JEE Main & Advanced Physics Electro Magnetic Induction Dynamic (Motional) EMI Due to Translatory Motion

(1) Consider a conducting rod of length l moving with a uniform velocity \[\vec{v}\] perpendicular to a uniform magnetic field \[\vec{B}\], directed into the plane of the paper. Let the rod be moving to the right as shown in figure. The conducting electrons also move to the right as they are trapped within the rod.

Conducting electrons experiences a magnetic force \[{{F}_{m}}=evB.\] So they move from P to Q within the rod. The end P of the rod becomes positively charged while end Q becomes negatively charged, hence an electric field is set up within the rod which opposes the further downward movement of electrons i.e. an equilibrium is reached and in equilibrium \[{{F}_{e}}={{F}_{m}}\] i.e. \[eE=evB\] or  \[E=vB\Rightarrow \] Induced emf \[e=El=Bvl\]  [\[E=\frac{V}{l}\]]

(2) If rod is moving by making an angle \[\theta \] with the direction of magnetic field or length. Induced emf \[e=Bvl\sin \theta \]

(3) Motion of conducting rod on an inclined plane : When conductor start sliding from the top of an inclined plane as shown, it moves perpendicular to it?s length but at an angle \[(90-\theta )\]with the direction of magnetic field.

Hence induced emf across the ends of conductor \[e=Bv\sin (90-\theta )l=Bvl\cos \theta \]

So induced current \[i=\frac{Bvl\cos \theta }{R}\] (Directed from Q to P).

The forces acting on the bar are shown in following figure. The rod will move down with constant velocity only if

\[{{F}_{m}}\cos \theta =mg\cos (90-\theta )\]\[=mg\sin \theta \]\[\Rightarrow \]\[Bil\cos \theta =mg\sin \theta \]

\[B\left( \frac{B{{v}_{T}}l\cos \theta }{R} \right)\,l\cos \theta =mg\sin \theta \]\[\Rightarrow \]\[{{v}_{T}}=\frac{mgR\sin \,\theta }{{{B}^{2}}{{l}^{2}}{{\cos }^{2}}\theta }\]