A magnetic dipole of moment \( 0.3 \, \text{A m}^2 \) is in a uniform field of \( 0.7 \, \text{T} \) at \( 30^\circ \).
**Axial versus equatorial** field comparison shows B_axial = 2 B_eq for same r. Using μ₀ = 4π×10⁻⁷ T·m/A, calculation involves r³ = (0.3)³ = 0.027 m³, so B = 10⁻⁷·m/r³ yields moment estimation. Torque is tau = m B sinθ . Given: m = 0.3 A m² , B = 0.7 T , θ = 30° , sin 30° = 0.5 . Substitute: tau = 0.3 × 0.7 × 0.5 = 0.105 N m . Substituting values gives 0.105 N m, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque relation τ = m B sinθ.
Ref: NCERT > Physics Book > Magnetism and Matter > Magnetic Field Due to Bar Magnet - Axial and Equatorial