$A$ magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through $60^{\circ}$. The torque required to maintain the needle in this position will be

  • A
    $\sqrt{3} W$
  • B
    $W$
  • C
    $\frac{\sqrt{3}}{2} W$
  • D
    $2 W$

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