JEE Main chapters · Class 12 · Physics

Electromagnetic Induction and Alternating Current: JEE Main Physics formulas and notes

Electromagnetic Induction and Alternating Current is one of the highest-scoring chapters in JEE Mains, usually giving 2 or 3 questions together. They are formula-based: the flux and the EMF, the force and power in a moving rod, an RL circuit at a given time, the rms and mean values, the impedance of an LCR circuit, resonance and Q, power factor and a transformer. Learn the formulas and practise the circuits; about 8 hours spread over 5 days is enough.

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Quick revision

• Φ = BA cosθ; ε = −N dΦ/dt; q = NΔΦ/R (independent of time); Lenz: opposes the change of flux
• Motional: ε = Blv; I = Blv/R; F = B²l²v/R; P = B²l²v²/R; rotating rod ε = ½Bωl²; coil ε = NBAω sin ωt
• ε = −L dI/dt; L = μ₀N²A/l; M = μ₀N₁N₂A/l; series L₁ + L₂ ± 2M; U = ½LI²; u = B²/2μ₀
• RL: τ = L/R; growth (E/R)(1 − e^−t/τ); decay I₀e^−t/τ; at t = 0 open, at t = ∞ a wire
• Eddy currents oppose the motion, heat the metal; laminate to cut them; motor back EMF Eb
• rms = peak/√2; half-cycle mean = 2·peak/π; full-cycle mean 0; rms of DC + sine = √(Id² + I₀²/2)
• XL = ωL (V leads I); XC = 1/ωC (I leads V); R in phase; Z = √(R² + (XL − XC)²); tanφ = (XL − XC)/R
• Resonance: ω₀ = 1/√LC, Z = R, cosφ = 1; Q = ω₀L/R = f₀/Δf; half-power at 0.707 of the peak
• P = VI cosφ = I²R; cosφ = R/Z; wattless I sinφ; LC: T = 2π√LC, UE = UB at q = Q₀/√2; transformer Vs/Vp = Ns/Np = Ip/Is

Formulas with solved examples

Magnetic flux and Faraday’s law

Magnetic flux

• Φ = ∫B·dA = BA cosθ for a uniform field
• θ is the angle between B and the normal to the loop (not the plane)
• unit: weber, 1 Wb = 1 T m²
• maximum when the plane is perpendicular to B, zero when the plane is parallel to B
• for N turns the flux linked is NΦ

How to apply. Draw the normal to the loop first and then find the angle between it and B. Many mistakes come from taking the angle with the plane.

Example. B = 0.5 T, loop 10 cm × 20 cm, B at 60° to the normal: Φ = 0.5 × 0.02 × cos 60° = 5 × 10⁻³ Wb.

Faraday’s law

• ε = −N(dΦ)/(dt)
• average EMF = −N(ΔΦ)/(Δt)
• an EMF is induced whenever the flux changes: B changes, A changes or θ changes
• EMF depends on the rate of change, not on the flux itself
• induced current I = (ε)/(R)

How to apply. Write Φ in terms of t, then differentiate. For an average value use the final minus the initial flux over the time taken.

Example. A coil of 100 turns: the flux through each turn falls from 0.02 Wb to 0.005 Wb in 0.1 s. ε = 100 × 0.015 ÷ 0.1 = 15 V.

Three ways to change the flux

• B changing: ε = −NA cosθ (dB)/(dt)
• area changing: ε = −NB cosθ (dA)/(dt)
• angle changing: ε = NBA sinθ (dθ)/(dt)
• a loop that is moved inside a uniform field without changing A or θ has no EMF
• a loop fully inside a uniform field, moving or not, has no EMF

How to apply. Ask which of B, A and θ varies and keep the other two as constants.

Example. B = 0.2 T falls to zero in 0.5 s through a loop of area 0.01 m² (θ = 0): ε = 0.01 × 0.2 ÷ 0.5 = 4 mV.

Induced electric field

• a changing B makes a non-conservative E field: ∮E·dl = −(dΦ)/(dt)
• circular region of radius R with (dB)/(dt) uniform: inside, E = (r)/(2)(dB)/(dt)
• outside, E = (R²)/(2r)(dB)/(dt)
• the field lines are closed circles
• an induced field is not due to charges, so it has no potential

How to apply. Take a circle of radius r through the point, equate E × 2πr to the rate of change of flux through it and solve. Remember that only the field region contributes outside.

Example. R = 10 cm, dB/dt = 0.02 T/s: at r = 5 cm inside, E = 0.05 ÷ 2 × 0.02 = 5 × 10⁻⁴ V/m. At r = 20 cm: E = 0.01 ÷ 0.4 × 0.02 = 5 × 10⁻⁴ V/m.

Lenz’s law and induced charge

Lenz’s law

• the induced current flows so as to oppose the change of flux that produces it
• flux growing: the induced field is opposite to B
• flux falling: the induced field is in the same direction as B
• it follows from conservation of energy
• the minus sign in Faraday’s law carries it

How to apply. First decide whether the flux through the loop is increasing or decreasing; then the induced field is opposite (or the same); then use the right-hand rule for the direction of the current.

Example. B pointing into the page grows through a loop. The induced field must point out of the page, so the current is anticlockwise.

