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MCQEasyJEE 2024Faraday's Laws of EMI

JEE Physics 2024 Question with Solution

A ceiling fan having 33 blades of length 80 cm80 \, \text{cm} each is rotating with an angular velocity of 1200 rpm1200 \, \text{rpm}. The magnetic field of Earth in that region is 0.5 G0.5 \, \text{G} and the angle of dip is 30∘30^\circ. The emf induced across the blades is Nπ×10−5 VN\pi \times 10^{-5} \, \text{V}. The value of NN is:

  • A

    1616

  • B

    2424

  • C

    3232

  • D

    4040

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: blade length ℓ=80 cm=0.8 m\ell = 80 \, \text{cm} = 0.8 \, \text{m}, angular velocity ω=1200 rpm\omega = 1200 \, \text{rpm}, Earth's magnetic field B=0.5 G=0.5×10−4 TB = 0.5 \, \text{G} = 0.5 \times 10^{-4} \, \text{T}, angle of dip δ=30∘\delta = 30^\circ.

Find: the value of NN in Nπ×10−5 VN\pi \times 10^{-5} \, \text{V}.

Only the vertical component of Earth's magnetic field is perpendicular to the horizontal plane of rotation. Therefore,

Bv=Bsin⁡δ=0.5×10−4×sin⁡30∘=0.5×10−4×12=14×10−4 TB_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}

Convert angular velocity from rpm to rad/s:

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

For a rotating blade, the induced emf is

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

with r=0.8 mr = 0.8 \, \text{m}.

Substitute the values:

ε=12×14×10−4×40π×(0.8)2\varepsilon = \frac{1}{2} \times \frac{1}{4} \times 10^{-4} \times 40\pi \times (0.8)^2 ε=12×14×10−4×40π×0.64=32π×10−5 V\varepsilon = \frac{1}{2} \times \frac{1}{4} \times 10^{-4} \times 40\pi \times 0.64 = 32\pi \times 10^{-5} \, \text{V}

Comparing with Nπ×10−5 VN\pi \times 10^{-5} \, \text{V}, we get N=32N = 32. Therefore, the correct option is C.

Detailed Computation

Given: L=0.8 mL = 0.8 \, \text{m}, BE=0.5×10−4 TB_E = 0.5 \times 10^{-4} \, \text{T}, δ=30∘\delta = 30^\circ, ω=1200 rpm\omega = 1200 \, \text{rpm}.

Find: the value of NN.

The vertical component of Earth's magnetic field is

BV=BEsin⁡(δ)B_V = B_E \sin(\delta) BV=(0.5×10−4)×sin⁡30∘B_V = (0.5 \times 10^{-4}) \times \sin 30^\circ BV=(0.5×10−4)×0.5=0.25×10−4 TB_V = (0.5 \times 10^{-4}) \times 0.5 = 0.25 \times 10^{-4} \, \text{T}

Now convert angular speed:

ω=1200 revmin×2π rad1 rev×1 min60 s=40π rad/s\omega = 1200 \, \frac{\text{rev}}{\text{min}} \times \frac{2\pi \, \text{rad}}{1 \, \text{rev}} \times \frac{1 \, \text{min}}{60 \, \text{s}} = 40\pi \, \text{rad/s}

Using the rotating rod formula,

E=12BVL2ω\mathcal{E} = \frac{1}{2} B_V L^2 \omega E=12(0.25×10−4)(0.8)2(40π)\mathcal{E} = \frac{1}{2} (0.25 \times 10^{-4}) (0.8)^2 (40\pi) E=12(0.25×10−4)(0.64)(40π)\mathcal{E} = \frac{1}{2} (0.25 \times 10^{-4}) (0.64) (40\pi)

Simplify step by step:

E=(0.5×0.25×0.64×40)π×10−4\mathcal{E} = (0.5 \times 0.25 \times 0.64 \times 40) \pi \times 10^{-4} E=(20×0.25×0.64)π×10−4\mathcal{E} = (20 \times 0.25 \times 0.64) \pi \times 10^{-4} E=(5×0.64)π×10−4\mathcal{E} = (5 \times 0.64) \pi \times 10^{-4} E=3.2π×10−4 V=32π×10−5 V\mathcal{E} = 3.2\pi \times 10^{-4} \, \text{V} = 32\pi \times 10^{-5} \, \text{V}

Hence, N=32N = 32 and the correct option is C.

Common mistakes

  • Using the total Earth's magnetic field instead of its vertical component is incorrect because the fan rotates in a horizontal plane. Only the component perpendicular to the plane, Bsin⁡δB\sin\delta, contributes to the motional emf.

  • Leaving angular velocity in rpm is incorrect because the emf formula requires ω\omega in rad/s. First convert 1200 rpm1200 \, \text{rpm} to 40π rad/s40\pi \, \text{rad/s}.

  • Using the blade length in centimetres directly is incorrect because SI units are required. Convert 80 cm80 \, \text{cm} to 0.8 m0.8 \, \text{m} before substitution.

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