Practice portal › Electromagnetic Induction › AC Generator
Asked in JEE Main 1st Feb 2nd Shift 2023 · Peak and instantaneous EMF
ω=(2π×500)/(60)=52.36 rad s⁻¹ and A=70×10⁻⁴ m².
ε₀=NBAω=600×0.4×70×10⁻⁴×52.36=88 V.
With the plane at 60° to the field, ε=ε₀ cos 60°=44 V.
Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.
Practise this with a timer