Magnet moving towards a coil

• the near face of the coil becomes the same pole as the approaching pole (repulsion)
• the near face becomes the opposite pole when the magnet moves away (attraction)
• the magnet always feels a force against its motion
• the work done by hand turns into heat in the coil

How to apply. Name the pole that faces the magnet by saying that the coil resists: it pushes the incoming pole away and holds on to the departing one.

Example. A north pole is pushed towards a coil: the face near it becomes a north pole, so the current is anticlockwise when seen from the magnet.

Induced charge

• q = ∫I dt = (NΔΦ)/(R)
• independent of the time taken
• the current is large for fast change and small for slow change but the charge is the same
• flipping a coil through 180° gives ΔΦ = 2BA
• q = (change of flux linked) ÷ (total resistance)

How to apply. Do not integrate the current: use q = NΔΦ/R straight away. Take R as the resistance of the whole circuit.

Example. A coil of 50 turns and area 0.01 m², with R = 5 Ω, is turned through 180° in B = 0.2 T (face perpendicular to B at the start). ΔΦ = 2 × 0.2 × 0.01 = 4 × 10⁻³ Wb and q = 50 × 4 × 10⁻³ ÷ 5 = 0.04 C.

Heat produced and the time dependence

• heat H = ∫(ε²)/(R)dt
• if ε is constant, H = (ε²)/(R)t
• I = (1)/(R)(dΦ)/(dt), so q = (1)/(R)ΔΦ and charge is the area under the I–t graph
• for a flux that changes uniformly the current is constant

How to apply. If Φ is given as a function of time, differentiate for ε, divide by R for I and integrate I² R for the heat.

Example. Φ = 3t² Wb in a loop of 2 Ω: ε = 6t volt, I = 3t ampere. The charge in the first 2 s is ∫3t dt = 6 C.

Motional EMF

Straight rod in a magnetic field

• ε = ∫(v × B)·dl = Blv for v ⊥ B ⊥ l
• general: ε = B l v sinθ with l, v, B at the angles that matter
• if the rod is slanted, use the effective length l′ perpendicular to v
• no EMF when v is parallel to B or when the rod is parallel to v
• the end where positive charge gathers is found from the direction of v × B

How to apply. Use only the components that are at right angles to each other. For a rod that is not straight, use the straight line between its ends.

Example. A rod of length 0.5 m moves at 10 m/s at right angles to B = 0.2 T and to its length: ε = 0.2 × 0.5 × 10 = 1 V.

Rod on rails: current, force and power

• I = (Blv)/(R)
• magnetic force F = BIl = (B²l²v)/(R) opposes the motion
• power to push at constant speed P = Fv = (B²l²v²)/(R) = I²R = ε I
• this equals the heat produced in R
• by Lenz’s law the force always opposes the velocity

How to apply. Find ε first, then I, then F. The mechanical power supplied equals the electrical power lost as heat, which is a check of every answer.

Example. B = 0.2 T, l = 0.5 m, v = 10 m/s, R = 2 Ω: I = 1 ÷ 2 = 0.5 A, F = 0.2 × 0.5 × 0.5 = 0.05 N and P = 0.05 × 10 = 0.5 W = I²R.

Rod released on rails (terminal speed)

• a rod of mass m on rails with an external force F: m(dv)/(dt) = F − (B²l²v)/(R)
• terminal speed vT = (FR)/(B²l²)
• for a rod slipping down a smooth incline of angle α: vT = (mgR sinα)/(B²l²cos²α) (field vertical)
• v = vT(1 − e^−t/τ) with τ = (mR)/(B²l²)

How to apply. At the terminal speed the force of the field equals the driving force; set them equal to find vT directly without solving the differential equation.

Example. m = 0.1 kg, l = 1 m, R = 1 Ω, B = 1 T, rod falling vertically on vertical rails (g = 10): vT = 0.1 × 10 × 1 ÷ 1 = 1 m/s.

Rotating rod and disc

• ε = (1)/(2)Bωl² = (1)/(2)Bl²(2πf) = BAf
• the speed of the free end is ωl and the mean speed is ωl/2
• the same holds for a disc or a spoke of radius l
• centre at O, potential difference between the axle and the rim of a disc is (1)/(2)Bωr²
• between two points at r₁ and r₂: ε = (1)/(2)Bω(r₂² − r₁²)
• if the rod rotates about a point not at its end, the two parts give opposite EMFs

How to apply. Use the average speed ωl/2 of the rod in ε = Bl(v average). For two parts of a rod about a pivot, subtract the two values.

Example. B = 0.1 T, l = 0.5 m, ω = 20 rad/s: ε = 0.5 × 0.1 × 20 × 0.25 = 0.25 V.

Rectangular loop entering a field

• while entering: ε = Bav (a the width cut by the field edge) and F = (B²a²v)/(R)
• fully inside a uniform field: ε = 0, no force
• while leaving: ε = Bav in the opposite sense
• duration of each stage = (b)/(v) (b the length of the loop along the motion)
• I–t and F–t graphs are rectangular pulses, with opposite signs for entering and leaving

How to apply. Divide the motion into entering, inside and leaving, and write the flux change for each part.

Example. A square loop of side 0.1 m and R = 0.5 Ω enters a field of 0.5 T at 2 m/s: ε = 0.5 × 0.1 × 2 = 0.1 V, I = 0.2 A, F = 0.5 × 0.2 × 0.1 = 0.01 N, for t = 0.05 s.

