Solution

The angle between the rod and the vertical (shown in green) is $\omega t$ at time $t$ after the initial momment. If there is current $I$ flowing radially outward through the rod, then the magnetic force on the rod is $F=IBL$. The angular speed is constant when counterclockwise moment due to $F$ equals clockwise moment due to gravity $mg$. Thus,

$$IBL\cdot\frac{L}{2}=mg\cdot\frac{L}{2}\sin(\omega t)\qquad\Rightarrow\qquad I=\frac{mg}{BL}\sin(\omega t).$$