← Physics (312) · Class 12

L-19: Electromagnetic Induction and Alternating Current

Physics — Class 12 · NIOS Code 312 · Module 5 · Source: 312_Physics_Eng_Lesson19.pdf

Electromagnetic Induction and Alternating Current

Generators and transformers make electricity available nationwide. This lesson covers electromagnetic induction, inductance, alternating current circuits, generators, and transformers.

NIOS objectives: Faraday's and Lenz's laws; eddy currents; self and mutual induction; AC/DC generators; R, L, C and LCR circuits; transformers and efficiency.

19.1 Electromagnetic Induction

A steady current produces a steady field, but changing magnetic flux through a coil induces emf and current. Observed when: switch opened/closed in neighbouring circuit; magnet moved relative to coil; coil moved relative to magnet.

Fig 19.1–19.2 — Electromagnetic Induction coil 1 coil 2 + G iron ring — flux linked (a) changing current in coil 1 induces momentary current in coil 2 N ring (b) moving magnet → induced current (direction reverses if magnet retreats)
Fig 19.1–19.2 — Induced emf only when magnetic flux through a coil changes

19.1.1 Faraday's Law

Magnetic flux: φB = ∫B·dS (weber, Wb = T·m²). 1 V = 1 Wb/s.

ε = − dφB / dt
Induced emf ∝ rate of change of flux linked with loop
For N closely wound turns: εr = −N dφB/dt
Negative sign = Lenz's law (opposition)

19.1.2 Lenz's Law

Induced current direction opposes the change causing it — consequence of energy conservation. Approaching magnet: induced field repels magnet; work done on magnet appears as electrical energy in the ring.

Fig 19.5 — Lenz's Law N magnet → conducting ring induced B opposes induced current creates field that repels approaching N-pole
Fig 19.5 — Induced magnetic field opposes the flux change (Lenz's law)

19.1.3 Eddy Currents

Induced closed-loop currents in bulk conductors (sheet/plate) when flux changes — look like eddies. Large heating → undesirable; reduced by lamination (insulated strips). Applications: induction furnaces, electric brakes on trains.

19.2 Inductance

19.2.1 Self-Inductance

φ = L I  |  ε = −L dI/dt
L = self-inductance (henry, H = Ω·s)
Back emf opposes change in current
Current through inductor cannot change instantaneously
Solenoid: L = μ₀N²A/l

19.2.2 LR Circuits

On closing switch, current rises gradually to ε₀/R. Time constant τ = L/R. Large L → spark on switching off (back emf). Inductance acts like electrical inertia.

19.2.3 Mutual Inductance

φ₂ = M I₁  |  ε₂ = −M dI₁/dt
M = mutual inductance (henry)
Changing current in one coil induces emf in nearby coil
Basis of transformers, ignition coils, chokes

19.3 Alternating Currents and Voltages

DC: unidirectional, steady magnitude. AC: magnitude and direction change periodically.

V = Vm cos ωt  |  Vrms = Vm/√2 ≈ 0.707 Vm
ω = 2πν (angular frequency)
India: Vm ≈ 310 V → Vrms ≈ 220 V at 50 Hz
Irms = Im/√2

19.3.1 Pure Resistor

V and I in phase. I = (Vm/R) cos ωt. Average power Pav = Irms² R = Vrms Irms.

19.3.2 Pure Capacitor

XC = 1/(ωC) = 1/(2πνC)
Capacitive reactance (Ω)
Current leads voltage by 90° (π/2)
XC decreases with frequency
Average power = 0 (energy stored/released, not dissipated)

19.3.3 Pure Inductor

XL = ωL = 2πνL
Inductive reactance (Ω)
Current lags voltage by 90°
XL increases with frequency
Average power = 0 (wattless current in pure L or C)

19.3.4 Series LCR Circuit

Z = √[R² + (XL − XC)²]
Z = impedance (Ω)
tan φ = (XL − XC)/R — phase angle φ
Irms = Vrms/Z
νr = 1/(2π√LC)  |  Zmin = R at resonance
At resonance XL = XC; circuit purely resistive
Maximum current; VL and VC cancel
Used in radio/TV tuning circuits
Pav = Vrms Irms cos φ  |  cos φ = R/Z
Power factor cos φ
Pure R or resonance: cos φ = 1, max power
Pure L or C: cos φ = 0, wattless current
Fig 19.12 & 19.22 — AC Waveforms and LCR AC DC R L C series LCR — Z, φ, resonance
Fig 19.12 — AC vs DC; series LCR circuit for impedance and resonance

19.4 Power Generator

Converts mechanical energy to electrical energy via electromagnetic induction.

  • AC generator (alternator): rotating coil in magnetic field; ε = NωAB sin ωt. Slip rings + brushes → alternating output. Armature, field magnet, slip rings, brushes.
  • DC generator (dynamo): Split-ring commutator reverses connections every half turn → unidirectional (pulsating) DC. Used in bicycles, car charging.
ε(t) = N ω A B sin ωt
φ(t) = AB cos ωt; ε = −dφ/dt
Rotating rectangular coil in uniform B
Fleming's right-hand rule for induced emf direction
Fig 19.27 — AC Generator armature coil N S R₁ R₂ slip rings brushes → AC load
Fig 19.27 — AC generator: slip rings give alternating emf in external circuit

19.5 Transformer

Static device transferring AC energy between windings by mutual induction. Primary connected to source; secondary to load. Laminated iron core reduces eddy currents.

