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I(A) – F(dyn) 800 260.40 dyne/A 700 - 600- F = IBLsin(4) sin(o) = 90° B = ? 500- 400 300 200 100 Al = 12(A) – 1(A) 0.0 0.5 1.0 1.5 2.0 2.5 3.0 I(A) F(dyne) AF = F2(dyne) – F1(dyne)

I(A) – F(dyn) 800 260.40 dyne/A 700 - 600- F = IBLsin(4) sin(o) = 90° B = ? 500- 400 300 200 100 Al = 12(A) – 1(A) 0.0 0.5 1.0 1.5 2.0 2.5 3.0 I(A) F(dyne) AF = F2(dyne) – F1(dyne)

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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find the B?

(ст) — F(dyn) 112.75 dyne/cm 1000 800 F = IBLsin(4) sin(ø) = 90° B = ? 600 400 Al = 12(cm) – 1(cm) 200 2 4 (ст) F(dyne) -Co AF = F2(dyne) – F1(dyne)
Transcribed Image Text:(ст) — F(dyn) 112.75 dyne/cm 1000 800 F = IBLsin(4) sin(ø) = 90° B = ? 600 400 Al = 12(cm) – 1(cm) 200 2 4 (ст) F(dyne) -Co AF = F2(dyne) – F1(dyne)
I(A) – F(dyn) 800 260.40 dyne/A 700 600 F = IBLsin(4) sin(4) = 90° B = ? 500 400 300 200 100 Al = 12(A) – 1(A) 0.0 0.5 1.0 1.5 2.0 2.5 3.0 I(A) F(dyne) AF = F2(dyne) - F1(dyne)
Transcribed Image Text:I(A) – F(dyn) 800 260.40 dyne/A 700 600 F = IBLsin(4) sin(4) = 90° B = ? 500 400 300 200 100 Al = 12(A) – 1(A) 0.0 0.5 1.0 1.5 2.0 2.5 3.0 I(A) F(dyne) AF = F2(dyne) - F1(dyne)
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