A $$20$$ cm long metallic rod is rotated with $$210$$ rpm about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field $$0.2$$ T parallel to the axis exists everywhere. The emf developed between the centre and the ring is _____ mV.
(Take $$\pi = \frac{22}{7}$$)

EMF developed when a rod rotates in a magnetic field:

$$\varepsilon = \frac{1}{2}B\omega L^2$$

Given: $$L = 20$$ cm $$= 0.2$$ m, $$B = 0.2$$ T, $$N = 210$$ rpm.

$$\omega = \frac{2\pi N}{60} = \frac{2\pi \times 210}{60} = 7\pi$$ rad/s

$$\varepsilon = \frac{1}{2} \times 0.2 \times 7\pi \times (0.2)^2 = \frac{1}{2} \times 0.2 \times 7\pi \times 0.04$$

$$= 0.028\pi \text{ V} = 0.028 \times \frac{22}{7} = 0.028 \times 3.1428... \approx 0.088 \text{ V} = 88 \text{ mV}$$

Using $$\pi = 22/7$$: $$\varepsilon = 0.028 \times 22/7 = 0.088$$ V $$= 88$$ mV.

The EMF developed is $$\mathbf{88}$$ mV.