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AP EAMCET Engineering 2018 Apr 23 Shift 2 Paper
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© examsnet.com
Question : 113
Total: 160
A coil of wire of radius ‘r’ has 600 turns and self inductance of 108 mH. The self inductance of a coil with same radius and 500 turns is _____
80 mH
75 mH
108 mH
90 mH
Validate
Solution:
The expression of the self inductance of coil is,
L
=
N
ϕ
B
I
L
=
N
I
µ
0
I
2
R
×
π
R
2
L
=
π
µ
0
N
R
2
L
∝
R
Thus,
L
2
L
1
=
N
1
N
2
Substitute values,
L
2
108
=
500
600
L
2
=
90
mH
© examsnet.com
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