A conducting rod PO of length l = 2 m is moving at a speed of 2 ms-1 making an angle of 30o with its length. A uniform magnetic field B = 2T exists in a direction perpendicular to the plane of motion. Then

Emf induced across the rod AB ise=Bℓvsin⁡θ=2×2×2×sin⁡30⇒e=4VFree electrons of the rod shift towards right due to forceF→=q(v→×B→)Thus, end P is at higher potentialor VP - VQ= 4V. Thus, choice (b) is correct.