Electromagnetic Induction and Alternating Currents

In a series LCR circuit the three voltmeters read $V_R=60$ V, $V_L=80$ V, $V_C=80$ V. What is the supply voltage?

Most add them: 220 V.

It is 60 V. $V_L$ and $V_C$ are exactly antiphase and cancel completely, leaving only $V_R$:

$V=\sqrt{V_R^2+(V_L-V_C{)}^2}=\sqrt{60^2+0}=60\text{V}$

Two chapters live under one title, and each has one idea holding it together:

• Induction responds to change, never to the field. A coil in an enormous steady field has zero emf. A coil in a feeble but rapidly changing one can have a large emf. Read Faraday's law as a statement about rates.
• AC is entirely phase. A resistor keeps current in step, an inductor makes it lag a quarter cycle, a capacitor makes it lead. Impedance, resonance, power factor and wattless current are all consequences of those three lines.
• Lenz's law is conservation of energy in disguise, not a separate rule.

The two halves meet at the transformer — mutual induction at mains frequency, and the sole reason AC won the current wars.

1. Magnetic Flux and Faraday's Law

$\Phi =BA\cos \theta ,\epsilon =-N\frac{d\Phi }{dt}$

Three things can change and each induces an emf: the field $B$, the area $A$, and the angle $\theta$ between $\vec{B}$ and the loop's normal. A generator changes $\theta$; a sliding rod changes $A$; a nearby switching circuit changes $B$.

An often-missed result concerns charge rather than current. Since $I=\epsilon /R$ and $q=\int Idt$:

$q=\frac{N\Delta \Phi }{R}$

The charge depends only on the total flux change and the resistance — never on how fast it happened. Move a magnet quickly and you get a large brief current; slowly, a small prolonged one; the charge delivered is identical.

Illustration 1

A square loop of side $a$ moves at constant speed $v$ out of a region of uniform field $B$, edge-first. Sketch the emf against time.

While the loop is fully inside, the flux is $B{a}^2$ and constant, so $\epsilon =0$ however strong the field is.

While it is leaving, the area inside the field shrinks at rate $av$:

$\epsilon =Bav\text{(constant, for the time }a/v)$

Once fully outside, $\epsilon =0$ again. A rectangular pulse — and the two zeroes are the point of the question.

2. Lenz's Law

The minus sign in Faraday's law: the induced current opposes the change that produced it.

Push a north pole toward a coil and the near face becomes a north pole, repelling it. Pull it away and that face becomes a south pole, attracting it back. Either way it resists you, so you must do work — and that work is exactly the electrical energy that appears.

Trap. Reversing the sign would break conservation of energy. An assisting current would accelerate the magnet, strengthening the current, accelerating it further — energy from nothing. Lenz's law is therefore derived, not an independent experimental law, and saying so earns the mark.

Illustration 2

A bar magnet is dropped through a horizontal copper ring. Compare its acceleration just above the ring and just below it.

Approaching, the flux through the ring rises, so the induced current opposes the approach and pushes the magnet up. Receding, the flux falls, so the current reverses and pulls the magnet back up.

$a<g\text{both above and below the ring}$

The retarding direction is the same on both sides even though the current reverses — which is why a magnet dropped down a copper pipe drifts down slowly while one dropped down a plastic pipe falls freely.

3. Motional EMF

$\epsilon =Bvl$

Two routes give this, and they agree exactly. From Faraday: the circuit's area sweeps at rate $lv$, so flux changes at $Blv$. From the Lorentz force: free charges in the rod feel $qvB$ and separate until their own electric field balances it. That agreement is a deep consistency requirement — and one of the hints that led to special relativity.

On rails closing a circuit of resistance $R$:

$I=\frac{Bvl}{R}\Rightarrow F_{retard}=BIl=\frac{B^2{l}^{2}v}{R}$

Mechanical power supplied is $Fv=B^2{l}^2{v}^2/R$, which is exactly $I^{2}R$. The bookkeeping closes to the joule, as Lenz's law demands.

Illustration 3

A rod of mass $m$ slides without friction down vertical rails of separation $l$ in a horizontal field $B$, the circuit having resistance $R$. Find its terminal velocity.

The retarding force grows with speed, so the rod stops accelerating when it balances gravity:

$mg=\frac{B^2{l}^2{v}_T}{R}\Rightarrow v_T=\frac{mgR}{B^2{l}^2}$

Exactly the structure of a falling body with viscous drag: a force proportional to $v$ always produces a terminal velocity. Cutting $R$ in half halves the terminal speed, which is how eddy-current brakes are tuned.

4. Eddy Currents

Currents circulating within the bulk of a conductor rather than round a defined loop, arising whenever a solid conductor sits in changing flux. They are dissipative — sometimes the point, sometimes a nuisance.

