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How does one understand this motional emf by invoking the Lorentz force acting on the free charge carriers of the conductor? Explain. |
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Answer» SOLUTION :Let `vecB=B(-hatk)` (in the negative z-DIRECTION) The rod PQ is MOVED towards right, i.e., in the y-direction. So, the charge carriers (electrons) in the rod moves also in the y-direction with a VELOCITY, `vecv=vhatj` As the charge of an electron is -e, the Lorentz magnetic force acting on it is `vecF=-ehatvxxvecB=-e[vhatjxxB(-hatk)]=evBhati` So, the electron DRIFT in the rod PQ is in the x-direction, i.e., from P to Q. Then, as per convention, the motional emf in PQ will be directed from Q to P. This is the direction obtained by applying Flemings right hand rule. 
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