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 1 of 2.
Amrit Kumar Mohanta said:
1 decade ago
Reluctance S=L/(MU*A)
Dan said:
1 decade ago
Formula for this one?
(1)
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
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)
Junaid said:
1 decade ago
If I'm not mistaken U=Uo.Ur why not user 1/Uo = 0.796x10^6.
Rajesh kumar patro said:
1 decade ago
I agree with all above formula.
Kannan said:
1 decade ago
It's nice.
Robin Bhowmick said:
9 years ago
Try to understand using your common sense and logic rather than memorising the formula.
Engineering is an art, it cannot be mastered only memorising the study.
Engineering is an art, it cannot be mastered only memorising the study.
Razin said:
9 years ago
The value of permeability is give not relative permeability.
Puspendra singh said:
9 years ago
Can anyone explain it by a easy calculation? please.
(1)
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