Derive an expression for (i) induced emf and (ii) induced current when a conductor of length $$I$$ is moved with a uniform velocity, normal to a uniform magnetic field $$B$$. Assume the resistance of conductor to be $$R$$.


Expression for Induced emf: We know that if a charge $$q$$ moves with velocity $$\overrightarrow { V } $$ in a magnetic field of strength $$\overrightarrow { B } $$, making an angle $$\theta$$ then magnetic Lorentz force 
$$F=q\ vB \sin \theta$$
If $$\overrightarrow { v } $$ and $$\overrightarrow { B } $$ mutually perpendicular, then $$\theta =90^o$$
$$F=q\ vB \sin 90^o=qvB$$
The directon of this force is perpendicular to both $$\overrightarrow { v } $$ and $$\overrightarrow { B } $$ and is given by Fleming's left hand rule.
Suppose a thin conductig rod $$PQ$$ is placed on two parallel metallic rails $$CD$$ and $$MN$$ in a magnetic field of strength $$\overrightarrow { B } $$ . The direction of magnetic field $$\overrightarrow { B } $$  is perpendicular to the plane of paper, downward. In fig $$\overrightarrow { B } $$ isrepresented by cross $$(\times )$$ marks. Suppose the rod is moving with velocity $$\overrightarrow { v } $$ , perpendicular to its own length, towards the right. We know that metallic conductors contain free electrons, which can move within the metal. As charge on electron, $$q=-e$$ therefore, each electron experiences a magnetic Lorents force, $$F_m=evB$$, whose direction, according to Fleming's left hand rule, will be from $$P$$ to $$Q$$ Thus the electrons are displaced from end $$P$$ toward end $$Q$$ Consequently the end $$P$$ of rod becomes positively charged and end $$Q$$ negatively charged. Thus a potential difference is produced between the ends of the conductor. This is the induced emf.
Due to induced emf, an electric field is produced in the conducting rod. The strength of this electric field
$$E=\dfrac{v}{l}$$        ...$$(i)$$
And its direction is from $$(+)$$ to $$(-)$$ charge, i.e., from $$p$$ to $$Q$$.
The force on a free electron due to this electric field, $$F_e=eE$$    ...$$(ii)$$
The direction of this force is from $$Q$$ to $$P$$which is opposite to that of electric field. Thus the emf produced opposes the motion of electrons caused due to Lorentz force. This is in accordance with Lenz's law. As the number of electrons at end becomes more and more, the magnitude of electric force $$Fe$$ goes on increasing, and a stage comes when electric force $$\overrightarrow {Fe}$$ and magnetic force $$\overrightarrow {Fm}$$ become equal and opposite. In this situation the potential difference produced across the ends of rod becomes constant. In this condition
$$eE=evB$$ or $$E=B_v$$   ...$$(iii)$$
$$\therefore$$ The potential difference produced,
$$V=EI=B\ v\ I\ Volt$$
Also the induced current $$I=\dfrac{V}{R}=\dfrac{Bvl}{R}$$ ampere


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