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Find the current in the wire for the configuration shown in figure. Wire PQ has negligible resistance. vecB, the magnetic field is coming out of the paper. theta is a fixed angle made by PQ travelling smoothly over two conducting parallel wires separated by a distance d. |
Answer» Solution : INDUCE emf in rod can be given by , `epsilon=Bvl` Here, `vecB, VECV` and `VECL`all should be mutually perpendicular and if it is not then we consider their perpendicular component.  Here `vecB` and `vecv` is perpendicular to each other but length of rod PQ is not perpendicular to `vecv` and `vecB`. But Isine component is perpendicular to both `vecv` and `vecB`. From figure , LSIN`theta`=perpendicular distance between two wire (d) `therefore lsintheta =d` `therefore epsilon=Bvd` Induced emf `epsilon=Bv(lsin theta) = Bvd` Induced current `I=epsilon IR` `=(Bvd)/R` Direction of this induced current is in clockwise direction according to Lenz.s LAW. |
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