Time taken for one complete revolution = time period =2s.
Therefore, ω=T2π=πrads−1
Now flux, ϕ=NABcos(ωt)
Therefore, induced EMF ϵ=∣−dtdϕ∣=NABωsin(ωt)
The maximum voltage generated, ϵmax=NABω
=200×0.2×0.01×π
=104π=52π volt
Hence, β=5.
JEE Main 2024 — Physics Electromagnetism
A coil of 200 turns and area 0.20m2 is rotated at half a revolution per second and is placed in uniform magnetic field of 0.01T perpendicular to axis of rotation of the coil. The maximum voltage generated in the coil is β2π volt. The value of β is ______.
Held on 1 Feb 2024 · Verified 6 Jul 2026.
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