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JEE Main 202430 JanuaryMorning Shift
Exam
JEE Main
Year
2024
Date
30 January
Shift
Morning Shift
Subject
Physics
Chapter
Magnetic Properties of Matter

JEE Main 2024 · 30 January · Morning Shift · Physics · Magnetic Properties of Matter

A ceiling fan having 3 blades of length 80cm each is rotating with...

A ceiling fan having 33 blades of length 80cm80cm each is rotating with an angular velocity of 1200rpm1200rpm. The magnetic field of earth in that region is 0.5G0 . 5G and angle of dip is 30°30 °. The emf induced across the blades is Nπ×10−5VN\pi \times {10}^{- 5}V. The value of NN is _______.

Correct answer

0

Step-by-step solution

Component of the magnetic field perpendicular to the plane of fan blades, Bv=Bsin30°=(0.5×10−4T)×12=14×10−4T{B}_{v} = Bsin30 ° = \left(0 . 5 \times {10}^{- 4}T\right) \times \dfrac{1}{2} = \dfrac{1}{4} \times {10}^{- 4}T

Angular frequency, ω=2π×f=2π60×1200rads−1=40πrads−1\omega = 2\pi \times f = \dfrac{2\pi}{60} \times 1200rad{s}^{- 1} = 40\pi rad{s}^{- 1}

Therefore, induced EMF

ε=12BVωl2=12×(14×10−4)×40π×(0.8)2ε = \dfrac{1}{2}{B}_{V}\omega{l}^{2} = \dfrac{1}{2} \times \left(\dfrac{1}{4} \times {10}^{- 4}\right) \times 40\pi \times {\left(0 . 8\right)}^{2}

=32π×10−5V= 32\pi \times {10}^{- 5}V

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