Vs/Vp = Ns/Np = k
Step-up: Ns > Np — voltage increases, current decreases
Step-down: Ns < Np — voltage decreases, current increases
Ideal: Pin = Pout; Ip/Is = Ns/Np
  • Cannot work on DC — needs changing flux for induction.
  • Losses: copper (I²R heating), eddy currents, hysteresis, flux leakage.
  • Power transmission: step up to ~330 kV (low I, less I²R loss); step down at substations to 220 V.
  • Efficiency η = (power output/power input) × 100% < 100%.
Fig 19.30–19.31 — Transformer primary Np secondary Ns AC in load laminated core · Vs/Vp = Ns/Np · step-up if Ns > Np
Fig 19.30–19.31 — Transformer couples windings magnetically, not electrically
         EM INDUCTION & AC — KEY FORMULAS
         ==================================
    Faraday/Lenz    :  ε = −dφB/dt = −N dφ/dt
    Self-inductance :  ε = −L dI/dt ;  L = μ₀N²A/l
    Mutual          :  ε₂ = −M dI₁/dt
    RMS             :  Vrms = Vm/√2
    Reactance       :  XC = 1/(ωC) ;  XL = ωL
    Impedance       :  Z = √[R²+(XL−XC)²]
    Resonance       :  νr = 1/(2π√LC)
    Power factor    :  Pav = Vrms Irms cos φ
    AC generator    :  ε = NωAB sin ωt
    Transformer     :  Vs/Vp = Ns/Np

Quick Revision

  • Induced emf when magnetic flux through a circuit changes.
  • Lenz's law: induced effects oppose the cause.
  • Eddy currents → lamination; self/mutual inductance in henry.
  • AC: R in phase; C leads 90°; L lags 90°; LCR resonance at 1/(2π√LC).
  • AC gen: slip rings; DC gen: commutator.
  • Transformer: AC only; step-up/down by turns ratio; high-V transmission saves power loss.
20 cards · click any card to flip
Electromagnetic induction
Emf induced when magnetic flux linked with a conductor/coil changes with time. Discovered by Faraday; basis of generators and transformers.
Faraday's law
ε = −dφB/dt. For N turns: ε = −N dφB/dt. 1 V = 1 Wb/s. Magnitude ∝ rate of flux change.
Lenz's law
Induced current opposes the flux change that caused it. Follows conservation of energy. Negative sign in ε = −dφ/dt.
Magnetic flux
φB = ∫B·dS (weber, Wb = T·m²). Number of field lines through a surface ∝ flux.
Eddy currents
Circulating induced currents in bulk metal. Cause heating — reduced by lamination. Used in induction furnaces and electric brakes.
Self-inductance
φ = LI; ε = −L dI/dt. Unit: henry (H). Back emf opposes current change. Solenoid L = μ₀N²A/l.
Mutual inductance
φ₂ = MI₁; ε₂ = −M dI₁/dt. Changing current in one coil induces emf in another. Basis of transformers.
LR circuit time constant
τ = L/R. Current rises gradually to ε/R. Large inductance → back emf spark on switch-off. Electrical inertia.
RMS values
Vrms = Vm/√2 ≈ 0.707 Vm. India: Vm ≈ 310 V → Vrms ≈ 220 V. Used for power calculations in AC.
Pure resistor (AC)
V and I in phase. Pav = Irms² R. Same heating as DC of value Irms through same R.
Capacitive reactance
XC = 1/(ωC). Current leads V by 90°. XC ↓ as frequency ↑. Average power = 0.
Inductive reactance
XL = ωL. Current lags V by 90°. XL ↑ as frequency ↑. Average power = 0 (wattless).
LCR impedance
Z = √[R² + (XL − XC)²]. tan φ = (XL − XC)/R. Irms = Vrms/Z.
Resonance (LCR)
νr = 1/(2π√LC) when XL = XC. Zmin = R; max current; used in radio tuning. cos φ = 1.
AC power factor
Pav = Vrms Irms cos φ; cos φ = R/Z. Pure L or C: cos φ = 0 (wattless). Resistive/resonant: cos φ = 1.
AC generator
ε = NωAB sin ωt. Slip rings + brushes → alternating output. Fleming's right-hand rule for emf direction.
DC generator (dynamo)
Split-ring commutator converts induced AC in armature to pulsating DC in external circuit. Bicycle/car dynamo.
Transformer turns ratio
Vs/Vp = Ns/Np. Step-up: Ns > Np. Ip/Is = Ns/Np (power conserved ideally). Works only on AC.
Transformer losses
Copper loss (I²R), eddy currents (laminated core), hysteresis, flux leakage. Efficiency < 100%.
High-voltage transmission
P = VI; loss = I²R = P²R/V². High V → low I → less transmission loss. Step-up at plant, step-down at substation.

Q1. Faraday's law of electromagnetic induction states that induced emf is proportional to:

Q2. Lenz's law is a consequence of the law of conservation of:

Q3. The SI unit of self-inductance is:

Q4. In India, household AC supply is approximately:

Q5. In a purely capacitive AC circuit, the current:

Q6. At resonance in a series LCR circuit:

Q7. An AC generator uses ______ to deliver alternating current to the external circuit:

Q8. A step-up transformer has:

Q9. A transformer cannot work on steady DC because:

Q10. For a transformer with 100 primary turns and 500 secondary turns, if primary voltage is 120 V, secondary voltage is:

ε = −dφ_B/dt
ε = −N dφ_B/dt
φ = L I  |  ε = −L dI/dt
L = μ₀ N² A / l
τ = L / R
φ₂ = M I₁  |  ε₂ = −M dI₁/dt
V_rms = V_m/√2  |  I_rms = I_m/√2
X_C = 1/(ωC) = 1/(2πνC)
X_L = ωL = 2πνL
Z = √[R² + (X_L − X_C)²]
ν_r = 1/(2π√LC)
P_av = V_rms I_rms cos φ
ε(t) = N ω A B sin ωt
V_s/V_p = N_s/N_p = k
I_p/I_s = N_s/N_p

1. Formulas & Definitions

Full Ch 19 study guide — Faraday/Lenz, inductance, AC circuits, generators, and transformers.

