Examsnet
Unconfined exams practice
Home
Exams
Banking Entrance Exams
CUET Exam Papers
Defence Exams
Engineering Exams
Finance Entrance Exams
GATE Exam Practice
Insurance Exams
International Exams
JEE Exams
LAW Entrance Exams
MBA Entrance Exams
MCA Entrance Exams
Medical Entrance Exams
Other Entrance Exams
Police Exams
Public Service Commission (PSC)
RRB Entrance Exams
SSC Exams
State Govt Exams
Subjectwise Practice
Teacher Exams
SET Exams(State Eligibility Test)
UPSC Entrance Exams
Aptitude
Algebra and Higher Mathematics
Arithmetic
Commercial Mathematics
Data Based Mathematics
Geometry and Mensuration
Number System and Numeracy
Problem Solving
Board Exams
Andhra
Bihar
CBSE
Gujarat
Haryana
ICSE
Jammu and Kashmir
Karnataka
Kerala
Madhya Pradesh
Maharashtra
Odisha
Tamil Nadu
Telangana
Uttar Pradesh
English
Competitive English
Certifications
Technical
Cloud Tech Certifications
Security Tech Certifications
Management
IT Infrastructure
More
About
Careers
Contact Us
Our Apps
Privacy
Test Index
GATE Electrical Engineering (EE) 2019 Solved Paper
Show Para
Hide Para
Share question:
© examsnet.com
Question : 47
Total: 65
The magnetic circuit shown below has uniform cross-sectional area and air gap of
0.2
c
m
. The mean path length of the core is
40
c
m
. Assume that leakage and fringing fluxes are negligible. When the core relative permeability is assumed to be infinite, the magnetic flux density computed in the air gap is 1 tesla. With same Ampere-turns, if the core relative permeability is assumed to be 1000 (linear), the flux density in tesla (round off to three decimal places) calculated in the air gap is __________.
Your Answer:
Validate
Solution:
length of the core
(
l
l
)
=
39.8
c
m
length of the air gap
(
l
d
)
=
0.2
c
m
We know
M
M
F
=
R
M
ϕ
N
I
=
R
M
ϕ
φ
=
N
I
R
M
where
R
M
=
reluctance
B
=
φ
A
B
=
N
I
R
M
A
B
α
1
R
M
Case-1 Permeability of the core
(
µ
c
)
=
∞
(
R
M
)
1
=
R
1
+
R
2
=
ℓ
c
µ
A
+
ℓ
a
µ
0
A
(
R
M
)
1
=
0
+
0.2
µ
0
A
Case-2
Permeability of the core
(
µ
c
)
=
1000
µ
0
(
R
M
)
2
=
R
1
+
R
2
=
39.8
10
3
µ
0
A
+
0.2
µ
0
A
(
R
M
)
2
=
1
µ
0
A
[
0.2398
]
B
1
B
2
=
(
R
M
)
2
(
R
M
)
1
B
2
=
(
R
M
)
1
(
R
M
)
2
=
0.2
0.2398
B
2
=
0.83
© examsnet.com
Go to Question:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
Prev Question
Next Question