Wire loop or coil rotating in a field

• Φ = NBA cos ωt
• ε = NBAω sin ωt
• peak value ε₀ = NBAω
• ε is zero when the plane is perpendicular to B, and maximum when the plane is parallel to B
• frequency f = (ω)/(2π)
• the same coil in the same field with half the speed gives half the peak EMF

How to apply. Write the angle at t = 0 first. If the plane is perpendicular to B at t = 0, ε goes as sin ωt.

Example. N = 100, B = 0.05 T, A = 0.02 m², ω = 100 rad/s: ε₀ = 100 × 0.05 × 0.02 × 100 = 10 V.

Self and mutual inductance

Self-inductance

• Φ (linked) = LI
• ε = −L(dI)/(dt)
• L depends on the shape, the number of turns and the core, not on I
• unit: henry, 1 H = 1 V s/A
• the induced EMF opposes the change of current, never the current itself
• L is the inertia of the circuit against change of current

How to apply. Write the flux linked as NΦ = LI to find L. In circuits use ε = L dI/dt as a drop across the inductor.

Example. L = 0.5 H and the current changes at 4 A/s: ε = 0.5 × 4 = 2 V.

Inductance of a solenoid

• L = μ₀n²Al = (μ₀N²A)/(l) (n turns per metre, N turns, area A, length l)
• with a core of relative permeability μr, multiply by μr
• L ∝ N²
• for a toroid L = (μ₀N²A)/(2πR)
• doubling the turns on the same length gives 4L

How to apply. Check the dependence first: L ∝ N² A / l for the same core. For a changed length at the same number of turns L ∝ 1/l.

Example. N = 1000, l = 0.5 m, A = 2 × 10⁻⁴ m²: L = 4π × 10⁻⁷ × 10⁶ × 2 × 10⁻⁴ ÷ 0.5 = 5.0 × 10⁻⁴ H = 0.5 mH.

Mutual inductance

• Φ₂ = MI₁
• ε₂ = −M(dI₁)/(dt)
• M = (μ₀N₁N₂A)/(l) for a coil wound over a long solenoid
• M₁₂ = M₂₁
• M = k√(L₁L₂), 0 ≤ k ≤ 1 (k = 1 means no flux leaks)
• depends on the geometry, the relative position and the core

How to apply. Remember that the EMF in the second coil comes from the changing current in the first. M is the same both ways, so you may choose the easier coil to calculate it.

Example. M = 0.2 H and the current in the primary changes at 5 A/s: ε in the secondary = 0.2 × 5 = 1 V.

Coils in series

• L = L₁ + L₂ + 2M if the fluxes add (same sense of winding)
• L = L₁ + L₂ − 2M if the fluxes oppose
• far apart coils (M = 0): L = L₁ + L₂
• parallel far apart: (1)/(L) = (1)/(L₁) + (1)/(L₂)
• equal coils joined to cancel the flux (k = 1): L = 0

How to apply. Decide whether the currents go the same way round both coils; this fixes the sign of 2M.

Example. L₁ = 2 H, L₂ = 3 H and M = 1 H: aiding, L = 2 + 3 + 2 = 7 H; opposing, L = 2 + 3 − 2 = 3 H.

Energy, RL circuits, eddy currents and generators

Energy stored in an inductor

• U = (1)/(2)LI²
• energy density u = (B²)/(2μ₀)
• for a solenoid U = u × (volume) = (B²)/(2μ₀)Al
• the energy comes from the battery while the current grows and is returned when it falls
• compare a capacitor, U = (1)/(2)CV²

How to apply. Use U = ½LI² for the total energy and u = B²/2μ₀ if the field is given. The two agree for a solenoid.

Example. L = 0.5 H and I = 2 A: U = 0.5 × 0.5 × 4 = 1 J.

Growth of current in an RL circuit

• I = (E)/(R)(1 − e^−t/τ) with τ = (L)/(R)
• at t = 0, I = 0 and (dI)/(dt) = (E)/(L) (the inductor acts like a break)
• at t = τ, I = 0.632 (E)/(R)
• at t = ∞ the inductor is a plain wire, I = (E)/(R)
• voltage across L: VL = E e^−t/τ
• time to reach half the final value: 0.693τ
• at the instant of closing, the full battery EMF appears across the inductor

How to apply. Write the three numbers: I at t = 0, at t = ∞ and the time constant. Then use the exponential only if a value at a given time is asked.

Example. E = 12 V, R = 4 Ω, L = 2 H: τ = 0.5 s, final I = 3 A and at t = 0.5 s, I = 3 × 0.632 = 1.90 A.

Decay of current in an RL circuit

• I = I₀e^−t/τ with τ = (L)/(R)
• after one time constant I = 0.368I₀
• after t = 0.693τ the current is half
• the energy (1)/(2)LI₀² is all turned into heat in the resistor
• the induced EMF can be much larger than E if the circuit is broken suddenly (a spark) because τ becomes very small

How to apply. Use the value of the current at the start as I₀, not the value of E/R unless the current had reached it.

Example. I₀ = 3 A and τ = 0.5 s: at t = 1 s, I = 3 e⁻² = 3 × 0.135 = 0.406 A.

Two-resistor RL circuits (switch just closed or long after)

• just after closing: the inductor branch carries no current, so treat it as open
• long after: the inductor has no EMF, so treat it as a wire (zero resistance)
• just after opening: the current in the inductor is the same as just before
• for a battery of E with R₁ in series with (R₂ in parallel with L): at t = 0 I = (E)/(R₁ + R₂) and at t = ∞ I = (E)/(R₁)

How to apply. Replace the inductor by an open circuit at t = 0 and by a wire at t = ∞, then solve the resistor circuit. The current in L cannot jump.

