A metallic rod of length $$l$$ is tied to a string of length $$2l$$ and made to rotate with angular speed $$\omega$$ on a horizontal table with one end of the string fixed. If there is a vertical magnetic field B in the region, the e.m.f. induced across the ends of the rod is:

When a conductor rotates in a magnetic field, the motional e.m.f. ($$d\varepsilon$$) induced in a small element $$dr$$ at a distance $$r$$ from the axis of rotation is:

$$d\varepsilon = Bv \, dr = B(\omega r) \, dr$$

Since the rod is attached to a string of length $$2l$$, its ends are at distances $$r_1 = 2l$$ and $$r_2 = 2l + l = 3l$$ from the fixed pivot point.

$$\varepsilon = \int_{2l}^{3l} B\omega r \, dr$$

$$\varepsilon = B\omega \left[ \frac{r^2}{2} \right]_{2l}^{3l}$$

$$\varepsilon = \frac{B\omega}{2} \left[ (3l)^2 - (2l)^2 \right]$$

$$\varepsilon = \frac{5B\omega l^2}{2}$$