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VITEEE 2015 Solved Paper
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© examsnet.com
Question : 28
Total: 120
A coil of area 5
c
m
2
having 20 turns is placed in a uniform magnetic field of
10
3
gauss. The normal to the plane of coil makes an angle 30° with the magnetic field. The flux through the coil is
6.67 x
10
−
4
wb
3.2 x
10
−
5
wb
5.9 x
10
−
4
wb
8.65 x
10
−
4
wb
Validate
Solution:
Area of the coil
A
=
5
c
m
2
=
5
×
1
0
−
4
m
2
Number of turns
N
=
2
0
Magnetic field
B
=
1
0
3
Gauss
=
1
0
−
1
T
Angle between magnetic field and normal to the plane of coil
θ
=
3
0
∘
Thus flux through the coil
φ
=
N
B
A
cos
θ
∴
φ
=
20
×
10
−
1
×
5
×
10
−
4
×
0.865
=
8.65
×
10
−
4
W
b
© examsnet.com
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