A ceiling fan having $$3$$ blades of length $$80 \text{ cm}$$ each is rotating with an angular velocity of $$1200 \text{ rpm}$$. The magnetic field of earth in that region is $$0.5 \text{ G}$$ and angle of dip is $$30°$$. The emf induced across the blades is $$N\pi \times 10^{-5} \text{ V}$$. The value of $$N$$ is ______.

The EMF induced in a conducting rod rotating about one end in a magnetic field is given by:

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

Here, $$B$$ is the component of the magnetic field perpendicular to the plane of rotation, $$\omega$$ is the angular velocity, and $$L$$ is the length of the rod (blade). In this problem, each blade has length $$L = 80 \text{ cm} = 0.8 \text{ m}$$, the fan rotates at $$\omega = 1200 \text{ rpm} = \frac{1200 \times 2\pi}{60} = 40\pi \text{ rad/s}$$, the magnitude of Earth's magnetic field is $$B = 0.5 \text{ G} = 0.5 \times 10^{-4} \text{ T}$$, and the angle of dip is $$\delta = 30°$$.

Since the fan rotates in the horizontal plane, the perpendicular (vertical) component of the magnetic field is

$$B_v = B \sin\delta = 0.5 \times 10^{-4} \times \sin 30° = 0.5 \times 10^{-4} \times 0.5 = 0.25 \times 10^{-4} \text{ T}$$

Substituting this into the expression for the induced EMF in one blade gives

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

$$\varepsilon = \frac{1}{2} \times 0.25 \times 10^{-4} \times 40\pi \times (0.8)^2$$

$$\varepsilon = \frac{1}{2} \times 0.25 \times 10^{-4} \times 40\pi \times 0.64$$

$$\varepsilon = \frac{1}{2} \times 0.25 \times 40 \times 0.64 \times \pi \times 10^{-4}$$

$$\varepsilon = \frac{1}{2} \times 6.4 \times \pi \times 10^{-4} = 3.2\pi \times 10^{-4} = 32\pi \times 10^{-5} \text{ V}$$

Comparing this result with the form $$N\pi \times 10^{-5}$$ shows that $$N = 32$$.