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Physics Question on Electromagnetic induction

A ceiling fan having 3 blades of length 80 cm each is rotating with an angular velocity of 1200 rpm. The magnetic field of earth in that region is 0.5 G and the angle of dip is 3030^\circ. The emf induced across the blades is Nπ×105VN \pi \times 10^{-5} \, \text{V}. The value of NN is \\_\\_\\_\\_\\_.

Answer

Step 1. Calculate the Effective Vertical Component of the Magnetic Field:

Given:

B=0.5G=0.5×104TB = 0.5 \, \text{G} = 0.5 \times 10^{-4} \, \text{T}

The vertical component of the magnetic field BvB_v, considering the angle of dip δ=30\delta = 30^\circ, is:

Bv=Bsinδ=0.5×104×sin30=0.5×104×12=14×104TB_v = B \sin \delta = 0.5 \times 10^{-4} \times \sin 30^\circ = 0.5 \times 10^{-4} \times \frac{1}{2} = \frac{1}{4} \times 10^{-4} \, \text{T}

Step 2. Convert Angular Velocity from rpm to rad/s:

Angular velocity ω\omega in rad/s is given by:

ω=2π×f=2π×120060=2π×20=40πrad/s\omega = 2 \pi \times f = 2 \pi \times \frac{1200}{60} = 2 \pi \times 20 = 40 \pi \, \text{rad/s}

Step 3. Determine the Radius of Rotation:

The length of each blade is =80cm=0.8m\ell = 80 \, \text{cm} = 0.8 \, \text{m}. Therefore, the effective radius rr of rotation is:

r=0.8mr = 0.8 \, \text{m}

Step 4. Calculate the Induced emf:

The emf ε\varepsilon induced across the tips of the blades (assuming the emf induced across two opposite ends) is given by:

ε=12Bvωr2\varepsilon = \frac{1}{2} B_v \omega r^2

Substituting the values:

ε=12×14×104×40π×(0.8)2\varepsilon = \frac{1}{2} \times \frac{1}{4} \times 10^{-4} \times 40 \pi \times (0.8)^2

Simplifying further:

ε=12×14×104×40π×0.64=32π×105V\varepsilon = \frac{1}{2} \times \frac{1}{4} \times 10^{-4} \times 40 \pi \times 0.64 = 32 \pi \times 10^{-5} \, \text{V}

Step 5. Conclude the Value of NN:

Comparing with Nπ×105VN \pi \times 10^{-5} \, \text{V}, we find:

N=32N = 32