AP EAMCET Engineering 23 Sep 2020 Shift 1 Paper

Given that, area, A=10m2
Magnetic field, B=0.3NA−1m−1
T=8s
When plane is perpendicular to field, then flux passing through it, φ=BA‌cos‌0∘=0.3×10×1=3 Wb
When plane is parallel to field, then flux passing through it, φ=BA‌cos‌90∘=0
For 0∘ to 90∘ position, average emf,
εavg=‌

∆φ

∆t

=‌

φ90∘−φ0∘

T 4

=‌

3−0

2

=‌

3

2

V
For 90∘ to 180∘ position,
savg=−‌

φ180∘−φ90∘

T 4

=−‌

0−3

2

=‌

3

2

V
Negative sign has been taken because flux associated to plane is decreasing.
Similarly, for 180∘ to 270∘ position,
εavg=−‌

φ270∘−φ180∘

T 4

=−‌

3−0

2

=‌

−3

2

V
For 270∘ to 360∘ position,
ε‌avg ‌=+‌

φ360∘−φ270∘

T 4

=+‌

0−3

2

=−‌

3

2

V