Question

8. A cast steel electromagnet has an air gap length of 3 mm and an iron path of length 40 cm.
Find the number of ampere-turns necessary to produce a flux density of 0.7 Wb/m2 in the gap
Neglect leakage and fringing. Assume ampere-turns required for air gap to be 70% of the total
ampere turns​

Answer 1

Explanation:

Given A cast steel electromagnet has an air gap length of 3 mm and an iron path of length 40 cm. Find the number of ampere-turns necessary to produce a flux density of 0.7 Wb/m2 in the gap

Neglect leakage and fringing. Assume ampere-turns required for air gap to be 70% of the total ampere turns

• Now we have
•   Flux density = 0.7 wb / m^2
•   Length of air gap = 3 mm = 3 x 10^-3 m  
• For the air gap ampere turns will be
•                                           = 1/4π x 10^-7 x flux density x length of gap
•                                          = 0.796 x 0.7 x 3 x 10^-3 x 10^6
•                                            = 1671 AT
• So from the b-h curve, the ampere turns per meter of flux path length for cast steel material will be 660 AT / m
• Now iron path of length is 40 cm = 0.4 m
• Therefore total ampere turns required for iron portion will be 660 x 0.4 = 264
• So the ampere turns by the electromagnet will be 1671 + 264 = 1935 AT

Reference link will be

https://brainly.in/question/2113717