At some location  the horizontal component of earth's magnetic field is $18\times {10}^{-6} \mathrm{T}.$ At this location, magnetic needle of length $0.12 \mathrm{m}$ and pole strength $1.8 \mathrm{Am}$ is suspended from its mid-point using a thread, it makes $45°$ angles with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is:

• A

  $1.8\times {10}^{-5} \mathrm{N}$
• B

  $3.6\times {10}^{-5} \mathrm{N}$
• Correct answer
  C

  $6.5\times {10}^{-5} \mathrm{N}$
• D

  $1.3\times {10}^{-5} \mathrm{N}$

Hence, the correct answer is option: 'C'.