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JEE Advanced 2020 Concept Recapitulation Test 2 Paper 1
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© examsnet.com
Question : 3
Total: 54
A thin insulated wire forms a plane spiral of
N
=
100
tight turns carrying a current
I
=
8
m
A
(milli ampere). The radii of inside and outside turns are
a
=
50
m
m
and
b
=
100
m
m
respectively. The magnetic induction at the center of the spiral is
5µT
7µT
8µT
10µT
Validate
Solution:
Let
n
be the number of turns per unit length along the radii of the spiral. Consider a ring of radii
x
and
x
+
d
x
.
Number of turns in the ring
=
n
d
x
∴
Magnetic field at the center due to this ring
=
µ
0
(
n
d
x
)
i
2
x
∴
B
(
total field
)
=
b
∫
a
µ
0
n
i
d
x
2
x
=
1
2
µ
0
n
i
ln
b
a
But
n
=
N
b
−
a
B
=
µ
0
N
i
2
(
b
−
a
)
ln
b
a
=
4
π
×
10
−
7
×
100
×
8
×
10
−
3
2
(
10
−
5
)
×
10
−
2
ln
10
5
=
6.9
×
10
−
6
T
=
6.9
µ
T
.
© examsnet.com
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