Unwanted eddy currents are suppressed by lamination: cores are built from thin insulated sheets rather than solid blocks, which breaks the circulating paths and cuts the loss sharply.

5. Self and Mutual Inductance

$N\Phi =LI\Rightarrow \epsilon =-L\frac{dI}{dt},L_{solenoid}={\mu }_0{n}^{2}Al$

An inductor opposes any change in the current through it. It is the electrical analogue of inertia: mass resists changes in velocity, inductance resists changes in current.

${\epsilon }_2=-M\frac{d{I}_1}{dt},M=k\sqrt{L_1{L}_2}$

• Perfect coupling is $k=1$, approached by winding both coils on a common iron core.
• Mutual inductance is symmetric — the same $M$ governs both directions. Not obvious, and a standard exam statement.
• Inductances combine like resistances when uncoupled: add in series, add reciprocally in parallel. Capacitance is the odd one out, because it relates charge to voltage rather than opposing a rate of change.

Illustration 4

A solenoid has 500 turns over 25 cm with cross-section 4 cm². Find $L$, and say what happens if the turns are doubled in the same length.

$n=\frac{500}{0.25}=2000{\text{m}}^{-1}$

$L={\mu }_0{n}^{2}Al=(4\pi \times 10^{-7})(2000{)}^2(4\times 10^{-4})(0.25)\Rightarrow L\approx 5\times 10^{-4}\text{H}$

Doubling the turns quadruples $L$, because it enters as $n^2$. Inductance is pure geometry — no current, no voltage, no frequency appears in it.

6. Energy Stored in an Inductor

$U=\frac{1}{2}L{I}^2,u=\frac{B^2}{2{\mu }_0}$

Set beside $\frac{1}{2}C{V}^2$ for a capacitor and $\frac{1}{2}m{v}^2$ for a moving mass, the pattern is plain: the inductor-current pairing plays the role of mass-velocity. The energy lives in the magnetic field, just as electric energy has density $\frac{1}{2}{\epsilon }_0{E}^2$.

Illustration 5

A 2 H inductor carries 3 A. The circuit is broken in 1 ms. Find the stored energy and the average emf across the switch.

$U=\frac{1}{2}(2)(3{)}^2=9\text{J},|\epsilon |=L\frac{\Delta I}{\Delta t}=\frac{2\times 3}{10^{-3}}=6000\text{V}$

Six kilovolts from a low-voltage circuit. That is the spark you see when an inductive circuit is switched off: the stored energy has to go somewhere, and a fast collapse means an enormous $LdI/dt$.

7. Alternating Current: Peak, RMS and Average

$I=I_0\sin \omega t$

Every AC meter reads RMS, and every quoted mains voltage is RMS. The 220 V Indian supply peaks at $220\sqrt{2}=311$ V — and insulation must withstand the peak, not the quoted value.

Illustration 6

A current $i=3+4\sin \omega t$ (amperes) flows through a resistor. Find its RMS value.

Not $\sqrt{3^2+4^2}/\sqrt{2}$, and not $7/\sqrt{2}$. Mean-square the two parts separately — the cross term averages to zero because $\sin \omega t$ does:

$⟨i^2⟩=3^2+\frac{4^2}{2}=9+8\Rightarrow i_{rms}=\sqrt{17}\approx 4.12\text{A}$

A DC offset contributes its full square; an AC component contributes half of its peak square. Superposed DC and AC is a standard trap and this is the whole method.

8. Phase Relationships in R, L and C

The mnemonic is CIVIL: in a Capacitor I leads V; V leads I in an L.

$X_L$ rises with frequency and $X_C$ falls with it. So an inductor blocks high frequencies and passes DC; a capacitor blocks DC and passes high frequencies.

Trap. Neither reactance dissipates anything. Over a full cycle an inductor returns all the energy it stored in its field and a capacitor returns all it stored on its plates. Reactance is measured in ohms but is not resistance — only $R$ consumes.

9. Impedance and the Series LCR Circuit

A phasor is a rotating arrow whose length is the peak value and whose angle is the phase; its vertical projection reproduces the instantaneous sinusoid. Its point is that adding two sinusoids of the same frequency becomes vector addition.

In a series circuit the current is common, so the current phasor is the natural reference.

$Z=\sqrt{R^2+(X_L-X_C{)}^2},\tan \varphi =\frac{X_L-X_C}{R}$

$X_L>X_C$ makes the circuit inductive and the current lags; $X_C>X_L$ makes it capacitive and the current leads.

Illustration 7

In a series LCR circuit the voltmeters read $V_R=40$ V, $V_L=90$ V, $V_C=60$ V. Find the supply voltage and the phase angle.

