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#primary coil

6 public questions tagged with this topic.

Mutual Induction and Mutual Inductance

A solenoid with mutual inductance 0.22 H has a current change of 7 A/s in the primary coil. What is the induced emf in t

**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) = 0.22 × 7 = 1.54 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 1.54 V follows, reflecting Faraday's law and Lenz's opposition.

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

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Mutual Induction and Mutual Inductance

A solenoid with mutual inductance 0.25 H has a current change of 6 A/s in the primary coil. What is the induced emf in t

**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 (dI/dt) = 0.25 × 6 = 1.5 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,

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

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Mutual Induction and Mutual Inductance

A solenoid with mutual inductance 0.15 H has a current change of 8 A/s in the primary coil. What is the induced emf in t

**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 (dI/dt) = 0.15 × 8 = 1.2 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 1.2 V follows, reflecting Faraday's law and Lenz's opposition.

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

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Magnetic Flux and Faraday's Laws of Induction

In a setup with two coils, if the primary coil’s current is constant DC, what will be the induced emf in the secondary c

**Faraday's first law** emf induced when flux linking coil changes, second law magnitude proportional to rate of change, e = -dΦ/dt, for N turns e = -N dΦ/dt, flux Φ = B A cosθ, change can be due to B change, A change, or θ change, all produce emf. A constant DC current produces a steady magnetic field, resulting in no change in magnetic flux through the secondary coil, so the induced emf is zero. 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 =

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

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AC Through Resistor - Phasor and Power

Why does an ideal transformer maintain constant power across its primary and secondary coils?

**AC through resistor** voltage and current in phase, φ=0°, I = V/R instantaneously, I(t)=I_peak sin ωt, V(t)=V_peak sin ωt, phasor diagram V and I same direction, power instantaneous P = V I = V_peak I_peak sin² ωt, average P_avg = V_rms I_rms = V_rms²/R, always positive, energy dissipated as heat. In an ideal transformer, there are no losses (e.g., resistance, flux leakage). Energy conservation dictates that input power ( V_p I_p ) equals output power ( V_s I_s ), so the power remains constant despite changes in voltage and current. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_

Ref: NCERT > Physics Book > Alternating Currents > AC Through Resistor - Phasor and Power

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Transformer, AC Generator and LC Oscillations

In a step-up transformer, how does the current in the secondary coil compare to the primary coil, assuming ideal conditi

**Transformer** works on mutual induction, V_s/V_p = N_s/N_p, I_s/I_p = N_p/N_s for ideal (power conserved V_p I_p = V_s I_s), step-up N_s>N_p V_s>V_p I_s V_p and N_s > N_p ), power is conserved ( V_p I_p = V_s I_s ). Since the secondary voltage is higher, the secondary current must be lower than the primary current ( I_s = I_p × (N_p/N_s) ), where (N_p/N_s) < 1 . Applying X_L = ωL, X_C = 1/ωC, Z =

Ref: NCERT > Physics Book > Alternating Currents > Transformer, AC Generator and LC Oscillations

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