The current density in a cylindrical wire of radius 4 mm is 4×10^6 A m^(-2). The current through the outer portion of the wire between radial distances R/2 and R is _____ π A.

Q86 / 90 · JEE Main 2022 (27 Jun Shift 1) · Physics

The current density in a cylindrical wire of radius $4 mm$ is $4 \times 10^{6} A m^{- 2}$ . The current through the outer portion of the wire between radial distances $\frac{R}{2}$ and $R$ is _____ $π A$ .

Correct answer: 48

Current density is defined as current per unit cross-section area. Therefore,

$i = J A = J \times π (R^{2} - \frac{R^{2}}{4})$

$\Rightarrow i = 4 \times 10^{6} \times π \times 16 \times 10^{- 6} \times \frac{3}{4}$

$\Rightarrow i = 48 π A$

Hence, the correct answer is 48.