A bar magnet with \( m = 2.8 \, \text{A m}^2 \) is at \( 0.4 \, \text{m} \) along its axis. What is \( B \)? (Take \( \m
**Magnetic field inside solenoid** with core B = μ₀ μ_r n I is uniform, direction along axis given by right-hand grip rule. For n = 2000 m⁻¹, μ_r = 400, B = 1.2 T, I = 1.2/(4π×10⁻⁷×400×2000) ≈ 1.19 A, showing modest current produces tesla-level field with high μ_r core. B = (μ₀/4π) (2m/r³) . Given: m = 2.8 A m² , r = 0.4 m , (μ₀/4π) = 10⁻⁷ . B = 10⁻⁷ × (2 × 2.8/(0.4)³) = 10⁻⁷ × (5.6/0.064) = 8.75 × 10⁻⁶ T . Substituting values gives 8.75 × 10⁻⁶ T, which matches expected magnitude for this magnetic configuration, confirming dipole
Ref: NCERT > Physics Book > Magnetism and Matter > Solenoid with Magnetic Core and Magnetic Properties
