question_answer

At some location on earth the horizontal component of earth’s magnetic field\[18\times {{10}^{-}}^{6}T\]. 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 \[45{}^\circ \] angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is-         [JEE Main 10-Jan-2019 Evening]

A) \[3.6\times {{10}^{-5}}\,N\]

B) \[1.8\times {{10}^{-5}}\text{ }N\]

C) \[1.3\times {{10}^{-5}}\,N\]

D) \[6.5\times {{10}^{-5}}\,N\]

Correct Answer: D

Solution :

[d]