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 $0.4 × 10^{-4}T$)
Answer & explanation
Correct answer: option 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 = $0.4×10^{-4}T$
$τ=MB\sin 90°$
$⇒M=\frac{τ}{B}=\frac{3×10^{-4}}{0.4×10^{-4}}=7.5Am^2$