Skip to content
Login Register

#energy stored

3 public questions tagged with this topic.

Energy Stored in Inductor and Magnetic Energy

A coil with \( L = 0.3 \, \text{H} \) has its current increased from 0 to 6 A in 0.6 s. What is the energy stored?

**Energy stored in inductor** U =½ L I² (J), L inductance (H), I current (A), resides in magnetic field, energy density u = B²/(2μ₀) (J/m³), B field (T), μ₀=4π×10⁻⁷ H/m. For solenoid B=μ₀ n I, volume V= A l, U = (B²/2μ₀) V =½ (μ₀ n² A l) I² =½ L I², consistent. W = (1/2) L I² = (1/2) × 0.3 × (6)² = 0.15 × 36 = 5.4 J . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L = μ₀ N²A/l, M = e/(dI/dt) and U

Ref: NCERT > Physics Book > Electromagnetic Induction > Energy Stored in Inductor and Magnetic Energy

View Details
Energy Stored in Inductor and Magnetic Energy

A coil with \( L = 0.1 \, \text{H} \) has its current increased from 0 to 4 A in 0.2 s. What is the energy stored?

**Energy in inductor** cannot change instantaneously because that would require infinite power, current through inductor continuous, voltage may jump, principle used in chokes, inductive kick, back emf, explaining why inductor opposes change in current. Energy: W = (1/2) L I² . W = (1/2) × 0.1 × (4)² = 0.05 × 16 = 0.8 J . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L = μ₀ N²A/l, M = e/(dI/dt) and U = ½ L I², result 0.8 J follows, reflecting Faraday's law and Lenz's opposition.

Ref: NCERT > Physics Book > Electromagnetic Induction > Energy Stored in Inductor and Magnetic Energy

View Details
Energy Stored in Inductor and Magnetic Energy

A coil with \( L = 0.5 \, \text{H} \) has its current increased from 0 to 6 A in 0.6 s. What is the energy stored?

**Energy stored in inductor** U =½ L I² (J), L inductance (H), I current (A), resides in magnetic field, energy density u = B²/(2μ₀) (J/m³), B field (T), μ₀=4π×10⁻⁷ H/m. For solenoid B=μ₀ n I, volume V= A l, U = (B²/2μ₀) V =½ (μ₀ n² A l) I² =½ L I², consistent. W = (1/2) L I² = (1/2) × 0.5 × (6)² = 0.25 × 36 = 9 J . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L = μ₀ N²A/l, M = e/(dI/dt) and U

Ref: NCERT > Physics Book > Electromagnetic Induction > Energy Stored in Inductor and Magnetic Energy

View Details