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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.

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter9: Multivariable Calculus
Section9.CR: Chapter 9 Review
Problem 57CR
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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) ? 

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.
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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