#solenoid

Lenz's Law, Eddy Currents and Applications

A solenoid of 800 turns per meter and area 0.013 m² has a current change from 9 A to 6 A in 0.2 s. What is the self-indu

**Eddy currents** are circulating currents induced in bulk conductor by changing flux, oppose motion, cause damping, heating, energy loss, minimized by laminating core into thin sheets insulated, increasing resistance, reducing eddy current magnitude, used in induction heating and braking. L = μ₀ n² A l , assume l = 1 m . L = 4π × 10⁻⁷ × (800)² × 0.013 × 1 = 0.01045 H . ε = L (Δ I/Δ t) = 0.01045 × (6 - 9/0.2) = 0.01045 × (-15) = 0.15675 V ≈ 0.157 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt =

Ref: NCERT > Physics Book > Electromagnetic Induction > Lenz's Law, Eddy Currents and Applications

View Details

Lenz's Law, Eddy Currents and Applications

A solenoid of 450 turns and length 0.9 m induces an emf of 1.8 V in a nearby coil when its current changes from 2 A to 5

**Lenz's law** induced current direction opposes change in flux causing it, e = -N dΦ/dt negative sign, conservation of energy. Magnet moved towards coil south pole first, approaching south pole increasing flux into coil with south polarity, coil face nearest magnet becomes south pole to repel, opposing approach, so face becomes south pole, repelling magnet. ε = M (Δ I/Δ t) . Δ I = 5 - 2 = 3 A , Δ t = 0.3 s . M = (ε/(Δ I/Δ t)) = (1.8/(3/0.3)) = (1.8/10) = 0.18 H . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt

Ref: NCERT > Physics Book > Electromagnetic Induction > Lenz's Law, Eddy Currents and Applications

View Details

Lenz's Law, Eddy Currents and Applications

A solenoid with mutual inductance 0.25 H has a current change of 4 A/s in the primary coil. What is the induced emf in t

**Lenz's law** induced current direction opposes change in flux causing it, e = -N dΦ/dt negative sign, conservation of energy. Magnet moved towards coil south pole first, approaching south pole increasing flux into coil with south polarity, coil face nearest magnet becomes south pole to repel, opposing approach, so face becomes south pole, repelling magnet. ε = M (dI/dt) = 0.25 × 4 = 1 V . 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

Ref: NCERT > Physics Book > Electromagnetic Induction > Lenz's Law, Eddy Currents and Applications

View Details

Energy Stored in Inductor and Magnetic Energy

A solenoid of 700 turns per meter and area 0.015 m² has a current drop from 8 A to 5 A in 0.3 s. What is the self-induce

**Magnetic energy density** u = B²/(2μ₀), for B=0.5 T, u=0.25/(2×4π×10⁻⁷)=0.25/(2.513×10⁻⁶)=99471 J/m³, large, but volume small, total energy moderate. Inductor stores energy in field, released when current interrupted causing spark. L = μ₀ n² A l , assume l = 1 m . L = 4π × 10⁻⁷ × (700)² × 0.015 × 1 = 0.00923 H . ε = L (Δ I/Δ t) = 0.00923 × (5 - 8/0.3) = 0.00923 × (-10) = 0.0923 V ≈ 0.092 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L

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

View Details

Energy Stored in Inductor and Magnetic Energy

A solenoid of 1000 turns/m and area 0.01 m² has \( \mu_r = 2 \). What is its self-inductance? (\( \mu_0 = 4\pi \times 10

**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 = μ_r μ₀ n² A l , assume l = 1 m . L = 2 × 4π × 10⁻⁷ × (1000)² × 0.01 × 1 = 0.02513 H ≈ 0.025 H . 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,

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

View Details

Energy Stored in Inductor and Magnetic Energy

A solenoid of 700 turns/m and area 0.008 m² has \( \mu_r = 1 \). What is its self-inductance? (\( \mu_0 = 4\pi \times 10

**Magnetic energy density** u = B²/(2μ₀), for B=0.5 T, u=0.25/(2×4π×10⁻⁷)=0.25/(2.513×10⁻⁶)=99471 J/m³, large, but volume small, total energy moderate. Inductor stores energy in field, released when current interrupted causing spark. L = μ_r μ₀ n² A l , assume l = 1 m . L = 1 × 4π × 10⁻⁷ × (700)² × 0.008 × 1 = 0.00492 H ≈ 0.005 H . 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 0.005 H 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 solenoid of 550 turns per meter and area 0.016 m² has a current drop from 7 A to 4 A in 0.3 s. What is the self-induce

**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. L = μ₀ n² A l , assume l = 1 m . L = 4π × 10⁻⁷ × (550)² × 0.016 × 1 = 0.00608 H . ε = L (Δ I/Δ t) = 0.00608 × (4 - 7/0.3) = 0.00608 × (-10) = 0.0608 V ≈ 0.061 V . Using

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

View Details

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

View Details

Mutual Induction and Mutual Inductance

A solenoid of 400 turns and length 0.5 m induces an emf of 0.8 V in a nearby coil when its current changes from 2 A to 4

**Solenoid carries steady current** second coil experiences emf only when current in solenoid changes because dΦ/dt ≠0 only when I changes, steady current gives constant flux, no induction, illustrating Faraday's law requires changing flux, not static field. ε = M (Δ I/Δ t) . Δ I = 4 - 2 = 2 A , Δ t = 0.2 s . M = (ε/(Δ I/Δ t)) = (0.8/(2/0.2)) = (0.8/10) = 0.08 H . 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

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

View Details

Mutual Induction and Mutual Inductance

A solenoid with mutual inductance 0.2 H has a current change of 5 A/s in the primary coil. What is the induced emf in th

**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. ε = M (dI/dt) = 0.2 × 5 = 1 V . 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 1 V follows, reflecting Faraday's law and Lenz's opposition.

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

View Details

Mutual Induction and Mutual Inductance

A solenoid of 350 turns and length 0.7 m induces an emf of 1.2 V in a nearby coil when its current changes from 1 A to 4

**Mutual inductance** M = N₂ Φ₂₁/I₁, flux linking coil 2 due to current in coil 1, emf in 2 e₂ = -M dI₁/dt, M = μ₀ N₁ N₂ A / l for coaxial solenoids, unit henry (H), same as self-inductance. For solenoid 500 turns length 1 m induces 1.5 V when current 0 to 3 A in 0.2 s, dI/dt=15 A/s, M = e/(dI/dt)=1.5/15=0.1 H. ε = M (Δ I/Δ t) . Δ I = 4 - 1 = 3 A , Δ t = 0.3 s . M = (ε/(Δ I/Δ t)) = (1.2/(3/0.3)) = (1.2/10) = 0.12 H . Using Φ = B

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

View Details

Mutual Induction and Mutual Inductance

A solenoid of 600 turns and length 0.8 m induces an emf of 2 V in a nearby coil when its current changes from 1 A to 4 A

**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. ε = M (Δ I/Δ t) . Δ I = 4 - 1 = 3 A , Δ t = 0.2 s . M = (ε/(Δ I/Δ t)) = (2/(3/0.2)) = (2/15) = 0.133 H ≈ 0.13 H . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L =

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

View Details