ε = −dφ_B/dt

Definition: Faraday's law — induced emf equals negative rate of change of magnetic flux.

Derivation

Flux φ_B = ∫B·dS (Wb); emf induced only when φ_B changes with time.

Variables

ε (V) · φ_B (Wb) · 1 V = 1 Wb/s

Why it works

Foundation of generators, transformers, and all electromagnetic induction.

Historical context

Faraday (1831); negative sign encodes Lenz's law.

Deep understanding

Steady B or steady φ → no emf. Moving magnet, changing current, or rotating coil all work.

2. Diagrams & Visuals

ε = −dφ/dt

Color-coded visual · step-by-step breakdown below

  1. Find flux φ_B through loop
  2. Compute rate of change dφ_B/dt
  3. ε = −dφ_B/dt
  4. Apply Lenz's law for direction

3. Solved Examples

Basic

Q: φ changes 0.02 Wb in 0.01 s.

Solution: ε=−2 V

Answer: 2 V magnitude

Intermediate

Q: Flux constant?

Solution: ε=0

Answer: No induction

Advanced

Q: Double rate of change?

Solution: ε doubles

Answer: 2× emf

Exam

Q: Faraday's law?

Solution: ε=−dφ/dt

Answer: Sec 19.1.1

ε = −N dφ_B/dt

Definition: Induced emf in coil of N closely wound turns.

Derivation

Each turn contributes; total emf is N times single-turn emf.

Variables

N = number of turns

Why it works

More turns → larger induced voltage in generators and transformers.

Historical context

Used in AC generator ε = NωAB sin ωt derivation.

Deep understanding

Flux φ_B is flux per turn (same through all turns if tightly wound).

2. Diagrams & Visuals

ε = −N dφ/dt

Color-coded visual · step-by-step breakdown below

  1. Flux per turn φ_B
  2. Multiply by N turns
  3. ε = −N dφ_B/dt

3. Solved Examples

Basic

Q: 1 turn: ε=2 V, N=500.

Solution: ε=1000 V

Answer: 1000 V

Intermediate

Q: Halve turns?

Solution: ε halves

Answer: N factor

Advanced

Q: vs single turn?

Solution: N× larger

Answer: Turns multiply

Exam

Q: Coil with N turns?

Solution: −N dφ/dt

Answer: Sec 19.1.1

φ = L I  |  ε = −L dI/dt

Definition: Self-inductance — flux linked with coil proportional to current.

Derivation

Changing current induces back emf opposing the change.

Variables

L (henry, H = Ω·s) · φ (Wb)

Why it works

Electrical inertia; prevents instantaneous current change; causes spark on switch-off.

Historical context

Back emf ε = −L dI/dt always opposes dI/dt.

Deep understanding

Current through inductor cannot jump instantly; energy stored in magnetic field.

2. Diagrams & Visuals

ε = −L dI/dt

Color-coded visual · step-by-step breakdown below

  1. Define flux φ = LI
  2. Differentiate: dφ/dt = L dI/dt
  3. Induced ε = −L dI/dt
  4. Opposes current change

3. Solved Examples

Basic

Q: L=2 H, dI/dt=5 A/s.

Solution: ε=−10 V

Answer: 10 V back emf

Intermediate

Q: L=0.5 H, dI/dt=2 A/s.

Solution: ε=−1 V

Answer: 1 V

Advanced

Q: Current constant?

Solution: dI/dt=0, ε=0

Answer: Steady DC

Exam

Q: Self-inductance back emf?

Solution: −L dI/dt

Answer: Sec 19.2.1

L = μ₀ N² A / l

Definition: Self-inductance of long solenoid.

Derivation

B = μ₀nI inside; flux φ = NBA = μ₀N²IA/l.

Variables

A = cross-section (m²) · l = length (m) · N = turns

Why it works

Design inductors, chokes, and estimate L for LR circuits.

Historical context

L ∝ N² — doubling turns quadruples inductance.

Deep understanding

Valid for long solenoid (l ≫ diameter); iron core increases L via μᵣ.

2. Diagrams & Visuals

L = μ₀N²A/l

Color-coded visual · step-by-step breakdown below

  1. Count turns N, area A, length l
  2. L = μ₀N²A/l
  3. Unit: henry

3. Solved Examples

Basic

Q: N=1000, A=10⁻⁴, l=0.5 m.

Solution: L≈0.25 mH

Answer: ~0.25 mH

Intermediate

Q: Double N?

Solution: L quadruples

Answer: N² dependence

Advanced

Q: Longer solenoid?

Solution: L decreases

Answer: L ∝ 1/l

Exam

Q: Solenoid inductance?

Solution: μ₀N²A/l

Answer: Sec 19.2.1

τ = L / R

Definition: Time constant of LR circuit — time for current to reach ~63% of final value.

Derivation

Exponential rise I(t) = (ε₀/R)(1−e^(−t/τ)); τ = L/R.

Variables

τ (s) · final current I = ε₀/R

Why it works

Predicts how fast current builds/decays; large L → slow response, spark on opening.

Historical context

Inductance acts like electrical inertia in switching circuits.

Deep understanding

After 5τ, current within ~1% of steady value.

2. Diagrams & Visuals

τ = L/R

Color-coded visual · step-by-step breakdown below

  1. Identify L and R
  2. τ = L/R
  3. Current rises over several τ
  4. Large L → back emf spark

3. Solved Examples

Basic

Q: L=4 H, R=2 Ω.

