A bar magnet when suspended horizontally and perpendicular to earth's field experiences a torque of $3× 10^{-4} Nm$. The Magnetic moment of the magnet would be: (take horizontal component of Earth's magnetic field at the place $0.4 × 10^{-4}T$)

$7.5 JT^{-1}$

The correct answer is Option (1) → $7.5 JT^{-1}$

Torque ($τ$) experienced by a Magnetic dipole (such as bar magnet) in a Magnetic field is -

$τ=MB\sin θ$

where,

M = Magnetic moment

$τ$ = Torque experienced = $3×10^{-4}Nm$

B = Horizontal component of earth magnetic field = $0.4×10^{-4}T$

$τ=MB\sin 90°$

$⇒M=\frac{τ}{B}=\frac{3×10^{-4}}{0.4×10^{-4}}=7.5Am^2$