The number of turns in the coil of an ac generator is 5000 and the area of the coil is 0.25 m . the coil is rotated at the rate of 100 cycles/sec in a magnetic field of 0.2t .the peak value of the emf generated is nearly
Answers
The peak value of the emf generated nearly is 157 kV .
Explanation:
No. of turns in the coil of an AC generator, N = 5000
Area of the coil, A = 0.25 m
The coil rotates at a rate of 100 cycles/sec i.e., frequency, ʋ = 100
Magnetic field, B = 0.2 T
The general formula for the induced emf is given by,
E = ωNBAsinωt ….. (i)
Where
ω = angular frequency = 2πʋ
sinωt = 1 …… [since for peak value of induced emf, value of ωNBAsinωt will be the maximum and maximum value of sinωt will be 1]
Let the peak value of the emf generated be denoted as “Eo”.
Thus, by substituting the given values in the formula (i), we get
The peak value of the emf generated is,
= (2πʋ)NBA * 1
= (2 * 3.14 * 100) * 5000 * 0.2 * 0.25 * 1
= 157000 V
= 157 kV
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Answer:
157kw
Explanation:
here is the answer for the question