Solution: τ=2 s

Answer: 2 s

Intermediate

Q: Double L?

Solution: τ doubles

Answer: 2τ

Advanced

Q: τ very small?

Solution: Fast switching

Answer: Small L/R

Exam

Q: LR time constant?

Solution: L/R

Answer: Sec 19.2.2

φ₂ = M I₁  |  ε₂ = −M dI₁/dt

Definition: Mutual inductance — flux in coil 2 due to current in coil 1.

Derivation

Changing I₁ induces emf in nearby coil 2 without electrical contact.

Variables

M (henry) · φ₂ (Wb) · I₁ (A)

Why it works

Basis of transformers, ignition coils, and coupled circuits.

Historical context

Energy transferred magnetically, not through wires.

Deep understanding

M depends on geometry, separation, and core material; M₁₂ = M₂₁.

2. Diagrams & Visuals

ε₂ = −M dI₁/dt

Color-coded visual · step-by-step breakdown below

  1. Current I₁ in primary coil
  2. Flux φ₂ = MI₁ linked with secondary
  3. ε₂ = −M dI₁/dt on secondary
  4. Requires changing I₁

3. Solved Examples

Basic

Q: M=0.5 H, dI₁/dt=4 A/s.

Solution: ε₂=−2 V

Answer: 2 V

Intermediate

Q: Steady I₁?

Solution: ε₂=0

Answer: DC steady: no induction

Advanced

Q: Transformer uses?

Solution: Mutual induction

Answer: AC only

Exam

Q: Mutual inductance?

Solution: ε₂=−M dI₁/dt

Answer: Sec 19.2.3

V_rms = V_m/√2  |  I_rms = I_m/√2

Definition: RMS (root mean square) values of AC voltage and current.

Derivation

Equivalent DC value giving same heating in a resistor.

Variables

V_m, I_m = peak values · √2 ≈ 1.414

Why it works

Meters and power ratings use RMS; India mains V_rms ≈ 220 V at 50 Hz.

Historical context

V = V_m cos ωt; India V_m ≈ 310 V → V_rms ≈ 220 V.

Deep understanding

ω = 2πν; ν = 50 Hz standard in India.

2. Diagrams & Visuals

V_rms = V_m/√2

Color-coded visual · step-by-step breakdown below

  1. Identify peak V_m or I_m
  2. Divide by √2
  3. Use RMS for power calculations

3. Solved Examples

Basic

Q: V_m=310 V.

Solution: V_rms≈220 V

Answer: 220 V

Intermediate

Q: I_m=1.414 A.

Solution: I_rms=1 A

Answer: 1 A

Advanced

Q: Peak vs RMS?

Solution: V_m=√2 V_rms

Answer: Factor √2

Exam

Q: India AC mains?

Solution: 220 V RMS

Answer: Sec 19.3

X_C = 1/(ωC) = 1/(2πνC)

Definition: Capacitive reactance — opposition to AC by a capacitor.

Derivation

I leads V by 90°; magnitude X_C = 1/ωC.

Variables

X_C (Ω) · ω = 2πν · C (F)

Why it works

High frequency → low X_C → capacitor passes AC more easily.

Historical context

Average power in pure C = 0 (energy stored/released, not dissipated).

Deep understanding

Current leads voltage by π/2; 'wattless' current component.

2. Diagrams & Visuals

I leads V by 90°

Color-coded visual · step-by-step breakdown below

  1. Frequency ν or ω
  2. X_C = 1/(ωC)
  3. Current leads V by 90°
  4. X_C ↓ as ν ↑

3. Solved Examples

Basic

Q: C=10 μF, ν=50 Hz.

Solution: X_C≈318 Ω

Answer: ~318 Ω

Intermediate

Q: Double frequency?

Solution: X_C halves

Answer: 1/ν

Advanced

Q: DC (ν=0)?

Solution: X_C→∞

Answer: Blocks DC

Exam

Q: Capacitive reactance?

Solution: 1/ωC

Answer: Sec 19.3.2

X_L = ωL = 2πνL

Definition: Inductive reactance — opposition to AC by an inductor.

Derivation

I lags V by 90°; magnitude X_L = ωL.

Variables

X_L (Ω) · L (H)

Why it works

High frequency → high X_L → inductor blocks AC more.

Historical context

Average power in pure L = 0 (wattless current).

Deep understanding

Choke coils use high X_L at line frequency to limit current.

2. Diagrams & Visuals

I lags V by 90°

Color-coded visual · step-by-step breakdown below

  1. Frequency ν or ω
  2. X_L = ωL
  3. Current lags V by 90°
  4. X_L ↑ as ν ↑

3. Solved Examples

Basic

Q: L=0.1 H, ν=50 Hz.

Solution: X_L≈31.4 Ω

Answer: ~31 Ω

Intermediate

Q: Double ν?

Solution: X_L doubles

Answer: ∝ ν

Advanced

Q: DC steady?

Solution: X_L=0 ideally

Answer: Short for DC

Exam

Q: Inductive reactance?

Solution: ωL

Answer: Sec 19.3.3

Z = √[R² + (X_L − X_C)²]

Definition: Impedance of series LCR circuit.

Derivation

Phasor addition of R and net reactance (X_L − X_C).

Variables

Z (Ω) · I_rms = V_rms/Z

Why it works

Generalizes Ohm's law for AC circuits with reactive elements.

Historical context

tan φ = (X_L − X_C)/R gives phase angle φ.

Deep understanding

X_L > X_C → inductive (I lags V); X_C > X_L → capacitive (I leads V).

