# electromagnetic relay question

Discussion in 'Homework Help' started by exidez, Nov 13, 2008.

1. ### exidez Thread Starter Member

Aug 22, 2008
26
0
this question has been bothering me as i do not have any solutions or answers! See attached jpg for the question.

It states the air gap has a flux intensity of 0.8 Telsa
so from the graph i can get H to be around 400 At/m for the cast steel arm

the length of the air gap is 1.5mm which is 1.5x10^-3 meters
H* l = N I
(H*L)/N = I
(400*1.5x10^-3)/1000 = 0.6 mA

is this correct? is all the other information useless??

i can easily work out part b. thats not problem

File size:
67.2 KB
Views:
12
File size:
42 KB
Views:
10
2. ### exidez Thread Starter Member

Aug 22, 2008
26
0
i found the solution!

you have to add all the components
NI = ƩΦR = ƩHL = ƩH(B/μ)

so the answer is
NI = HL (cast steel) + HL (sheet steel) + H(B/μ) (air gap)
NI = (0.12*500) + (0.06*250) + (1.5x10^-3 * [ 0.8 / 4∏*10^-7 ] )
NI = 1030
1030 / 1000 = I = 1.03 A

it always seems so simple AFTER you know the answer!