• # question_answer On a smooth plane inclined at $30{}^\circ$ with the horizontal, a thin current carrying metallic rod is placed parallel to the horizontal ground. The plane is located in a uniform magnetic field of 0.15 T in the vertical direction. For what value of current can the rod remain stationary? The mass per unit length of the rod is $0.03\,kg{{m}^{-1}}$

Let a rod PQ is horizontally placed on an inclined plane as shown in figure. Inclined plane

 Forces act on the current carrying rod PQ are: (i) Weight (mg) (vertically downward)             (1) (ii) Horizontal force, $F=IBl$ (due to magnetic field B) Resolving mg and BIl along and perpendicular to inclined plane.
For rod to be stationary,             $Mg\,\sin \theta =BIl\cos \theta$             ?(i) If l is the length of rod and m as mass per unit length. i.e.        $\frac{M}{l}=m$ $\Rightarrow$   M = ml $\therefore$      $(ml)\,g\,sin\theta =BIlcos\theta$ or         $I=\frac{mg\,\tan \theta }{B}=\frac{0.03\times 9.8\,\tan \,30{}^\circ }{0.15}$ $\Rightarrow$   I = 1.132 A

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