A short bar magnet of magnetic dipole moment $10^{-5} A\, m^2$ is kept in a uniform magnetic field $3×10^{-4}T$ at an angle of 30°. The torque acting on it is:

$1.5×10^{-9} Nm$

The correct answer is Option (1) → $1.5×10^{-9} Nm$

The torque $(τ)$ acting on a magnetic field is -

$τ=MB\sin θ$

M, Magnetic Dipole Moment = $10^{-5}Am^2$

B, Magnetic field = $3×10^{-4}T$

$θ=30°$

$∴τ=(10^{-5})(3×10^{-4})(\sin(30°))$

$=1.5×10^{-9} Nm$