2. Diagrams & Visuals

Z = √(R²+(XL−XC)²)

Color-coded visual · step-by-step breakdown below

  1. Find R, X_L, X_C
  2. Net reactance X_L − X_C
  3. Z = √[R²+(X_L−X_C)²]
  4. I_rms = V_rms/Z

3. Solved Examples

Basic

Q: R=30, X_L=40, X_C=10 Ω.

Solution: Z=50 Ω

Answer: 50 Ω

Intermediate

Q: X_L = X_C?

Solution: Z=R

Answer: Resonance

Advanced

Q: Only resistor?

Solution: Z=R

Answer: φ=0

Exam

Q: LCR impedance?

Solution: √[R²+(XL−XC)²]

Answer: Sec 19.3.4

ν_r = 1/(2π√LC)

Definition: Resonant frequency of LC circuit — X_L = X_C.

Derivation

Set X_L = X_C → ω_r = 1/√LC → ν_r = ω_r/2π.

Variables

ν_r (Hz) · at resonance Z_min = R

Why it works

Maximum current at resonance; used in radio/TV tuning.

Historical context

Circuit purely resistive at resonance; V_L and V_C cancel.

Deep understanding

Below ν_r: capacitive; above ν_r: inductive.

2. Diagrams & Visuals

ν_r = 1/(2π√LC)

Color-coded visual · step-by-step breakdown below

  1. Identify L and C
  2. ν_r = 1/(2π√LC)
  3. At ν_r: Z = R, max current
  4. cos φ = 1

3. Solved Examples

Basic

Q: L=1 H, C=1 μF.

Solution: ν_r≈159 Hz

Answer: ~159 Hz

Intermediate

Q: L×C constant?

Solution: ν_r unchanged

Answer: Product sets ν_r

Advanced

Q: At resonance power?

Solution: Max, cosφ=1

Answer: Purely resistive

Exam

Q: LC resonance?

Solution: 1/(2π√LC)

Answer: Sec 19.3.4

P_av = V_rms I_rms cos φ

Definition: Average power in AC circuit; cos φ is power factor.

Derivation

Only in-phase (resistive) component delivers net power.

Variables

cos φ = R/Z · φ = phase angle

Why it works

Utilities care about power factor; pure L/C have cos φ = 0.

Historical context

P_av = I_rms² R = V_rms²/R when purely resistive.

Deep understanding

Wattless current in L or C stores energy but averages zero power.

2. Diagrams & Visuals

P = V_rms I_rms cos φ cos φ = R/Z

Color-coded visual · step-by-step breakdown below

  1. Find V_rms, I_rms, φ
  2. cos φ = R/Z
  3. P_av = V_rms I_rms cos φ
  4. Resonance: cos φ = 1

3. Solved Examples

Basic

Q: V_rms=220, I_rms=2, cosφ=0.8.

Solution: P=352 W

Answer: 352 W

Intermediate

Q: Pure inductor?

Solution: cosφ=0, P=0

Answer: Wattless

Advanced

Q: At resonance?

Solution: cosφ=1, max P

Answer: Z=R

Exam

Q: AC average power?

Solution: V_rms I_rms cosφ

Answer: Sec 19.3.4

ε(t) = N ω A B sin ωt

Definition: EMF induced in rotating coil AC generator.

Derivation

Flux φ = AB cos ωt; ε = −dφ/dt = NωAB sin ωt.

Variables

N turns · A area · B field · ω angular speed

Why it works

Mechanical rotation → alternating emf; basis of power stations.

Historical context

Slip rings + brushes → AC output; commutator → DC (pulsating).

Deep understanding

Fleming's right-hand rule for induced emf direction.

2. Diagrams & Visuals

ε = NωAB sinωt

Color-coded visual · step-by-step breakdown below

  1. Flux φ = AB cos ωt
  2. Differentiate: ε = NωAB sin ωt
  3. Peak emf = NωAB
  4. Slip rings for AC

3. Solved Examples

Basic

Q: Peak ε=310 V.

Solution: V_rms≈220 V

Answer: AC mains scale

Intermediate

Q: Double ω?

Solution: Peak ε doubles

Answer: ∝ ω

Advanced

Q: DC generator?

Solution: Commutator

Answer: Unidirectional

Exam

Q: AC generator emf?

Solution: NωAB sinωt

Answer: Sec 19.4

V_s/V_p = N_s/N_p = k

Definition: Transformer voltage ratio equals turns ratio.

Derivation

Same flux links both windings; induced emf ∝ turns.

Variables

k = transformation ratio · p = primary, s = secondary

Why it works

Step-up for transmission (high V, low I); step-down for domestic use.

Historical context

Works on AC only — needs changing flux. Cannot work on steady DC.

Deep understanding

Step-up: N_s > N_p; step-down: N_s < N_p.

2. Diagrams & Visuals

Np Ns Vs/Vp = Ns/Np

Color-coded visual · step-by-step breakdown below

  1. Count N_p, N_s
  2. V_s/V_p = N_s/N_p
  3. Step-up or step-down
  4. Laminated core reduces eddy currents

3. Solved Examples

Basic

Q: Np=100, Ns=1000.

Solution: Vs=10 Vp

Answer: Step-up 10×

Intermediate

Q: Ns/Np=0.1?

Solution: Step-down

Answer: 220 V → 22 V

Advanced

Q: DC on primary?

Solution: No steady induction

Answer: AC only

Exam

Q: Transformer ratio?

Solution: Vs/Vp=Ns/Np

Answer: Sec 19.5

I_p/I_s = N_s/N_p

Definition: Ideal transformer current ratio (inverse of voltage ratio).

Derivation

Conservation of power: P_in = P_out → V_p I_p = V_s I_s.

