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Part (a): The magnetic flux linkage in a single turn of the inductor can be calculated as, p = [B·ds Given that, B = B (ŷ2 +23) sin cot 6 = [ B. ds = [(B₁ (ŷ2 + 23) sin cot). 2ds $ = 37a²B₁ sin at

Part (a): The magnetic flux linkage in a single turn of the inductor can be calculated as, p = [B·ds Given that, B = B (ŷ2 +23) sin cot 6 = [ B. ds = [(B₁ (ŷ2 + 23) sin cot). 2ds $ = 37a²B₁ sin at

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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I want to see the steps on how they did the intergration that lead them to get the final result. I have been doing the math and can't seem to get that result. I step by step procces would be beneficial. thanks!

Part (a): The magnetic flux linkage in a single turn of the inductor can be calculated as, p = [B·ds Given that, B = B (ŷ2 +23) sin cot 6 = [ B. ds = [(B₁ (ŷ2 + 23) sin cot). 2ds $ = 37a²B₁ sin at
Transcribed Image Text:Part (a): The magnetic flux linkage in a single turn of the inductor can be calculated as, p = [B·ds Given that, B = B (ŷ2 +23) sin cot 6 = [ B. ds = [(B₁ (ŷ2 + 23) sin cot). 2ds $ = 37a²B₁ sin at
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