Example. E = 10 V, R₁ = 5 Ω in series with the parallel pair R₂ = 5 Ω and L: at t = 0, I = 10 ÷ 10 = 1 A; at t = ∞, I = 10 ÷ 5 = 2 A.

Eddy currents

• closed loops of current induced in a bulk conductor by a changing flux or by its motion in a field
• they oppose the cause (Lenz) and heat the metal
• uses: induction furnace, magnetic braking of trains, damping of galvanometer and balance pointers, energy meters
• reduced by laminating the core (thin insulated sheets), which cuts the loop area
• a plate with slits falls faster through a magnet than a solid plate

How to apply. Say what they do in one line: they brake the motion and produce heat. For a falling magnet or plate through a tube, the speed reaches a terminal value.

Example. A solid copper plate and a plate with slits swing into a magnet: the solid one stops sooner, since larger loops carry larger eddy currents.

AC generator and back EMF

• coil of N turns, area A, rotating at ω in field B: ε = NBAω sin ωt
• mechanical energy turns into electrical energy
• in a motor the rotating coil produces a back EMF opposing the supply: I = (E − Eb)/(R)
• starting current is large because Eb = 0 at rest
• the back EMF is largest at full speed, so the current falls

How to apply. Treat the motor as a source (E) and a resistor (R) with an opposing source Eb in series.

Example. E = 220 V, R = 4 Ω and Eb = 200 V: I = 20 ÷ 4 = 5 A. At rest the current would be 220 ÷ 4 = 55 A.

AC values: peak, rms and average

Alternating current and its values

• i = I₀ sin ωt, v = V₀ sin ωt
• ω = 2πf = (2π)/(T)
• I rms = (I₀)/(√2) = 0.707I₀
• mean over a half cycle = (2I₀)/(π) = 0.637I₀
• the mean over a full cycle is zero
• form factor = (rms)/(mean of half cycle) = 1.11
• the rms is the value of the steady DC that gives the same heat in a resistor
• ammeters and voltmeters for AC read rms

How to apply. Use rms for any heating, power or meter reading, the half-cycle average for rectified charge, and the peak for insulation or breakdown.

Example. A mains voltage of 220 V (rms) has the peak 220 × 1.414 = 311 V. A current of peak 10 A has rms 7.07 A and half-cycle mean 6.37 A.

Reading a time function

• from v = V₀ sin(ωt + φ): V₀ is the peak, ω the angular frequency, φ the phase
• f = (ω)/(2π)
• V rms = (V₀)/(√2)
• time to reach the first peak from the zero crossing = (T)/(4)
• for v = 311 sin 100πt: f = 50 Hz and the rms is 220 V
• in sin and cos forms, cos ωt = sin(ωt + π/2)

How to apply. Compare the given expression with I₀ sin(ωt + φ) term by term. Do not multiply the peak by 0.707 twice.

Example. v = 311 sin(100πt): ω = 100π so f = 50 Hz and T = 20 ms; V rms = 311 ÷ 1.414 = 220 V.

Rms of other waveforms

• rms = √(mean of i²)
• pure sine: (I₀)/(√2)
• square wave of height ±I₀: I₀
• half-wave rectified sine: (I₀)/(2)
• DC plus a sine, I = Id + I₀ sin ωt: √(Id² + (I₀²)/(2))
• triangular or sawtooth wave of peak I₀: (I₀)/(√3)
• rms over time of a pulse of I for a fraction τ of the period: I√τ

How to apply. Square the current, average over a whole period and take the root. Sine squared averages to 1/2.

Example. I = 3 + 4 sin ωt: Irms = √(9 + 8) = √17 = 4.12 A (the mean is 3 A).

AC through R, L and C; phasors

Phasors and phase

• a phasor is a rotating vector of length equal to the peak value; its projection on the vertical axis is the instantaneous value
• all phasors turn anticlockwise at ω
• the angle between two phasors is their phase difference
• resistor: v and i in phase
• inductor: v leads i by 90°
• capacitor: i leads v by 90°

How to apply. Memory aid: for L the current lags (V leads I); for C the current leads. Draw I along the axis for a series circuit, V along the axis for a parallel one.

Example. For a pure inductor with i = I₀ sin ωt, v = V₀ sin(ωt + 90°) = V₀ cos ωt.

Inductive reactance

• XL = ωL = 2πfL
• V₀ = I₀XL
• XL is zero for DC and grows with frequency
• current lags the voltage by π/2
• average power over a cycle is zero
• energy is stored and returned each quarter cycle
• an inductor lets low frequencies through and blocks high ones (choke)

How to apply. Take XL straight from the frequency: if f doubles, XL doubles. Use rms values on both sides if you want rms current.

Example. L = 0.1 H at 50 Hz: XL = 2π × 50 × 0.1 = 31.4 Ω. On 220 V the rms current is 220 ÷ 31.4 = 7.0 A.

Capacitive reactance

• XC = (1)/(ωC) = (1)/(2πfC)
• V₀ = I₀XC
• XC is infinite for DC and falls as f rises
• current leads the voltage by π/2
• average power over a cycle is zero
• a capacitor passes high frequencies and blocks DC (a blocking capacitor)

How to apply. The reactance goes the opposite way to that of the inductor: if f doubles, XC halves.