Variables

Ideal: no losses · η < 100% in real transformers

Why it works

Step-up increases V but decreases I — reduces I²R transmission losses.

Historical context

High-V transmission (~330 kV) with low current saves power.

Deep understanding

Losses: copper heating, eddy currents, hysteresis, flux leakage.

2. Diagrams & Visuals

Ip/Is = Ns/Np P_in ≈ P_out (ideal)

Color-coded visual · step-by-step breakdown below

  1. Voltage ratio k = N_s/N_p
  2. Ideal: I_p/I_s = N_s/N_p
  3. Step-up: I_s < I_p
  4. Efficiency η = P_out/P_in

3. Solved Examples

Basic

Q: Step-up 10× voltage?

Solution: I_s = I_p/10

Answer: Current down

Intermediate

Q: P_p=1000 W, ideal.

Solution: P_s=1000 W

Answer: No loss

Advanced

Q: Why high-V lines?

Solution: Less I²R loss

Answer: Transmission

Exam

Q: Transformer current?

Solution: Ip/Is=Ns/Np

Answer: Sec 19.5

5. Special Features & Extras

Complete study guide for Electromagnetic Induction and Alternating Current.

Exam Tips & Tricks

  • Induction needs change — steady flux or steady current → no emf.
  • Lenz's law — induced effects oppose the cause; negative sign in ε = −dφ/dt.
  • 1 V = 1 Wb/s — unit check for Faraday problems.
  • AC phases: R in phase · C: I leads V by 90° · L: I lags V by 90°.
  • Resonance: X_L = X_C → Z = R, max current, cos φ = 1.
  • RMS: V_rms = V_m/√2 — India mains 220 V RMS, 50 Hz.
  • Transformer: AC only; Vs/Vp = Ns/Np; step-up lowers current.

Common Student Mistakes

  • Using Faraday's law when flux is not changing
  • Confusing peak V_m with RMS V_rms (220 V is RMS)
  • Adding R, X_L, X_C directly instead of phasor/Z formula
  • Forgetting cos φ in AC power calculations
  • Expecting transformer to work on steady DC
  • Wrong phase: capacitor leads vs inductor lags

Memory Aids & Mnemonics

Faraday + Lenz: "Change flux → emf; emf fights the change"
Reactance: "C fights high f weakly (X_C↓)" · "L fights high f strongly (X_L↑)"
ELI the ICE man: E (emf/L) leads I in L; I leads E in C
Transformer: "More turns upstairs = step-up voltage, downstairs current"

Which Formula When?

  • Changing flux through loop? → ε = −dφ/dt (×N for coil)
  • Current changing in inductor? → ε = −L dI/dt
  • Coupled coils / transformer? → ε₂ = −M dI₁/dt
  • AC voltage/current rating? → V_rms = V_m/√2
  • Capacitor in AC? → X_C = 1/ωC
  • Inductor in AC? → X_L = ωL
  • Series LCR? → Z = √[R²+(X_L−X_C)²]
  • Tuning / resonance? → ν_r = 1/(2π√LC)
  • AC power? → P_av = V_rms I_rms cos φ
  • Rotating coil generator? → ε = NωAB sin ωt
  • Transformer? → V_s/V_p = N_s/N_p

QUICK REFERENCE — Ch 19 EM Induction & AC

ε = −dφ_B/dtε = −N dφ_B/dtφ = L I  |  ε = −L dI/dtL = μ₀ N² A / lτ = L / Rφ₂ = M I₁  |  ε₂ = −M dI₁/dtV_rms = V_m/√2  |  I_rms = I_m/√2X_C = 1/(ωC) = 1/(2πνC)X_L = ωL = 2πνLZ = √[R² + (X_L − X_C)²]ν_r = 1/(2π√LC)P_av = V_rms I_rms cos φε(t) = N ω A B sin ωtV_s/V_p = N_s/N_p = kI_p/I_s = N_s/N_p

Units: φ in weber (Wb) · L, M in henry (H) · Z, X in Ω · 1 V = 1 Wb/s

Key: ε=−dφ/dt · X_C=1/ωC · X_L=ωL · ν_r=1/(2π√LC) · Vs/Vp=Ns/Np

Tip: Draw phasor diagram for LCR — R on x-axis, reactance on y-axis, Z is hypotenuse.

PYQ — Previous Year Questions

Extracted from NIOS Physics (312) board exam papers in your PDF. Chapter L19 — Electromagnetic Induction and Alternating Current only. Use Model Answer for marking points; Explanation for concept clarity.

Chapter 19 — Electromagnetic Induction and Alternating Current (L19)

8 questions · Section B (short/long) · Sources: 312/TUS/104A, 68/ESS/1-312-A, Marking Scheme

Section B — Short Answer (2–3 marks)

PYQ1. State Faraday's laws of electromagnetic induction and explain them with the help of an example.

3 marks · Section B Q40 · 312/TUS/104A

Model Answer

First law: Magnitude of induced emf in a circuit is proportional to the rate of change of magnetic flux linked with it: |ε| ∝ |dΦ/dt|.

Second law: For a coil of N turns: ε = −N dΦ/dt (minus sign = Lenz's law).

Example: Moving a magnet toward a coil increases flux → induced current opposes the increase (Lenz). Relative motion between coil and field is essential.

Explanation

Faraday's laws unify motional and transformer emfs (L19 §19.1). Flux change can be by varying B, area, or orientation. Fleming's right-hand rule gives induced current direction.

PYQ2. Show that magnetic energy required to build up the current I in a coil of self inductance L is given by ½ LI².

