$F_{m a g}^\rightarrow=i \int(\overrightarrow{d l} \times \vec{B})=i L B \sin (90-\theta)=i L B \cos \theta$

$F_{m a g}^{\rightarrow}$ up the inclined plane.

Component of weight of $E F$ down the inclined plane: $m g \sin \theta$

For $E F$ to be in equilibrium, $i L B \cos \theta=m g \sin \theta$

$\therefore i L B=m g \tan \theta$

For $\vec{F}_{m a g}$ to be up the inclined plane, $i$ needs to flow from $E$ to $F$.