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#changing current

6 public questions tagged with this topic.

Energy Stored in Inductor and Magnetic Energy

A solenoid of 500 turns per meter and area 0.01 m² has a current change from 3 A to 0 A in 0.5 s. What is the self-induc

**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. L = μ₀ n² A l , assume l = 1 m . L = 4π × 10⁻⁷ × (500)² × 0.01 × 1 = 0.00314 H . ε = L (Δ I/Δ t) = 0.00314 × (0 - 3/0.5) = 0.01884 V ≈ 0.019 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt,

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

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Mutual Induction and Mutual Inductance

A coil is wound tightly around a core material. If the current through it changes rapidly, the induced emf opposing this

**Mutual inductance calculation** M = e₂/(dI₁/dt), for 200 turns length 0.5 m nearby coil e=0.5 V dI=2 A dt=0.2 s dI/dt=10 A/s, M=0.5/10=0.05 H, depends on geometry, orientation, number of turns, area, separation, coupling coefficient k = M/√(L₁ L₂) ≤1. This is self-induction, where a changing current in a coil induces an emf that opposes the change, proportional to the coil’s self-inductance. 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 Self-induction follows, reflecting Faraday's law and Lenz's opposition.

Ref: NCERT > Physics Book > Electromagnetic Induction > Mutual Induction and Mutual Inductance

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Self-Induction and Self-Inductance

A solenoid of 600 turns per meter and area 0.012 m² has a current change from 4 A to 1 A in 0.2 s. What is the self-indu

**Energy stored in inductor** U =½ L I², L inductance, I current, energy in magnetic field, density u = B²/(2μ₀), B=μ₀ n I inside solenoid, U = (B²/2μ₀)×volume, illustrating equivalence of circuit and field energy. L = μ₀ n² A l , assume l = 1 m . L = 4π × 10⁻⁷ × (600)² × 0.012 × 1 = 0.00543 H . ε = L (Δ I/Δ t) = 0.00543 × (1 - 4/0.2) = 0.00543 × (-15) = 0.08145 V ≈ 0.081 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v =

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

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Self-Induction and Self-Inductance

A solenoid of 850 turns per meter and area 0.017 m² has a current drop from 10 A to 7 A in 0.25 s. What is the self-indu

**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 = μ₀ n² A l , assume l = 1 m . L = 4π × 10⁻⁷ × (850)² × 0.017 × 1 = 0.01541 H . ε

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

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Self-Induction and Self-Inductance

A solenoid of 650 turns per meter and area 0.014 m² has a current change from 6 A to 3 A in 0.25 s. What is the self-ind

**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 = μ₀ n² A l , assume l = 1 m . L = 4π × 10⁻⁷ × (650)² × 0.014 × 1 = 0.00744 H . ε

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

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Magnetic Flux and Faraday's Laws of Induction

A solenoid carries a steady current. A second coil placed nearby experiences an emf only when what happens to the curren

**Flux change example** coil 150 turns area 0.06 m² B 0.14 T drops to zero in 0.3 s, ΔΦ = B A =0.14×0.06=0.0084 Wb per turn, ΔΦ/Δt=0.028 Wb/s, e=150×0.028=4.2 V, illustrating calculation from B and area. Emf is induced in the second coil only when the magnetic flux changes, which occurs when the current in the solenoid changes, not when it remains steady. 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 The current increases steadily

Ref: NCERT > Physics Book > Electromagnetic Induction > Magnetic Flux and Faraday's Laws of Induction

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