Given here, N=600, A=70×10−4m2, and B=0.4T
Angular speed of coil is ω=602πn=60500×2π=6100πrads−1
Instantaneous emf in coil is given by ϵ=−Ndtdϕ, where, magnetic flux ϕ=BAcosωt and ωt=30∘ is angle between A and B.
So, ϵ=−Ndtd(BAcosωt)=−N(−BAωsinωt)=NBAωsinωt
Putting the values, we get
ϵ=600×70×10−4×0.4×6100π×21
=44V