#current increase

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

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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

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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

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Solenoid, Toroid and Ampere's Law

What happens to the magnetic field inside a solenoid when the current through it is doubled?

**Field outside long solenoid** considered negligible because magnetic lines are concentrated inside and return path spreads over large area outside, making B_out ≈ 0. This justifies using solenoid for uniform field experiments, with n = 950 m⁻¹, I = 1.4 A giving B = 4π×10⁻⁷×950×1.4 = 1.67×10⁻³ T. The magnetic field inside a solenoid is proportional to the current, given by B = μ₀ n I . If the current I is doubled, the magnetic field B also doubles. Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ =

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Solenoid, Toroid and Ampere's Law

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