A bar magnet produces a field of \( 5 \times 10^{-6} \, \text{T} \) at \( 0.5 \, \text{m} \) on its equatorial line. Wha
**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) (m/r³) , so m = (B r³/(μ₀/4π)) . Given: B = 5 × 10⁻⁶ T , r = 0.5 m , (μ₀/4π) = 10⁻⁷ . m = (5 × 10⁻⁶ × (0.5)³/10⁻⁷) = (5 × 10⁻⁶ × 0.125/10⁻⁷) = 6.25 A m² . Substituting values gives 6.25 A m², which matches expected
Ref: NCERT > Physics Book > Magnetism and Matter > Solenoid with Magnetic Core and Magnetic Properties
