Question 26.10: The magnitude J of the current density in a certain lab wire......

Assuming \vec{J} is directed along the wire (with no radial flow) we integrate, starting with Eq. 26-4,

i=\int \vec{J} \cdot d \vec{A}           (26-4)

i=\int|\vec{J}| d A=\int_{9 R / 10}^R\left(k r^2\right) 2 \pi r d r=\frac{1}{2} k \pi\left(R^4-0.656 R^4\right)

where k=3.0 \times 10^8 and SI units are understood. Therefore, if R = 0.00200 m, we obtain i=2.59 \times 10^{-3}\, A .