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

Question

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

Infinity Learn · Accepted Answer

ε$$=$$∫v⇀×B⇀.dl⇀⇒ϵ$$=$$∫[(lsin$$θ$$)ω×$$B(-j$$∧$$)$$].[dlsinθ$$(-i$$∧$$)+dlcos$$θ$$(-j$$∧$$)$$]⇒ε$$=$$∫Bωl $$sin$$θi∧.dl $$sin$$θ$$-i$$∧$$+dl$$ $$cos$$θ$$-j$$∧$$=-B$$ω$$sin2$$θ∫$$0Lldl=-12B$$ω$$sin2$$θ$$L2$$⇒ϵ$$=Vb-Va=-12B$$ω$$L2sin2$$θNegative 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.$$eab+ebc+eca=0but$$ $$ebc=12B$$ωL $$sin$$θ$$2$$,$$eca=0$$ putting the values, we $$geteab=-12B$$ωL $$sin$$θ$$2$$

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

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