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Electromagnetic Induction question

2024 · 30 Jan · Shift 1 · Q90
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  5. /2024 · 30 Jan · Shift 1 · Q90

Electromagnetic Induction question

2024 · 30 Jan · Shift 1 · Q90

JEE MainPhysicsElectromagnetic InductionNumerical+4 / −1
A ceiling fan having 3 blades of length 80 cm80 \mathrm{~cm}80 cm each is rotating with an angular velocity of 1200 rpm\mathrm{rpm}rpm. The magnetic field of earth in that region is 0.5G0.5 \mathrm{G}0.5G and angle of dip is 30∘30^{\circ}30∘. The emf induced across the blades is Nπ×10−5 V\mathrm{N} \pi \times 10^{-5} \mathrm{~V}Nπ×10−5 V. The value of N\mathrm{N}N is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 32

  1. Relevant magnetic field component

A horizontal ceiling fan rotates in a horizontal plane about a vertical axis.

For motional emf in a rotating blade, only the component of magnetic field parallel to the axis of rotation contributes.

Given:

  • Earth's magnetic field: B=0.5 G=0.5×10−4 T=5×10−5 TB = 0.5\,\text{G} = 0.5 \times 10^{-4}\,\text{T} = 5 \times 10^{-5}\,\text{T}B=0.5G=0.5×10−4T=5×10−5T
  • Angle of dip: δ=30∘\delta = 30^\circδ=30∘

The vertical component is

Bv=Bsin⁡δ=5×10−5⋅sin⁡30∘=5×10−5⋅12=2.5×10−5 TB_v = B \sin \delta = 5 \times 10^{-5} \cdot \sin 30^\circ = 5 \times 10^{-5} \cdot \frac{1}{2} = 2.5 \times 10^{-5}\,\text{T}Bv​=Bsinδ=5×10−5⋅sin30∘=5×10−5⋅21​=2.5×10−5T
  1. Angular velocity

The fan rotates at 1200 rpm1200\,\text{rpm}1200rpm.

So,

1200 rpm=120060=20 rev/s1200\,\text{rpm} = \frac{1200}{60} = 20\,\text{rev/s}1200rpm=601200​=20rev/s

Hence angular speed,

ω=2πf=2π×20=40π rad/s\omega = 2\pi f = 2\pi \times 20 = 40\pi\,\text{rad/s}ω=2πf=2π×20=40πrad/s
  1. Emf induced across one blade

For a rod of length lll rotating about one end in a magnetic field component along the axis,

ε=12Bωl2\varepsilon = \frac{1}{2} B \omega l^2ε=21​Bωl2

Here,

  • l=80 cm=0.8 ml = 80\,\text{cm} = 0.8\,\text{m}l=80cm=0.8m
  • B=Bv=2.5×10−5 TB = B_v = 2.5 \times 10^{-5}\,\text{T}B=Bv​=2.5×10−5T

Therefore,

ε=12(2.5×10−5)(40π)(0.8)2\varepsilon = \frac{1}{2}(2.5 \times 10^{-5})(40\pi)(0.8)^2ε=21​(2.5×10−5)(40π)(0.8)2

Now,

(0.8)2=0.64(0.8)^2 = 0.64(0.8)2=0.64

So,

ε=12(2.5×10−5)(40π)(0.64)\varepsilon = \frac{1}{2}(2.5 \times 10^{-5})(40\pi)(0.64)ε=21​(2.5×10−5)(40π)(0.64) =(1.25×10−5)(40π)(0.64)= (1.25 \times 10^{-5})(40\pi)(0.64)=(1.25×10−5)(40π)(0.64) =(1.25×40×0.64)π×10−5= (1.25 \times 40 \times 0.64)\pi \times 10^{-5}=(1.25×40×0.64)π×10−5 =32π×10−5 V= 32\pi \times 10^{-5}\,\text{V}=32π×10−5V
  1. About the 3 blades

The emf asked is across the blades; each blade has the same induced emf between center and tip. The number of blades does not multiply the emf across one blade.

Thus,

ε=Nπ×10−5 V\varepsilon = N\pi \times 10^{-5}\,\text{V}ε=Nπ×10−5V

with

N=32N = 32N=32
  1. Comparison with stored answer

Stored correct answer = 323232

This matches our derived result.

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