Why does Hm have minus x in front of it. How did they do integral?

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter6: The Trigonometric Functions
Section6.6: Additional Trigonometric Graphs
Problem 77E
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Why does Hm have minus x in front of it. How did they do integral?
Given
Ĥ = ŷHm cos(wt – Bz)
J=0, o=0.
Now according to ampere's law in differential
form
V ×Ħ ( j= 0)
пош
Hm Cos(wt – Bz)
= -êHmß sin(wt – Bz)
-âHmB sin(wt – B2)
→ D = |[-êHm3 sin(wt – Bz)]dt
= D = ê-
Hm3
cos(wt – Bz)
HmB
-cos(wt – B2)
= - = r
Eo
Transcribed Image Text:Given Ĥ = ŷHm cos(wt – Bz) J=0, o=0. Now according to ampere's law in differential form V ×Ħ ( j= 0) пош Hm Cos(wt – Bz) = -êHmß sin(wt – Bz) -âHmB sin(wt – B2) → D = |[-êHm3 sin(wt – Bz)]dt = D = ê- Hm3 cos(wt – Bz) HmB -cos(wt – B2) = - = r Eo
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