Example. C = 100 μF at 50 Hz: XC = 1 ÷ (2π × 50 × 10⁻⁴) = 31.8 Ω.

Variation with frequency

• R is independent of f
• XL ∝ f is a straight line through the origin
• XC ∝ (1)/(f) is a hyperbola
• they cross at f₀ = (1)/(2π√(LC)), where XL = XC
• for f < f₀ the circuit is capacitive; for f > f₀ it is inductive
• a bulb in series with a capacitor gets brighter if f rises; with an inductor, dimmer

How to apply. Compare ωL and 1/ωC first to decide whether the net reactance is inductive or capacitive.

Example. L = 1 mH and C = 10 μF: f₀ = 1592 Hz. At 50 Hz XL = 0.31 Ω and XC = 318 Ω, so the circuit is capacitive.

Series LCR circuit and resonance

Impedance of series circuits

• LR: Z = √(R² + XL²), tanφ = (XL)/(R), voltage leads
• CR: Z = √(R² + XC²), tanφ = (XC)/(R), current leads
• LCR: Z = √(R² + (XL − XC)²), tanφ = (XL − XC)/(R)
• I = (V)/(Z)
• V² = VR² + (VL − VC)²
• the voltages across L or C can be larger than the supply

How to apply. Draw the phasor triangle: R along the axis, XL up, XC down. The hypotenuse is Z, and the angle is φ.

Example. R = 30 Ω, XL = 70 Ω, XC = 30 Ω: Z = √(900 + 1600) = 50 Ω and φ = tan⁻¹(40/30) = 53°. On 100 V rms, I = 2 A.

Resonance

• XL = XC at ω₀ = (1)/(√(LC)), f₀ = (1)/(2π√(LC))
• Z = R is the minimum, I = (V)/(R) is the maximum
• V and I are in phase, cosφ = 1
• VL = VC but they are opposite and cancel
• the voltage across L (or C) is Q times the supply
• energy goes back and forth between L and C, while R takes the power

How to apply. At resonance the circuit behaves as a pure resistor. Use this to find the current, the power and the voltage across each part.

Example. L = 1 mH and C = 10 μF: ω₀ = 1 ÷ √(10⁻⁸) = 10⁴ rad/s and f₀ = 1592 Hz. With R = 2 Ω and E = 10 V rms, I = 5 A.

Quality factor and bandwidth

• Q = (ω₀L)/(R) = (1)/(ω₀CR) = (1)/(R)√((L)/(C))
• Q = (ω₀)/(Δω) = (f₀)/(f₂ − f₁) where f₁ and f₂ are the half-power points (I falls to 0.707 of the maximum)
• Δω = (R)/(L)
• the voltage across L or C at resonance is QE
• a smaller R gives a larger Q, a sharper and taller peak, and a narrower band
• Q has no unit

How to apply. Q describes the sharpness. To find the bandwidth, Δf = f₀/Q, and use it for the half-power points f₀ ± Δf/2 (for large Q).

Example. L = 1 mH, C = 10 μF, R = 2 Ω: Q = 10 ÷ 2 = 5 and the bandwidth = 1592 ÷ 5 = 318 Hz. At E = 10 V the voltage across the inductor is 50 V.

Same current for two frequencies

• a current I below the peak is found at two frequencies f₁ and f₂, with f₁f₂ = f₀²
• so f₀ = √(f₁f₂)
• the circuit is capacitive at the lower one and inductive at the upper
• the impedance and the power are the same at both

How to apply. Use the symmetry in ω₁ω₂ = 1/LC to find L or C from the two frequencies.

Example. The same current flows at 400 Hz and at 900 Hz: f₀ = √(400 × 900) = 600 Hz.

Tuning by changing C or L

• f₀ ∝ (1)/(√(LC))
• to double f₀, reduce C to a quarter
• to halve f₀, make L four times
• a radio picks one station when its f₀ equals the station frequency
• for XL > XC (inductive) raise C to approach resonance; for XC > XL lower C

How to apply. Use f₀ ∝ 1/√(LC) to scale. Check which of XL and XC is larger before deciding which way to move.

Example. A radio tunes from 500 kHz to 1500 kHz by changing C: the ratio of the capacitances = (1500/500)² = 9.

Power, LC oscillations and the transformer

Power and power factor

• P(avg) = VrmsIrms cosφ = Irms²R
• cosφ = (R)/(Z) = power factor, between 0 and 1
• apparent power VrmsIrms (VA)
• pure L or C: φ = 90°, so P = 0
• at resonance cosφ = 1
• a low power factor means more current for the same power and larger line loss

How to apply. Find Z first, then cosφ = R/Z, and then use P = Vrms² R/Z² or Irms² R. Use rms values only.

Example. R = 30 Ω, XL = 40 Ω, V = 100 V rms: Z = 50 Ω, I = 2 A, cosφ = 0.6, P = 2 × 100 × 0.6 = I²R = 4 × 30 = 120 W.

Wattless current

• the current has a component Icosφ in phase with V (does the work) and a component Isinφ at 90° (wattless)
• wattless current = Irms sinφ
• for a pure inductor or capacitor the whole current is wattless
• the average power in the wattless part is zero

How to apply. Resolve the current along the voltage and at right angles to it. Only the first part carries power.

