CUET Preparation Today
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 $0.4 × 10^{-4}T$) |
$7.5 JT^{-1}$ $5.7 JT^{-1}$ $0.75 JT^{-1}$ $0.57 JT^{-1}$ |
$7.5 JT^{-1}$ |
The correct answer is Option (1) → $7.5 JT^{-1}$ |
