The current density J through a plane area has a magnitude 6 Am^-2. If J makes a angle of 60° with the area vector of magnitude 3 m² , the magnitude of the current through the area is

The current is given by 

                             $I={\int }_{ }\mathit{J}\mathbf{·}\mathit{A}$

If J is constant throughout the area, then

                            I=J$·$A

where A is the area vector. Thus

                      I=JA cos $\theta$

                       =$6\times 3 \cos {60}^0$= 9A