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:
Answer & explanation
Correct answer: option 1
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 = $3×10^{-4}T$
$θ=30°$
$∴τ=(10^{-5})(3×10^{-4})(\sin(30°))$
$=1.5×10^{-9} Nm$