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JEE Main 10 Jan 2019 Shift 2 Solved Paper
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© examsnet.com
Question : 8
Total: 90
At some location on earth the horizontal component of earth's magnetic field is 18 ×
10
–
6
T. At this location, magnetic needle of length 0.12 m and pole strength 1.8 Am is suspended from its mid-point using a thread, it makes 45° angle with horizontalin equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is :
3.6 ×
10
–
5
N
6.5 ×
10
–
5
N
1.3 ×
10
–
5
N
1.8 ×
10
–
5
N
Validate
Solution:
µBsin 45° = F
l
2
sin 45°
F = 2µB
© examsnet.com
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