Electromagnetic Induction

JEE Main level, numerical answer, medium

80 of 153

Question 80 of 153

A coil of 200200 turns and area 0.20m20.20{m}^{2} is rotated at half a revolution per second and is placed in uniform magnetic field of 0.01T0.01T perpendicular to axis of rotation of the coil. The maximum voltage generated in the coil is 2πβ\dfrac{2\pi }{\beta } volt. The value of β\beta is ______.

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Answer: 5

Time taken for one complete revolution = time period =2s=2s.
Therefore, ω=2πT=πrads−1\omega =\dfrac{2\pi }{T}=\pi rad{s}^{-1}
Now flux, ϕ=NABcos(ωt)\phi =NABcos(\omega t)
Therefore, induced EMF ϵ=∣−dϕdt∣=NABωsin⁡(ωt)\epsilon =\left|-\dfrac{d\phi }{dt}\right|=NAB\omega \sin (\omega t)
The maximum voltage generated, ϵmax⁡=NABω{\epsilon }_{\max }=NAB\omega
=200×0.2×0.01×π=200\times 0.2\times 0.01\times \pi
=4π10=2π5=\dfrac{4\pi }{10}=\dfrac{2\pi }{5} volt
Hence, β=5\beta =5.

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