Physics · Electromagnetic Induction

JEE Main 2024 — 30 January, Shift 1 — Question 55

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 angle of dip is $30^{\circ}$ . The emf induced across the blades is $N \pi \times 10^{-5} \mathrm{~V}$ . The value of N is $\_\_\_\_\_\_\_$ .

Answer: 32

Numerical answer — enter this value.

Step-by-step solution

$\quad B_{v}=B \sin 30=\frac{1}{4} \times 10^{-4}$

$\omega=2 \pi \times f=\frac{2 \pi}{60} \times 1200 \mathrm{rad} / \mathrm{s}$

$\varepsilon=\frac{1}{2} B_{V} \omega \ell^{2}$ $=32 \pi \times 10^{-5} \mathrm{~V}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 30 January, Shift 1
Subject: Physics
Chapter: Electromagnetic Induction
Topic: Motional EMF