A coil of 200 turns and area $$0.20 \text{ m}^2$$ is rotated at half a revolution per second and is placed in uniform magnetic field of 0.01 T perpendicular to axis of rotation of the coil. The maximum voltage generated in the coil is $$\frac{2\pi}{\beta}$$ volt. The value of $$\beta$$ is ______.

A coil of 200 turns and area $$0.20 \text{ m}^2$$ is rotated at half a revolution per second in a uniform magnetic field of 0.01 T perpendicular to the axis of rotation. We need to find $$\beta$$ where the maximum voltage is $$\frac{2\pi}{\beta}$$ V.

The EMF induced in a rotating coil in a magnetic field is:

$$\varepsilon = NAB\omega\sin(\omega t)$$

The maximum EMF is:

$$\varepsilon_{\max} = NAB\omega$$

The coil rotates at $$\frac{1}{2}$$ revolution per second, so:

$$\omega = 2\pi \times \frac{1}{2} = \pi \text{ rad/s}$$

$$\varepsilon_{\max} = 200 \times 0.20 \times 0.01 \times \pi$$

$$= 200 \times 0.002 \times \pi$$

$$= 0.4\pi$$

$$= \frac{2\pi}{5}$$

Comparing $$\frac{2\pi}{5}$$ with $$\frac{2\pi}{\beta}$$:

$$\beta = 5$$

The answer is 5.