Question: The following Figure shows a simplified rotor and stator for a de motor. The mean path length of the rotor is 5 cm, and its cross-sectional area may be assumed to be equal to the cross-sectional area of the stator. Each air gap between the rotor and the stator is 0.5 cm wide, and the cross- sectional area of each air gap (including fringing) is 27 cm2. The iron of the core has a relative pemeability of 2000, and there are 1000 turns of wire on the core. If the current in the wire is adjusted to be 2 A, what will the resulting flux density in the air gaps be?
Question: The following Figure shows a simplified rotor and stator for a de motor. The mean path length of the rotor is 5 cm, and its cross-sectional area may be assumed to be equal to the cross-sectional area of the stator. Each air gap between the rotor and the stator is 0.5 cm wide, and the cross- sectional area of each air gap (including fringing) is 27 cm2. The iron of the core has a relative pemeability of 2000, and there are 1000 turns of wire on the core. If the current in the wire is adjusted to be 2 A, what will the resulting flux density in the air gaps be?
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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Transcribed Image Text:Question: The following Figure shows a simplified rotor and stator for a de
motor. The mean path length of the rotor is 5 cm, and its cross-sectional area
may be assumed to be equal to the cross-sectional area of the stator. Each air
gap between the rotor and the stator is 0.5 cm wide, and the cross- sectional
area of each air gap (including fringing) is 27 cm2. The iron of the core has a
relative permeability of 2000, and there are 1000 turns of wire on the core. If
the current in the wire is adjusted to be 2 A, what will the resulting flux density
in the air gaps be?
10 cm
13 cm
N= 1,000 turns
1, =5 cm-
4= 0.5 cm
13 cm
10 cm
- 30 cm-
3 cm
Depth = 5 cm
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