Example. I = 2 A and cosφ = 0.6: sinφ = 0.8, wattless current = 1.6 A and the useful component = 1.2 A.

LC oscillations

• ω = (1)/(√(LC)), T = 2π√(LC)
• q = Q₀ cos ωt, I = −Q₀ω sin ωt
• total energy ½LI₀² = (Q₀²)/(2C) is constant (no resistance)
• UE = (q²)/(2C), UB = ½Li²
• charge and current are 90° out of phase: UE is maximum when i = 0
• UE = UB when q = (Q₀)/(√2)
• the energies oscillate at a frequency 2f
• with R, the oscillations are damped

How to apply. Treat it like a spring and mass: q is the position, i the velocity, 1/C the spring constant and L the mass. Conserve the energy.

Example. L = 1 mH, C = 10 μF, Q₀ = 100 μC: ω = 10⁴ rad/s, I₀ = Q₀ω = 1 A and the energy = Q₀²/2C = 5 × 10⁻⁴ J = ½LI₀².

Time and energy sharing in LC

• the capacitor discharges fully in (T)/(4)
• equal energy in L and C at t = (T)/(8) (when q = Q₀/√2)
• if the capacitor has Q₀ and the switch is closed at t = 0, then q = Q₀ cos ωt
• when the energy of L is three times that of C, q = Q₀/2
• the magnetic energy is largest when q = 0

How to apply. Write q = Q₀cos ωt and set the fraction of energy; do not use the period as a rule of thumb for energy ratios other than 1 : 1.

Example. Q₀ = 100 μC, C = 10 μF: UE at q = 50 μC is (50 × 10⁻⁶)² ÷ 2 × 10⁻⁵ = 1.25 × 10⁻⁴ J, a quarter of the total, so UB = 3.75 × 10⁻⁴ J.

Ideal transformer

• (Vs)/(Vp) = (Ns)/(Np) = (Ip)/(Is)
• step-up: Ns > Np, Vs > Vp, Is < Ip
• step-down: Ns < Np
• ideal: Vs Is = Vp Ip (no loss of power)
• works only on AC, because it needs a changing flux
• the frequency is unchanged
• it transfers energy through mutual induction

How to apply. Take the turns ratio first, get Vs, then use the power equality to find the current.

Example. Np = 1000, Ns = 100, Vp = 220 V: Vs = 22 V. If the secondary delivers 100 W, then Is = 100/22 = 4.5 A.

Efficiency and losses

• η = (output power)/(input power) = (VsIs)/(VpIp)
• losses: copper loss I²R in the windings, iron loss (eddy currents and hysteresis) in the core, flux leakage and humming
• remedy: laminated core of soft iron, thick wire for the large current, windings on top of each other
• efficiency is usually 95 % to 99 %

How to apply. Input = output ÷ η. Then Ip = input power ÷ Vp.

Example. Output 100 W at η = 90 %: input = 111 W and Ip = 111 ÷ 220 = 0.505 A.

Transmission of power

• line current I = (P)/(V) (at unit power factor)
• loss = I²R = (P²R)/(V²)
• a higher voltage cuts the loss with the square
• step up at the power station and step down near the user
• fractional loss = (PR)/(V²)

How to apply. Write the loss as P²R/V² and compare two voltages: ten times the voltage gives one hundredth of the loss.

Example. P = 100 kW, line R = 2 Ω: at 2 kV, I = 50 A and loss = 5 kW (5 %). At 20 kV, I = 5 A and loss = 50 W (0.05 %).

Lamp, heater and choke on AC

• bulb of rating (P, V): R = (V²)/(P), e.g. 100 W, 220 V: R = 484 Ω
• a choke (inductor) in series dims the lamp without much loss of power, but a resistor wastes power as heat
• DC through an inductor: only R matters
• for the same rms, the lamp is equally bright on DC and AC

How to apply. Find the resistance of the lamp from its rating and then treat it as R in the LCR formulas.

Example. A 100 W, 220 V lamp is in series with an ideal coil on a 220 V supply. For half the power (50 W) the current falls by √2, so Z = 484 × 1.414 = 684 Ω and XL = 484 Ω.

Most asked question types

How do I solve “Flux through a tilted loop” questions in JEE Main?

A loop of given area is in a uniform field at some angle. Find the flux, or the change of flux when the loop is turned. 1. Find θ between B and the normal. 2. Φ = BA cosθ. 3. Subtract final from initial flux for the change. Example: 0.5 T, 0.02 m², 60° to the normal: 5 × 10⁻³ Wb.

How do I solve “EMF from a changing B or a changing area” questions in JEE Main?

B changes with time, or a coil is pulled out of a field. Find the average or the instantaneous EMF. 1. Write Φ = NBA cosθ. 2. ε = −dΦ/dt (or ΔΦ/Δt). 3. Put the given numbers; keep units in SI. Example: 100 turns, flux falls by 0.015 Wb in 0.1 s: 15 V.

How do I solve “Direction of the induced current (Lenz’s law)” questions in JEE Main?

A magnet moves near a coil, or the field in a loop grows or falls. Find the direction of the current or the face that becomes a pole. 1. Is the flux growing or falling? 2. The induced field opposes the change. 3. Right-hand rule gives the current. Example: B into the page growing: anticlockwise current.

How do I solve “Charge that flows when the flux changes” questions in JEE Main?

