Skip to content
Login Register

#current decay

3 public questions tagged with this topic.

Lenz's Law, Eddy Currents and Applications

A coil of self-inductance 0.6 H has its current decreased from 4 A to 1 A in 0.15 s. What is the magnitude of the induce

**Solenoid second coil** experiences emf only when current in solenoid changes because flux linkage changes only then, steady current gives constant Φ, dΦ/dt=0, no emf, when current changes, dΦ/dt ≠0, emf induced, illustrating Faraday's law requirement of changing flux. ε = L (Δ I/Δ t) . Δ I = 1 - 4 = -3 A , Δ t = 0.15 s . ε = 0.6 × (-3/0.15) = 0.6 × (-20) = -12 V , magnitude = 12 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 > Lenz's Law, Eddy Currents and Applications

View Details
Energy Stored in Inductor and Magnetic Energy

A coil of self-inductance 1.0 H has its current decreased from 5 A to 2 A in 0.15 s. What is the magnitude of the 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. ε = L (Δ I/Δ t) . Δ I = 2 - 5 = -3 A , Δ t = 0.15 s . ε = 1.0 × (-3/0.15) = 1.0 × (-20) = -20 V , magnitude = 20 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

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

View Details
Self-Induction and Self-Inductance

A solenoid of 500 turns per meter and area 0.02 m² has a current drop from 6 A to 3 A in 0.25 s. What is the self-induce

**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 × (500)² × 0.02 × 1 = 0.00628 H . ε = L (Δ I/Δ t) = 0.00628 × (3 - 6/0.25) = 0.00628 × (-12) = 0.07536 V ≈ 0.075 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

View Details