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 \[0.03\,kg{{m}^{-1}}\]

Answer:

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