f=50 Hz, voltage peak is 32 V, thus the induced voltage can be expressed as e = 32 cos(100) do As e N- there is Thus dt و 1 >= Sedt N = | | fedt = 1 [32 cos(100m)dt = 11 12 32 1 -sin(100) = 0.0085sin(100) Wb 12 100π The flux density is simply this divided by the area $ 0.0085sin(100) = 0.85 sin(100) Wb/m², or T (Tesla) B = A 0.01 If you derived an equivalent expression with cos() instead of sin(), that's fine. It just depends whether you assume a voltage expression based on sin() or cos() at the very start of your solution.
f=50 Hz, voltage peak is 32 V, thus the induced voltage can be expressed as e = 32 cos(100) do As e N- there is Thus dt و 1 >= Sedt N = | | fedt = 1 [32 cos(100m)dt = 11 12 32 1 -sin(100) = 0.0085sin(100) Wb 12 100π The flux density is simply this divided by the area $ 0.0085sin(100) = 0.85 sin(100) Wb/m², or T (Tesla) B = A 0.01 If you derived an equivalent expression with cos() instead of sin(), that's fine. It just depends whether you assume a voltage expression based on sin() or cos() at the very start of your solution.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Hi, Can you help me to understand a step in the following derivate? I don't understand the part I highlighted in yellow. Why is 1/100pi ? Once is derived shouldn't be 32 x 100pi x -sin(100 pi t) ?
Transcribed Image Text:f=50 Hz, voltage peak is 32 V, thus the induced voltage can be expressed as
e = 32 cos(100)
do
As e N- there is
Thus
dt
و
1
>= Sedt
N
= | | fedt = 1 [32 cos(100m)dt =
11 12
32
1
-sin(100) = 0.0085sin(100) Wb
12 100π
The flux density is simply this divided by the area
$ 0.0085sin(100) = 0.85 sin(100) Wb/m², or T (Tesla)
B =
A
0.01
If you derived an equivalent expression with cos() instead of sin(), that's fine. It just depends
whether you assume a voltage expression based on sin() or cos() at the very start of your
solution.
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