A solenoid of 750 turns/m and area 0.02 m² has \( \mu_r = 1 \). What is its self-inductance? (\( \mu_0 = 4\pi \times 10^
**Self-induction** emf induced in coil due to change in its own current, e = -L dI/dt, L self-inductance (H), L = μ₀ N² A / l for solenoid, N turns, A area (m²), l length (m), μ₀=4π×10⁻⁷ H/m. For solenoid 650 turns/m means n=650, A=0.014 m², L = μ₀ n² A l? Actually per unit length? For length l, N=n l, L= μ₀ n² A l, if l=1 m, L=4π×10⁻⁷×650²×0.014=7.43×10⁻³ H, dI/dt=(3-6)/0.25=-12 A/s, e= -L×(-12)=0.089 V. L = μ_r μ₀ n² A l , assume l = 1 m . L = 1 × 4π × 10⁻⁷ × (750)² × 0.02 × 1 = 0.01414
Ref: NCERT > Physics Book > Electromagnetic Induction > Self-Induction and Self-Inductance