The motional electromotive force (emf) induced in a conducting rod rotating in a magnetic field is derived from the Lorentz force. For a small element of length $dr$ at a distance $r$ from the axis, the induced emf is $d\varepsilon = Bv dr = B(\omega r) dr$ .

1. Identify the Conducting Segment: The stick has a total length of $3l$. The 'upper one third' refers to the segment furthest from the pivot (the outer end). Therefore, the conducting copper coating extends from radius $r_1 = 2l$ to $r_2 = 3l$. The remaining part (0 to $2l$) is wood (insulator) and contributes no motional emf because it lacks free charge carriers to separate and create a potential difference.

2. Calculate EMF: Integrate the differential emf over the limits of the conducting segment: $\varepsilon = \int_{2l}^{3l} B \omega r \, dr$ $\varepsilon = B \omega \left[ \frac{r^2}{2} \right]_{2l}^{3l}$ $\varepsilon = \frac{1}{2} B \omega \left( (3l)^2 - (2l)^2 \right)$ $\varepsilon = \frac{1}{2} B \omega (9l^2 - 4l^2)$ $\varepsilon = \frac{5 B \omega l^2}{2}$

Thus, the potential difference generated across the whole stick is effectively the potential difference across the conducting copper segment.