# 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 horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is: (a) 3.6×10−5 N (b) 6.5×10−5 N (c) 1.3×10−5 N (d) 1.8×10−5 N

## Question ID - 50256 :- 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 horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is: (a) 3.6×10−5 N (b) 6.5×10−5 N (c) 1.3×10−5 N (d) 1.8×10−5 N

3537

Answer Key : (b) -

Bsin45°=Fsin45°

F=2B

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