2 marks · Section B Q34 · Marking Scheme (68/ESS/1-312-A)

Model Answer

Induced back-emf: e = −L di/dt

Work to increase current: dW = |e| i dt = L i di

W = ∫₀I L i di = ½ LI²

Energy stored in the magnetic field of the inductor.

Explanation

Marking scheme derivation: oppose growing current → external source does work against back-emf. Same form as capacitor energy ½CV² (L19 §19.2).

Section B — Long Answer (5 marks)

PYQ3. A metallic rod of length l is rotated with frequency ν. One end is hinged at the centre and the other at the circumference of a circular metallic ring. The ring rotates about an axis through the centre, normal to its plane, in a uniform magnetic field B parallel to the axis. (a) Obtain an expression for the e.m.f. induced between the centre and the ring. (b) Given rod resistance R, how much power will be generated?

5 marks · Section B Q42 · 312/TUS/104A

Model Answer

(a) Angular speed ω = 2πν. Element dr at radius r has speed v = ωr.

Motional emf dε = Bv dr = Bωr dr. Integrate 0 to l:

ε = ½ B ω l²

(b) Power dissipated: P = ε²/R = B²ω²l⁴/(4R)

Explanation

Disc-type generator / homopolar arrangement (L19 §19.1). B ⊥ motion → ε = ∫B(v×dl). Power = I²R = ε²/R when only rod resistance loads the circuit.

PYQ4. A device X is connected across an AC source V = V₀ sin ωt. The current is I = I₀ sin(ωt + π/2). (a) Identify X and write its reactance. (b) Draw graphs of voltage and current vs time for one cycle. (c) Draw the phasor diagram for X.

5 marks · Section B Q42 (OR) · 312/TUS/104A

Model Answer

(a) X = capacitor. Current leads voltage by 90° (π/2).

Reactance: XC = 1/(ωC)

(b) V and I sinusoids; I reaches maximum π/2 earlier than V.

(c) Phasor diagram: current phasor 90° ahead of voltage phasor.

Explanation

I = I₀ sin(ωt + π/2) with V = V₀ sin ωt ⇒ I leads V by 90° — capacitive AC behaviour (L19 §19.3). Inductor would lag by π/2.

PYQ5. (a) State Lenz's law. "Lenz's law is a consequence of the principle of conservation of energy." Justify this statement. (b) Deduce an expression for the mutual inductance of two long coaxial solenoids having different radii and different numbers of turns.

5 marks · Section B Q43 · Marking Scheme (68/ESS/1-312-A)

Model Answer

(a) Lenz's law: Induced current opposes the cause producing it (change of flux).

Energy justification: Without opposition, a small push would accelerate motion → unlimited energy. Mechanical work must be done against induced effects → energy conserved.

(b) Outer solenoid S₁: field B₁ = μ₀n₁I₁. Flux through inner S₂: Φ₂ = N₂B₁A₁.

Mutual inductance: M = N₂Φ₂/I₁ = μ₀n₁N₂πr₁²/l

Explanation

Marking scheme: magnet approach/recession examples for Lenz. M defined by Φ₂ = MI₁. Coaxial geometry uses B of outer on inner cross-section (L19 §19.2).

PYQ6. Using a phasor diagram for a series LCR circuit connected to an AC source E = Em cos ωt, derive the relation for current in the circuit and the expression for resonance frequency. Draw a plot showing variation of peak current (Im) with frequency of the AC source.

5 marks · Section B Q43 (OR) · Marking Scheme (68/ESS/1-312-A)

Model Answer

From phasor diagram: V = √[(IR)² + (IXL − IXC)²]

⇒ I = Em/√[R² + (XL − XC)²] where XL = ωL, XC = 1/(ωC)

Resonance: XL = XC → ωL = 1/(ωC) → ω₀ = 1/√(LC)

Im vs ω: maximum at ω₀ (sharp peak when R small).

Explanation

Series LCR impedance Z = √(R² + (ωL − 1/ωC)²). At resonance Zmin = R → maximum current (L19 §19.4).

PYQ7. A transformer has 200 turns in primary and 1000 in secondary. Primary and secondary resistances are 0.4 Ω and 2 Ω. Secondary power output is 13.5 kW at 1200 V; efficiency is 90%. Calculate: (i) input voltage, (ii) input power, (iii) currents in primary and secondary, (iv) power lost in primary, (v) Is power loss 10% of input power? Explain.

5 marks · Section B Q43 · 68/ESS/1-312-A (Marking Scheme)

Model Answer

Pout = 13.5 kW, η = 0.90 → Pin = 15 kW

(i) Vp = Vs × Np/Ns = 1200 × 200/1000 = 240 V

(iii) Is = Pout/Vs = 13500/1200 = 11.25 A; Ip = Pin/Vp = 62.5 A

(iv) Copper loss ≈ Ip²Rp = 62.5² × 0.4 ≈ 1562 W (plus Is²Rs)

(v) Total loss ≈ 1.5 kW = 10% of input — efficiency 90% ⇒ losses are 10% of input power.

Explanation

Ideal transformer: Vp/Vs = Np/Ns, Pin = Pout/η. Real transformer has copper and iron losses (L19 §19.5).

PYQ8. An inductor of 60 mH, a capacitor of 50 μF and a resistor of 20 Ω are connected in series across an AC source of peak voltage 210 V and angular frequency 400 rad s⁻¹. Calculate: (i) impedance, (ii) rms current, (iii) rms voltages across L, C and R, (iv) Is the current forward-moving or backward-moving relative to voltage?

