$A$ magnetic needle lying parallel to a magnetic field is turned through $60^{\circ}$. The work done on it is $W$. The torque required to maintain the magnetic needle in the position mentioned above is

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

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