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

$\epsilon =\int \left(\stackrel{\rightharpoonup }{v}\times \stackrel{\rightharpoonup }{B}\right).\stackrel{\rightharpoonup }{dl}$

$\Rightarrow ϵ=\int [(lsin\theta )\omega \times B(-\stackrel{\wedge }{j})].[dlsin\theta (-\stackrel{\wedge }{i})+dlcos\theta (-\stackrel{\wedge }{j})]$

$\Rightarrow \epsilon =\int \left[B\omega l \sin \theta \stackrel{\wedge }{i}\right].\left[dl \sin \theta \left(-\stackrel{\wedge }{i}\right)+dl \cos \theta \left(-\stackrel{\wedge }{j}\right)\right]$

$=-B\omega {sin}^2\theta {\int }_0^{L}ldl=-\frac{1}{2}B\omega {sin}^2\theta L^2$

$\Rightarrow ϵ=V_b-V_a=-\frac{1}{2}B\omega L^2{sin}^2\theta$

Negative 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.

$e_{ab}+e_{bc}+e_{ca}=0$

but $e_{bc}=\frac{1}{2}B\omega {\left(L \sin \theta \right)}^2,e_{ca}=0$ putting the values, we get

$e_{ab}=-\frac{1}{2}B\omega {\left(L \sin \theta \right)}^2$