5 marks · Section B Q43 (OR) · 68/ESS/1-312-A (Marking Scheme)

Model Answer

XL = ωL = 400 × 0.06 = 24 Ω; XC = 1/(ωC) = 1/(400 × 50×10⁻⁶) = 50 Ω

(i) Z = √[R² + (XL−XC)²] = √[400 + 676] ≈ 32.8 Ω

(ii) Irms = (210/√2)/Z ≈ 4.5 A

(iii) VL = IrmsXL, VC = IrmsXC, VR = IrmsR

(iv) XC > XL → capacitive → current leads voltage (forward-moving relative to voltage).

Explanation

Board numerical on series LCR at ω = 400 rad/s — below resonance (ω₀ ≈ 577 rad/s for these L, C) so circuit is capacitive-dominated.

Problem Solving — L19 EMI and Alternating Current

Six problems spanning this chapter’s NIOS syllabus. Every question is built from the notes and formula sheet: solve with equations first, then read the formal textbook-style write-up, the easy explanation, and the topic in depth (formulas, meaning, exam tips). Explanations open by default.

Question 1 of 6Flux

Loop area 0.02 m² in 0.50 T field, normal parallel to B. Flux? If rotated to θ=90°, new flux?

Φ = B A cosθ

Solution — step by step with formulas

  1. Φ₁ = 0.010 Wb.
  2. Φ₂ = 0.

Final answer: 0.010 Wb then 0

Formulas used in this problem

Φ = B A cosθ

Textbook formal language

Flux is surface integral of B; for uniform B, BA cosθ.

Working formula set for this problem: Φ = B A cosθ. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Max flux when face is square-on to field; zero when edge-on.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Magnetic flux

Weber is SI unit of flux.

Link to chapter notes (L19 — Magnetic flux): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: Φ = B A cosθ. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write Φ = B A cosθ before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 2 of 6Faraday

Flux through a coil of 50 turns falls from 0.02 Wb to 0 in 0.10 s. Average emf?

ε = −dΦ/dt

Solution — step by step with formulas

  1. ε = N ΔΦ/Δt = 50×0.02/0.10 = 10 V (magnitude).

Final answer: |ε| = 10 V

Formulas used in this problem

ε = −dΦ/dt

Textbook formal language

Induced emf equals negative rate of change of flux linkage.

Working formula set for this problem: ε = −dΦ/dt. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Flux vanishing quickly makes larger voltage; 50 turns multiplies it.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Faraday’s law

Minus sign: Lenz’s law direction.

Link to chapter notes (L19 — Faraday’s law): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: ε = −dΦ/dt. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write ε = −dΦ/dt before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 3 of 6Lenz

North pole of a magnet approaches a loop. What is the polarity of the face toward the magnet and why?

Solution — step by step with formulas

  1. Face becomes north (repels approaching N) to oppose increase of flux toward loop.

Final answer: Facing side acts as N pole (opposes approach)

Textbook formal language

Induced current opposes the change in flux that produces it (energy conservation).

Working formula set for this problem: (see solution steps). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Nature fights the change—approaching north is pushed back by an induced north.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Lenz’s law

Explains motor/generator energy balance.

Link to chapter notes (L19 — Lenz’s law): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: (see solution steps). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write (see solution steps) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 4 of 6Motional emf

Rod length 0.20 m moves at 5.0 m·s⁻¹ ⟂ to B = 0.40 T (B, ℓ, v mutually perpendicular). Emf?

ε = B ℓ v

Solution — step by step with formulas

  1. ε = 0.40×0.20×5.0 = 0.40 V.

Final answer: ε = 0.40 V

Formulas used in this problem

ε = B ℓ v

Textbook formal language

Charges in the rod experience magnetic force, separating until E balances vB.

Working formula set for this problem: ε = B ℓ v. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Moving metal in field builds a voltage Bℓv across its ends.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Motional emf

Used in rail generators and eddy-current braking discussions.

Link to chapter notes (L19 — Motional emf): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: ε = B ℓ v. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write ε = B ℓ v before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 5 of 6AC

AC voltage V = 311 sin(ωt) volt. Find peak and rms values.

I_rms = I₀/√2
V_rms = V₀/√2

Solution — step by step with formulas

  1. V₀ = 311 V.
  2. V_rms = 311/√2 ≈ 220 V.

Final answer: V₀ = 311 V; V_rms ≈ 220 V

Formulas used in this problem

I_rms = I₀/√2
V_rms = V₀/√2

Textbook formal language

RMS value is the effective DC equivalent for power in a resistor.

Working formula set for this problem: I_rms = I₀/√2; V_rms = V₀/√2. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Household ~220 V is rms; the wave peaks near 311 V.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — AC peak and rms

Average of sin over cycle is zero; use rms for power.

Link to chapter notes (L19 — AC peak and rms): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: I_rms = I₀/√2; V_rms = V₀/√2. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write I_rms = I₀/√2; V_rms = V₀/√2 before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 6 of 6Transformer

Step-down transformer 220 V to 12 V, N_p = 1100. Find N_s (ideal).

V_s/V_p = N_s/N_p
I_p/I_s = N_s/N_p (ideal)

Solution — step by step with formulas

  1. N_s = N_p (V_s/V_p) = 60 turns.

Final answer: N_s = 60

Formulas used in this problem

V_s/V_p = N_s/N_p
I_p/I_s = N_s/N_p (ideal)

Textbook formal language

Ideal transformer voltage ratio equals turns ratio; power in ≈ power out.

Working formula set for this problem: V_s/V_p = N_s/N_p; I_p/I_s = N_s/N_p (ideal). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Fewer secondary turns give lower voltage.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Transformer

Real transformers have flux leakage and copper/iron losses.

Link to chapter notes (L19 — Transformer): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: V_s/V_p = N_s/N_p; I_p/I_s = N_s/N_p (ideal). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write V_s/V_p = N_s/N_p; I_p/I_s = N_s/N_p (ideal) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).