A coil is flipped, removed from a field or the field switched off. Find the charge that passes through the circuit. 1. q = NΔΦ/R. 2. Find the change of flux (2BA for a flip). 3. The time does not matter. Example: 50 turns, flip in 0.2 T, 0.01 m², 5 Ω: 0.04 C.

How do I solve “EMF in a moving rod” questions in JEE Main?

A rod moves in a field at some speed and orientation. Find the EMF, which end is positive, or the potential at a point. 1. Find the parts of v, l and B that are at right angles. 2. ε = Blv. 3. The direction comes from v × B. Example: 0.5 m, 10 m/s, 0.2 T: 1 V.

How do I solve “Rod on rails: current, force and power” questions in JEE Main?

A rod is pulled on rails joined by a resistor. Find the current, the force needed at constant speed and the power. 1. ε = Blv, I = ε/R. 2. F = BIl = B²l²v/R. 3. P = Fv = I²R. Example: 1 V and 2 Ω: 0.5 A, 0.05 N, 0.5 W.

How do I solve “Terminal speed of a rod or a loop” questions in JEE Main?

A rod slides under gravity or under a constant force on rails. Find the final constant speed. 1. At the terminal speed the force of the field equals the driving force. 2. B²l²v/R = F. 3. vT = FR/(B²l²). Example: 0.1 kg, 1 m, 1 Ω, 1 T, falling: 1 m/s.

How do I solve “Rotating rod or disc” questions in JEE Main?

A rod or a metal disc rotates in a perpendicular field. Find the EMF between the axle and the rim. 1. Mean speed is ωl/2. 2. ε = ½Bωl². 3. Between r₁ and r₂ subtract the squares. Example: 0.1 T, 0.5 m, 20 rad/s: 0.25 V.

How do I solve “Loop entering or leaving a field” questions in JEE Main?

A rectangular loop moves into a region of field. Find the EMF, the current and the force in each stage, or choose the right graph. 1. EMF only while the flux changes. 2. ε = Bav while entering or leaving. 3. Zero when fully inside. Example: 0.1 m loop, 0.5 T, 2 m/s: 0.1 V for 0.05 s.

How do I solve “Self-inductance of a solenoid” questions in JEE Main?

The number of turns, area and length are given. Find L, or see how L changes when N, l or the core change. 1. L = μ₀N²A/l. 2. L ∝ N² and L ∝ 1/l. 3. An iron core multiplies L by μr. Example: 1000 turns, 0.5 m, 2 × 10⁻⁴ m²: 0.5 mH.

How do I solve “Mutual inductance and coupled coils” questions in JEE Main?

Two coils are coupled. Find M, the EMF in the secondary, or the inductance of two coils in series. 1. ε₂ = −M dI₁/dt. 2. M = k√(L₁L₂). 3. L = L₁ + L₂ ± 2M. Example: 2 H and 3 H with M = 1 H: 7 H or 3 H.

How do I solve “Energy stored in an inductor” questions in JEE Main?

L and I are given, or the field in a solenoid. Find the energy or the energy per unit volume. 1. U = ½LI². 2. u = B²/2μ₀. 3. Compare U with a capacitor’s ½CV². Example: 0.5 H, 2 A: 1 J.

How do I solve “RL circuit: current at a given time” questions in JEE Main?

A battery is connected to an RL circuit, or the battery is removed. Find the current at some time, the time constant or the time for a fraction. 1. τ = L/R. 2. Growth (E/R)(1 − e^−t/τ), decay I₀e^−t/τ. 3. Half value at 0.693τ. Example: 12 V, 4 Ω, 2 H, t = 0.5 s: 1.90 A.

How do I solve “Switch just closed, or long after (inductor in a circuit)” questions in JEE Main?

A circuit has resistors and an inductor. Find the current in some branch at t = 0 or at t = ∞. 1. At t = 0 treat L as open. 2. At t = ∞ treat L as a wire. 3. Solve as a resistor circuit. Example: 10 V, 5 Ω and (5 Ω ‖ L): 1 A at t = 0, 2 A at t = ∞.

How do I solve “Peak, rms and mean values” questions in JEE Main?

A current or voltage function is given. Find the rms, the mean over a half cycle, or the heat produced. 1. rms = peak/√2. 2. Half-cycle mean = 2·peak/π. 3. For other waves, average the square. Example: 3 + 4 sin ωt: rms = 4.12 A.

How do I solve “Reactance, phase and the current in L-only or C-only circuits” questions in JEE Main?

An AC source is connected to a pure inductor or a pure capacitor. Find the reactance, the current and the phase. 1. XL = ωL, XC = 1/ωC. 2. I = V/X. 3. L: V leads. C: I leads. Example: 0.1 H, 50 Hz: 31.4 Ω, 7.0 A at 220 V.

How do I solve “Impedance and phase angle of LR, CR and LCR circuits” questions in JEE Main?

R, L, C and f are given. Find Z, I, the phase angle and the voltage across each part. 1. Find XL and XC. 2. Z = √(R² + (XL − XC)²). 3. tanφ = (XL − XC)/R. Example: 30 Ω, 70 Ω, 30 Ω: Z = 50 Ω, φ = 53°.

How do I solve “Resonance, Q factor and bandwidth” questions in JEE Main?

An LCR circuit is tuned to resonance. Find f₀, the current at resonance, Q, the bandwidth or the voltage across L. 1. f₀ = 1/(2π√LC). 2. Q = ω₀L/R, Δf = f₀/Q. 3. VL = QE at resonance. Example: 1 mH, 10 μF, 2 Ω: f₀ = 1592 Hz, Q = 5, Δf = 318 Hz.

