Questions

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

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a
12B ω L2
b
12B ω L2 tan2θ
c
12B ω L2 cos2θ
d
12B ω L2 sin2θ

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detailed solution

Correct option is D

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