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AP EAMCET Engineering 2018 Apr 24 Shift 1 Paper
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© examsnet.com
Question : 111
Total: 160
Two circular loops of diameters 0.6 cm and 40 cm are kept coaxially with a separation of 15 cm between their centers. If a current 2 A flows through the smaller loop, the flux linked with the bigger loop is (approximately)
9.0 ×
10
−
11
Wb
0.9 ×
10
−
11
Wb
1.8 ×
10
−
11
Wb
2.7 ×
10
−
11
Wb
Validate
Solution:
The expression of the magnetic field intensity due to small loop at location of larger loop is,
B
1
=
µ
0
I
r
2
2
x
3
Substitute values,
B
1
=
(
4
π
×
10
−
7
)
(
2
)
(
0.3
×
10
−
2
)
2
2
(
15
×
10
−
2
)
3
B
1
=
36
π
15
3
×
10
−
9
T
The expression of the flux linked with larger loop is,
ϕ
2
=
B
1
A
2
Substitute values,
ϕ
2
=
86
π
15
3
×
10
−
19
×
π
(
20
×
10
−
2
)
2
ϕ
2
=
0.42
×
10
−
11
W
b
Thus no option is correct.
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