A rod of length $$60$$ cm rotates with a uniform angular velocity $$20$$ rad s$$^{-1}$$ about its perpendicular bisector, in a uniform magnetic field $$0.5$$ T. The direction of magnetic field is parallel to the axis of rotation. The potential difference between the two ends of the rod is _____ V.

We need to find the potential difference between the two ends of a rod rotating about its perpendicular bisector in a magnetic field parallel to the axis of rotation.

Rod length = 60 cm = 0.6 m, angular velocity $$\omega = 20$$ rad/s, magnetic field $$B = 0.5$$ T. The rod rotates about its perpendicular bisector, and the magnetic field is parallel to the axis of rotation.

When a rod rotates about its center (perpendicular bisector), the EMF induced between the center and one end is:

$$\varepsilon = \frac{1}{2}B\omega\left(\frac{L}{2}\right)^2 = \frac{B\omega L^2}{8}$$

Both halves of the rod generate EMF from center to their respective ends. The EMF from center to each end is the same in magnitude. Since both halves sweep in opposite directions relative to the center, the induced EMF from one end to center equals that from center to the other end.

End 1 is at higher potential than center by $$\frac{B\omega(L/2)^2}{2}$$, and end 2 is also at higher potential than center by the same amount.

So the potential difference between the two ends = 0.

Alternatively: By symmetry, both ends are at the same potential (both are at the same distance from the axis). The rod is symmetric about its center, so the motional EMF from center to end 1 equals the motional EMF from center to end 2. Both ends are at the same potential.

The potential difference between the two ends is $$0$$ V.

Therefore, the answer is 0.