(a)
Integral (B.dA)
B.dA cos (wt)
B.cos wt integral (dA)
B cos 2ft .
so,
= B cos ((2ft)
_____________________
(b)
at t = 0.121 s
= 0.21 * * 0.35^2 cos ( 2 * 5.5 * 0.121)
= 0.081 Wb
_____________________
(c)
emf = d / dt
emf = 2 f B sin (2ft)
___________________________
(d)
at t = 0.121 s
emf = 0.2 V
________________
(e)
I = emf / R
I = 2 f B sin (2ft) / R
______________
(f)
I = 0.018 Amps
(11%) Problem 8: A circular wire loop of radius r = 0.35 m and resistance R=...
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