Question

A hollow copper wire with an inner diameter of 0.50 mm and an outer diameter of 1.8 mm carries a current of 15

Answer 1

The current density is equal to the current intensity divided by the cross-sectional area through which the current passes:
$J= \frac{I}{A}$

The inner radius of the wire is 
$r_i = \frac{0.50 mm}{2}=0.25 mm$
while the outer radius is
$r_o = \frac{1.80 mm}{2}=0.90 mm$
Therefore the cross-sectional area of the wire is
$A= \pi (r_o^2 - r_i^2)=\pi ((0.9mm)^2-(0.25 mm)^2)=2.35 mm^2$
$= 2.35 \cdot 10^{-6} m^2$

So the current density in the wire is
$J= \frac{I}{A}= \frac{15 A}{2.35 \cdot 10^{-6} m^2}=6.38 \cdot 10^6 A/m^2$