A current of 1.5 amperes in a 800 turns coil produces1.5 x 10-5 weber magnetized flux in each turn.
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Explanation:
Magnetic field B=5×10
−5
T
Number of turns in coil N=800
Area of coil A=0.05m
2
Time taken to rotate Δt=0.1s
Initial angle θ
1
=0
o
Final angle θ
2
=90
o
change in magnetic flux Δϕ
=NBAcos90
o
−BAcos0
o
=−NBA
=−800×5×10
−5
×0.05
=−2×10
−3
weber
e=−
Δt
Δϕ
=
0.1s
−(−)2×10
−3
Wb
=0.02V
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