A conducting rod moving ion a π shaped conductor:-
A conducting rod of length (l) is moving smoothly on the parallel rails of a π shaped conductor with a constant velocity (v). As an ext-agency is pulling the rod to the right,: there is a rate of change of area. lxv of the loop as a result the rate of change of magnetic flux is Blv
\(therefore varepsilon =left| frac{dphi }{dt} right|=Blv\)Sign of EMI using lenz’s law:- Here the cause the right. The induced EMF establishes a current I in anti clock wise direction, so that the FB acting on the rod is to the left This, opposing the motion of the rod is accordance with lenz’s law
Alternate method Direction of EMI:- As the rod is moving to the right
\(varepsilon =-frac{d{{phi }_{B}}}{dt}\) \(varepsilon =-left( -frac{d{{phi }_{B}}}{dt} right)\)= + ve
Current is anticlockwise.
\(varepsilon =Blv\)In this case 3 vectors Blv one mutually ┴on, the EMI induced is maximum.
