Question

$AB$ 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 :

• A. $i$ must flow from $E$ to $F$
• B. $Bil=mg\tan \theta$
• C. $Bil=mg\sin \theta$
• D. $Bil=mg$

Answer 1

Answer: (A), (B)

The correct options are
A $i$ must flow from $E$ to $F$
B $Bil=mg\tan \theta$

Force on EF,
$\overrightarrow{F_{mag}}=i\int (\overrightarrow{dl}\times \overrightarrow{B})=iLB\sin (90-\theta )=iLB\cos \theta$
$\overrightarrow{F_{mag}}$ up the inclined plane.
Component of weight of $EF$ down the inclined plane: $mg\sin \theta$
For $EF$ to be in equilibrium, $iLB\cos \theta =mg\sin \theta$
$\therefore iLB=mg\tan \theta$
For ${\overrightarrow{F}}_{mag}$ to be up the inclined plane, $i$ needs to flow from $E$ to $F$.