Skip to content
Login Register

#inductance calculation

2 public questions tagged with this topic.

Induced EMF Due to Change in Magnetic Field

A solenoid of 400 turns per meter and area 0.015 m² has a current drop from 5 A to 2 A in 0.3 s. What is the self-induce

**Field decreasing to zero** induces emf trying to maintain field, current direction such that its field adds to original. For 150 turns area 0.06 m² B 0.14 T to zero in 0.3 s, e=150×0.06×0.14/0.3=4.2 V, as earlier, showing linear dependence on N, A, ΔB/Δt. L = μ₀ n² A l , assume l = 1 m . L = 4π × 10⁻⁷ × (400)² × 0.015 × 1 = 0.003016 H . ε = L (Δ I/Δ t) = 0.003016 × (2 - 5/0.3) = 0.003016 × (-10) = 0.03016 V ≈ 0.03 V . Using Φ = B A cosθ, e = -N dΦ/dt

Ref: NCERT > Physics Book > Electromagnetic Induction > Induced EMF Due to Change in Magnetic Field

View Details
Self-Induction and Self-Inductance

A solenoid of 950 turns/m and area 0.012 m² has \( \mu_r = 1 \). What is its self-inductance? (\( \mu_0 = 4\pi \times 10

**Self-inductance of solenoid** L = μ₀ N² A / l, N total turns, A cross-section, l length, for N=650 turns per meter means n=650 m⁻¹, if length 1 m N=650, A=0.014, L=4π×10⁻⁷×650²×0.014/1=0.00743 H, self-induced emf magnitude L |dI/dt|, dI/dt=12 A/s, e=0.089 V, opposes change. L = μ_r μ₀ n² A l , assume l = 1 m . L = 1 × 4π × 10⁻⁷ × (950)² × 0.012 × 1 = 0.0136 H ≈ 0.014 H . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L = μ₀

Ref: NCERT > Physics Book > Electromagnetic Induction > Self-Induction and Self-Inductance

View Details