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AIIMS 26 May 2019 Morning Question Paper with solutions for online practice
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© examsnet.com
Question : 4
Total: 146
For a toroid N = 500, radius = 40 cm, and area of cross section = 10 cm 2 . Find inductance
125 µH
250 µH
0.00248 H
zero
Validate
Solution:
If we flow a current i, then
B
=
µ
0
N
i
2
π
R
ϕ
sell
=
(
B
×
A
)
×
N
=
(
µ
0
N
i
2
π
R
)
A
×
N
⇒
ϕ
set
=
(
µ
0
N
2
A
2
π
R
)
i
and
ϕ
self
=
L
i
L
=
µ
0
N
2
A
2
π
R
=
µ
0
4
π
2
N
2
A
R
⇒
L
=
(
10
−
7
)
×
2
×
(
500
)
2
×
10
×
(
10
−
2
)
2
0.4
=
125
µ
H
© examsnet.com
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