A current is uniformly distributed across the cross-section of cylindrical wire of radius R = 4 mm with current density $J = 2.5 × 10^5\, Am^{-2}$. The amount of current flowing through the outer portion of the wire between radial distance R/2 and R is

9.42 A

The correct answer is Option (3) → 9.42 A

Given:

Radius of wire: $R = 4 \, \text{mm} = 0.004 \, \text{m}$

Current density: $J = 2.5 \times 10^5 \, \text{A/m}^2$

Outer portion: $r = R/2$ to $R$

Current through area: $I = J \cdot A$

Area of annular portion:

$A = \pi (R^2 - (R/2)^2) = \pi (R^2 - R^2/4) = \pi (3 R^2 / 4) = \frac{3 \pi R^2}{4}$

Substitute $R = 0.004$ m:

$A = \frac{3 \pi (0.004)^2}{4} = \frac{3 \pi \cdot 16 \times 10^{-6}}{4} = \frac{48 \pi \times 10^{-6}}{4} = 12 \pi \times 10^{-6} \, \text{m}^2$

Current through this portion:

$I = J \cdot A = 2.5 \times 10^5 \cdot 12 \pi \times 10^{-6} = 3 \pi \, \text{A} \approx 9.42 \, \text{A}$

Answer: $I \approx 9.42 \, \text{A}$