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.
Shir Ahmad said:
8 years ago
Rf= Lavg /permeability* area.
So that after the analysis it find 1111.
So that after the analysis it find 1111.
Vishnu said:
8 years ago
R = pL/A, S = L/uA.
p=Resistivity u = Permeability.
So,
S = .07/(0.014*4500* 10^-6) = 1111 At/Wb.
p=Resistivity u = Permeability.
So,
S = .07/(0.014*4500* 10^-6) = 1111 At/Wb.
(1)
Naresh said:
8 years ago
1111 At/Wb.
Akshaya said:
8 years ago
S=l/ (uoura).
uo=means permittivity ur=permeability.
uo=means permittivity ur=permeability.
(2)
Swetha said:
7 years ago
Key point is;
Resistivity is inversely proportional to permeability.
Resistance = length/ua.
Resistivity is inversely proportional to permeability.
Resistance = length/ua.
(2)
Rutu said:
5 years ago
Well said @Swetha.
(1)
Gabrielle said:
2 years ago
Reluctance = length/Uo * Ur * Area.
Given:
Permeability (U) = 4500u Wb/Atxm.
Find Ur (relative permeability):
Ur=Permeability (U) /Free Space Permeability (Uo).
Ur =4500×10^-6 / 4pix10^-7 = 3580.98.
Reluctance = 0.07 / (4pix10^-7) (3580.98) (0.014).
Reluctance = 1111.11 At/Wb.
Given:
Permeability (U) = 4500u Wb/Atxm.
Find Ur (relative permeability):
Ur=Permeability (U) /Free Space Permeability (Uo).
Ur =4500×10^-6 / 4pix10^-7 = 3580.98.
Reluctance = 0.07 / (4pix10^-7) (3580.98) (0.014).
Reluctance = 1111.11 At/Wb.
(4)
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