A rod of length L rotates in the form of a conical pendulum with an angular velocity ω about its axis as shown in figure. The rod makes an angle θ with the axis. The magnitude of the motional emf developed across the two ends of the rod is

ε=∫v⇀×B⇀.dl⇀⇒ϵ=∫[(lsinθ)ω×B(-j∧)].[dlsinθ(-i∧)+dlcosθ(-j∧)]⇒ε=∫Bωl sinθi∧.dl sinθ-i∧+dl cosθ-j∧=-Bωsin2θ∫0Lldl=-12Bωsin2θL2⇒ϵ=Vb-Va=-12BωL2sin2θNegative sign indicates that end b will be negative w.r.t. a.Alternate Method:Consider two more conductors ac and bc. This completes a closed loop. The net emf induced in this closed loop should be zero, as net flux through this loop always remains zero.eab+ebc+eca=0but ebc=12BωL sinθ2,eca=0 putting the values, we geteab=-12BωL sinθ2