Electrical Engineering - Magnetism and Electromagnetism - Discussion
Discussion Forum : Magnetism and Electromagnetism - General Questions (Q.No. 6)
6.
What is the reluctance of a material that has a length of 0.07 m, a cross-sectional area of 0.014 m2, and a permeability of 4,500 
Wb/At × m?
Wb/At × m?Discussion:
17 comments Page 2 of 2.
Kannan said:
1 decade ago
It's nice.
Rajesh kumar patro said:
1 decade ago
I agree with all above formula.
Junaid said:
1 decade ago
If I'm not mistaken U=Uo.Ur why not user 1/Uo = 0.796x10^6.
Chandraprakash borkar said:
1 decade ago
Use this farmula
R=l/uA Here l= length,u permeability, a= Area
R= .07/(0.0014*4500*10^6)
R=l/uA Here l= length,u permeability, a= Area
R= .07/(0.0014*4500*10^6)
Daniel said:
1 decade ago
Lets relate Reluctance with Resistance,
R = pL/A, S = L/uA.
p=Resistivity u = Permeability
so,
S = .07/(0.014*4500* 10^-6) = 1111 At/Wb
R = pL/A, S = L/uA.
p=Resistivity u = Permeability
so,
S = .07/(0.014*4500* 10^-6) = 1111 At/Wb
Dan said:
1 decade ago
Formula for this one?
(1)
Amrit Kumar Mohanta said:
1 decade ago
Reluctance S=L/(MU*A)
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