Given data

• Length of rod: $L = 60\text{ cm} = 0.6\text{ m}$
• Angular velocity: $\omega = 20\,\text{rad s}^{-1}$
• Magnetic field: $B = 0.5\,\text{T}$
• The rod rotates about its perpendicular bisector.
• Magnetic field is parallel to the axis of rotation.

Physical idea For a rotating conductor in a magnetic field, motional emf is produced because charges experience magnetic force $\vec F = q(\vec v \times \vec B).$

Since the rod rotates in a plane perpendicular to the axis, and $\vec B$ is along the axis, the force on charges is along the rod.

Symmetry of the rod The rod rotates about its midpoint. So each half of the rod has length $\frac{L}{2} = 0.3\text{ m}.$

The midpoint is the axis of rotation. At a distance $r$ from the center, the speed is $v = \omega r.$

The motional emf developed between the center and one end is $\varepsilon_{\text{half}} = \int_0^{L/2} B\omega r\,dr$ $= B\omega \left[\frac{r^2}{2}\right]_0^{L/2}$ $= \frac{1}{2}B\omega\left(\frac{L}{2}\right)^2.$

Potential of the two ends Both halves are identical and symmetrically placed about the center. Hence the emf from the center to each end is equal in magnitude.

Therefore, both ends acquire the same potential relative to the center.

So the potential difference between the two ends is $V_{AB} = 0.$

Final answer $\boxed{0\text{ V}}$

Comparison with stored answer Stored correct answer = $0$

This matches the derived answer.