At some location on earth the horizontal components of earth's magnetic field is $18 \times 10^{-} 6 T .$ 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 $45^{0}$ angle with horizontal in equilibrium to keep this needle horizontal, the vertical force that should be applied at one of its ends is:

3.6 \times 10^{-} 5 N

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6.5 \times 10^{-} 5 N

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1.3 \times 10^{-} 5 N

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1.8 \times 10^{-} 5 N

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Solution

The correct option is D $6.5 \times 10^{-} 5 N$
At $45^{o}, B_{H} = B_{V}$

$\frac{F l}{2} = M B_{V} = m \times l \times B_{V}$

$F = \frac{2 m l B_{V}}{l} = 3.6 \times 18 \times 10^{- 6}$

$F = 6.5 \times 10^{- 5} N$

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The correct option is D 6.5×10−5NAt 45o,BH=BVFl2=MBV=m×l×BVF=2mlBVl=3.6×18×10−6F=6.5×10−5N

The correct option is D $6.5 \times 10^{-} 5 N$
At $45^{o}, B_{H} = B_{V}$

$\frac{F l}{2} = M B_{V} = m \times l \times B_{V}$

$F = \frac{2 m l B_{V}}{l} = 3.6 \times 18 \times 10^{- 6}$

$F = 6.5 \times 10^{- 5} N$