mm is `1.0 xx 10^(6) A//m^(2)`. The current through the outer portion of the wire between radial distances `(r )/(2)` and r is `x pi` A, where x is

Explanation

To solve the problem, we need to find the current flowing through the outer portion of the cylindrical wire between the radial distances \( \frac{r}{2} \) and \( r \). Here are the steps to arrive at the solution:

### Step 1: Understand the Given Data
- Radius of the wire, \( r = 4 \, \text{mm} = 4 \times 10^{-3} \, \text{m} \)
- Current density, \( J = 1.0 \times 10^{6} \, \text{A/m}^2 \)

### Step 2: Define the Area for Current Calculation
We need to find the current flowing through the annular region between \( r/2 \) and \( r \). The area \( A \) of this annular region can be calculated as the difference between the area of the outer circle (radius \( r \)) and the area of the inner circle (radius \( r/2 \)).

\[ A = \pi r^2 - \pi \left(\frac{r}{2}\right)^2 \]

### Step 3: Calculate the Areas
Calculating the area of the outer circle:
\[ A_{\text{outer}} = \pi r^2 = \pi (4 \times 10^{-3})^2 = \pi (16 \times 10^{-6}) = 16\pi \times 10^{-6} \, \text{m}^2 \]

Calculating the area of the inner circle:
\[ A_{\text{inner}} = \pi \left(\frac{r}{2}\right)^2 = \pi \left(2 \times 10^{-3}\right)^2 = \pi (4 \times 10^{-6}) = 4\pi \times 10^{-6} \, \text{m}^2 \]

Now, subtract the inner area from the outer area:
\[ A = A_{\text{outer}} - A_{\text{inner}} = (16\pi \times 10^{-6}) - (4\pi \times 10^{-6}) = 12\pi \times 10^{-6} \, \text{m}^2 \]

### Step 4: Calculate the Current
The current \( I \) through the annular region can be calculated using the formula:
\[ I = J \cdot A \]
Substituting the values:
\[ I = (1.0 \times 10^{6} \, \text{A/m}^2) \cdot (12\pi \times 10^{-6} \, \text{m}^2) = 12\pi \, \text{A} \]

### Step 5: Identify \( x \)
From the problem, we know that the current through the outer portion of the wire is given as \( x\pi \, \text{A} \). From our calculation, we found that:
\[ I = 12\pi \, \text{A} \]
Thus, we can identify \( x \):
\[ x = 12 \]

### Final Answer
The value of \( x \) is \( 12 \).

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