How do I solve “Power, power factor and wattless current” questions in JEE Main?

The voltage and the circuit are given. Find the average power, cosφ or the wattless current. 1. cosφ = R/Z. 2. P = VrmsIrms cosφ = I²R. 3. Wattless current = I sinφ. Example: 30 Ω, 40 Ω, 100 V: cosφ = 0.6, P = 120 W.

How do I solve “LC oscillations: frequency and energy” questions in JEE Main?

A charged capacitor is connected to an inductor. Find the frequency, the maximum current or the energy or charge at some moment. 1. ω = 1/√(LC). 2. I₀ = Q₀ω. 3. Total energy Q₀²/2C is constant. Example: 1 mH, 10 μF, 100 μC: ω = 10⁴ rad/s, I₀ = 1 A, 5 × 10⁻⁴ J.

How do I solve “Transformer: voltage, current and efficiency” questions in JEE Main?

The turns ratio, the input voltage and the load are given. Find the secondary voltage, the currents and the efficiency. 1. Vs/Vp = Ns/Np = Ip/Is. 2. Input = output ÷ η. 3. Ip = input power ÷ Vp. Example: 1000 : 100 turns, 220 V: 22 V. 100 W at 90 %: 0.505 A.

How do I solve “Power transmission and line losses” questions in JEE Main?

Power is sent through a line of resistance R at some voltage. Find the line current, the loss or the effect of the voltage. 1. I = P/V. 2. Loss = I²R = P²R/V². 3. Loss falls as 1/V². Example: 100 kW, 2 Ω: 5 kW at 2 kV, 50 W at 20 kV.

Common myths and the truth

• Myth: The induced EMF depends on the flux through the loop. Truth: It depends on the rate of change of flux. A large constant flux gives no EMF.
• Myth: Lenz’s law says the induced field is opposite to the applied field. Truth: It opposes the change of flux. When the flux falls, the induced field is in the same direction as B.
• Myth: A heavy magnet falls freely through a copper tube. Truth: Eddy currents in the tube oppose the fall, so the magnet reaches a small terminal speed.
• Myth: The inductor stops the current. Truth: It resists a change of current. A steady current passes freely.
• Myth: The current in an RL circuit reaches E/R in a time τ. Truth: After τ it has reached only 63 % of its final value. It takes several τ to settle.
• Myth: AC of 220 V has a maximum of 220 V. Truth: 220 V is the rms value. The peak is 311 V.
• Myth: A capacitor in AC is an open circuit. Truth: It passes AC with a reactance 1/ωC. It blocks only DC.
• Myth: At resonance the voltage across L and C is zero. Truth: Each is Q times the supply, and they are 180° apart so they cancel in the sum.
• Myth: A pure inductor consumes power. Truth: The average power is zero. It stores and returns the energy each quarter cycle.
• Myth: A transformer can increase both the voltage and the power. Truth: A step-up transformer raises V and lowers I; the output power is never more than the input.

Mistakes to avoid

• Mixing up f and ω: XL = 2πfL and XC = 1/(2πfC).
• The angle in Φ = BA cosθ is with the normal, not with the plane.
• Using the peak where the rms is needed in P = VI cosφ.
• Taking the current in an inductor as jumping at t = 0; it cannot change suddenly.
• Adding R, XL and XC directly: Z is the vector sum, with XL and XC in opposite directions.
• Saying that L, C or R at resonance have zero voltage; VL = VC = QE, only the sum is zero.
• Taking the average of an AC over a full cycle as 0.637 of the peak (that is for a half cycle).
• The transformer needs AC; with DC the secondary shows no steady voltage.

Simulations for Electromagnetic Induction and Alternating Current

• Eddy-current braking — Rotate a conducting disc through a magnetic brake and follow the loss of rotational energy.
• Magnetic flux through a tilted coil — Tilt a coil in a uniform magnetic field and see the flux follow cos θ.
• Magnet through a coil — Move a magnet past a coil and watch the induced current reverse.
• Sliding rod generator — Move a conducting rod across a uniform magnetic field.
• Mutual induction between coils — Drive a changing current in one coil and measure the emf induced in its neighbour.
• Current growth in an RL circuit — Close a circuit and watch an inductor oppose the rise in current.
• EMF of a rotating rod — Spin a metal rod about one end in a uniform magnetic field. The free electrons are pushed outward.
• Self-inductance of a solenoid — Change the turns, size and core of a solenoid and find its inductance and stored energy.
• AC generator — Rotate a coil and track the alternating voltage it produces.
• AC power and phase lag — Compare voltage, current, and instantaneous power in a series resistor–inductor circuit.
• Energy exchange in an LC circuit — Watch a charged capacitor and an inductor exchange electric and magnetic energy.
• LR circuit: phasors and phase lag — Drive a resistor and inductor with AC. The current lags the voltage by the phase angle φ.
• Reactance versus frequency — Sweep the frequency and compare inductive and capacitive reactance.
• Series RLC resonance — Sweep frequency and find the peak current.
• RMS and average values of waveforms — Compare sine, square and triangle waves of the same peak. The rms value sets the heating effect.
• Ideal transformer — Change the turn ratio and compare primary and secondary RMS voltages.

More: Class 12 Physics simulations · all JEE Main chapters.