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?
1111 At/Wb
111 At/Wb
11 At/Wb
1 At/Wb
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
17 comments Page 1 of 2.

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.
(4)

Rutu said:   5 years ago
Well said @Swetha.
(1)

Swetha said:   7 years ago
Key point is;

Resistivity is inversely proportional to permeability.

Resistance = length/ua.
(2)

Akshaya said:   8 years ago
S=l/ (uoura).

uo=means permittivity ur=permeability.
(2)

Naresh said:   8 years ago
1111 At/Wb.

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.
(1)

Shir Ahmad said:   8 years ago
Rf= Lavg /permeability* area.

So that after the analysis it find 1111.

Puspendra singh said:   9 years ago
Can anyone explain it by a easy calculation? please.
(1)

Razin said:   9 years ago
The value of permeability is give not relative permeability.

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.


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