$AB$ and $CD$ are smooth parallel rails, separated by a distance $l$, and inclined to the horizontal at an angle $\theta$ . $A$ uniform magnetic field of magnitude $B$, directed vertically upwards, exists in the region. $EF$ is a conductor of mass $m$, carrying a current $i$. For $EF$ to be in equilibrium,
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Force on $EF$
$F_{m a g}^\rightarrow=i \int(\overrightarrow{d l} \times \vec{B})=i L B \sin (90-\theta)=i L B \cos \theta$
$F_{m a g}^{\rightarrow}$ 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 $F$.
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