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 Online Paper (Held On 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 :

\[{{B}_{H}}={{B}_{Earth}}\text{ }\cos \,45{}^\circ \] \[{{B}_{Earth}}={{B}_{H}}\sqrt{2}\] \[{{Z}_{ab{{t}_{0}}}}=0\] \[M{{B}_{Earth}}\sin \,45{}^\circ -F\times 0.06=0\] \[1.8\times 0.12\text{ }{{B}_{H}}\frac{1}{\sqrt{2}}\times \sqrt{2}-F\times 0.06=0\] \[F=1.8\times 2\times {{B}_{H}}\] \[=\,\,\,3.6\times 18\times {{10}^{-6}}\] \[=\text{ }64.8\times {{10}^{-6}}\] \[=\text{ }6.5\times {{10}^{-5}}\,N\]