A bar magnet has a magnetic moment of  ${\text{100\hspace{0.33em}Am}}^2$. The magnet is suspended in a magnetic field of  ${\text{0.60\hspace{0.33em}NA}}^{\text{-1}}{\text{m}}^{\text{-1}}$. The torque required to rotate the magnet from its equilibrium position through an angle of  ${\text{30}}^0$ will be:

• A

  30 N-m

• B

  $30\sqrt{3}$N-m

• C

  60 N-m

• D

  $60\sqrt{3}$N-m