1. Given data

• Radius of wire: $r = 4.0\,\text{mm} = 4.0 \times 10^{-3}\,\text{m}$
• Current density: $J = 1.0 \times 10^6\,\text{A/m}^2$

We need the current through the outer annular portion of the wire from radial distance $r/2$ to $r$.

2. Area of the outer annulus

The current density is uniform, so current through a region is: $I = J \times A$

The required area is the area of a ring: $A = \pi r^2 - \pi \left(\frac{r}{2}\right)^2$

$A = \pi r^2 - \pi \frac{r^2}{4} = \frac{3}{4}\pi r^2$

3. Substitute $r = 4.0 \times 10^{-3}$ m

First, $r^2 = (4.0 \times 10^{-3})^2 = 16 \times 10^{-6} = 1.6 \times 10^{-5}\,\text{m}^2$

So, $A = \frac{3}{4}\pi (1.6 \times 10^{-5})$

$A = 1.2 \times 10^{-5} \pi\,\text{m}^2$

4. Compute current

$I = J A = (1.0 \times 10^6)(1.2 \times 10^{-5}\pi)$

$I = 12\pi\,\text{A}$

Thus, comparing with $x\pi$ A, $x = 12$

5. Comparison with stored answer

Stored correct answer = 12, which matches the derived answer.