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Physics · Electromagnetic Induction

JEE Main 2024 — 1 February, Shift 2 — Question 57

A coil of 200 turns and area 0.20 m20.20 \mathrm{~m}^{2} is rotated at half a revolution per second and is placed in uniform magnetic field of 0.01 T perpendicular to axis of rotation of the coil. The maximum voltage generated in the coil is 2πβ\frac{2 \pi}{\beta} volt. The value of β\beta is

Answer: 5

Numerical answer — enter this value.

Step-by-step solution

Given:N=200 turnsA=0.20 m2ω=12 rev/s=π rad/sB=0.01 TVmax⁡=2πβ volt\begin{aligned} \text{Given:} \quad &N = 200 \text{ turns} \\ &A = 0.20 \,\text{m}^2 \\ &\omega = \frac{1}{2}\,\text{rev/s} = \pi\,\text{rad/s} \\ &B = 0.01 \,\text{T} \\ &V_{\max} = \frac{2\pi}{\beta} \text{ volt} \end{aligned} Step 1: Magnetic flux through coil:Φ=NBAcos⁡(ωt)\begin{aligned} \text{Step 1: Magnetic flux through coil:} \quad \Phi &= NBA\cos(\omega t) \end{aligned} Step 2: Induced EMF:ε=−dΦdt=−ddt(NBAcos⁡(ωt))=NBAωsin⁡(ωt)\begin{aligned} \text{Step 2: Induced EMF:} \quad \varepsilon &= -\frac{d\Phi}{dt} \\ &= -\frac{d}{dt}(NBA\cos(\omega t)) \\ &= NBA\omega\sin(\omega t) \end{aligned} Step 3: Maximum EMF:εmax⁡=NBAω=200×0.01×0.20×π=0.4π V\begin{aligned} \text{Step 3: Maximum EMF:} \quad \varepsilon_{\max} &= NBA\omega \\ &= 200 \times 0.01 \times 0.20 \times \pi \\ &= 0.4\pi \,\text{V} \end{aligned} Step 4: Given 2πβ=0.4π:2πβ=0.4πβ=5\begin{aligned} \text{Step 4: Given } \frac{2\pi}{\beta} = 0.4\pi: \quad \frac{2\pi}{\beta} &= 0.4\pi \\ \beta &= 5 \end{aligned} β=5\boxed{\beta = 5}

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Exam
JEE Main 2024
Paper
1 February, Shift 2
Subject
Physics
Chapter
Electromagnetic Induction
Topic
Magnetic Flux, Faraday's Law and Lenz's Law

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