$V=\sqrt{40^2+(90-60{)}^2}=\sqrt{1600+900}=50\text{V}$

$\tan \varphi =\frac{30}{40}\Rightarrow \varphi =36.9^{\circ },\text{current lagging}$

The three readings sum to 190 V arithmetically, against a 50 V supply. Nothing is wrong with the meters — the readings are phasors at different angles.

10. Resonance

$X_L=X_C\Rightarrow {\omega }_0=\frac{1}{\sqrt{LC}},f_0=\frac{1}{2\pi \sqrt{LC}}$

At resonance the impedance falls to its minimum of $R$ alone, the current is maximum, and $\varphi =0$ so the circuit is purely resistive. A series LCR at resonance is an acceptor circuit — how a radio picks one station out of many.

$Q=\frac{{\omega }_{0}L}{R}=\frac{1}{R}\sqrt{\frac{L}{C}}$

High $Q$ means low resistance, a narrow peak and better selectivity. It is also why $V_L$ and $V_C$ can each reach $Q$ times the supply voltage at resonance — while cancelling each other exactly.

Illustration 8

A series LCR circuit carries 2 A at resonance. Find the current at a frequency where $X_L-X_C=R$.

$Z=\sqrt{R^2+R^2}=R\sqrt{2}\Rightarrow I=\frac{I_{max}}{\sqrt{2}}=1.41\text{A}$

Power goes as $I^2$, so this is the half-power point. The two frequencies where it occurs bound the resonance curve's bandwidth, and their separation is ${\omega }_0/Q$ — the formal meaning of "sharp".

11. Power in AC Circuits

$P_{avg}=V_{rms}{I}_{rms}\cos \varphi ,\cos \varphi =\frac{R}{Z}$

The power factor $\cos \varphi$ separates apparent power from real power. At $\varphi =\pi /2$ — a pure inductor or capacitor — the average power is exactly zero. Current flows and no energy is consumed: wattless current.

A choke coil exploits this, limiting current in an AC circuit almost without dissipating anything, which a resistor can never do.

Illustration 9

A factory load draws 10 A at 220 V with power factor 0.6. Find the real power, and the current after correction to unity power factor at the same real power.

$P=(220)(10)(0.6)=1320\text{W}\Rightarrow I^{\prime }=\frac{1320}{220}=6\text{A}$

Forty per cent less current for identical useful power. The cables, switchgear and transformers all carried that extra 4 A for nothing — which is why factories install capacitor banks and why utilities penalise a poor power factor.

Illustration 10

A lamp rated 60 W at 10 V is to run from a 100 V, 50 Hz supply. Compare dropping the extra 90 V with a resistor against dropping it with a choke.

The operating current is $I=60/10=6$ A either way.

With a resistor, the voltages add arithmetically because everything is in phase:

$R=\frac{90}{6}=15\Omega \Rightarrow P_{wasted}=I^{2}R=(36)(15)=540\text{W}$

Nine times the lamp's own consumption, thrown away as heat.

With a choke, the lamp and choke voltages are $90^{\circ }$ apart, so they add as phasors rather than as numbers:

$V_L=\sqrt{100^2-10^2}=99.5\text{V}\Rightarrow X_L=\frac{99.5}{6}=16.6\Omega$

$L=\frac{X_L}{2\pi f}=\frac{16.6}{314}=53\text{mH},P_{wasted}=0$

The choke drops nearly 100 V while consuming nothing, because its current is wattless. This is the practical payoff of the whole phase discussion, and it is why a fluorescent fitting contains a choke rather than a resistor.

12. AC Generator and Transformer

$\epsilon =NBA\omega \sin \omega t$

Trap. The emf is maximum when the coil's plane is parallel to the field — where the flux is zero. Flux and its rate of change peak a quarter cycle apart, so maximum emf coincides with zero flux, not maximum flux.

Illustration 11

A 200-turn coil of area 100 cm² rotates at 50 Hz in a field of 0.1 T. Find the peak emf, and the emf at the instant the coil's plane is perpendicular to the field.

$\omega =2\pi (50)=314{\text{rad s}}^{-1}$

${\epsilon }_0=NBA\omega =(200)(0.1)(10^{-2})(314)\Rightarrow {\epsilon }_0=62.8\text{V}$

With the plane perpendicular to $\vec{B}$, the coil's normal lies along $\vec{B}$, so $\cos \theta =1$ and the flux is at its maximum. A maximum is a stationary point:

$\frac{d\Phi }{dt}=0\Rightarrow \epsilon =0$

Maximum flux, zero emf. The emf peaks a quarter turn later, where the flux passes through zero fastest — the same quarter-cycle offset that runs through the whole AC half of this chapter.

A transformer is mutual inductance at mains frequency:

$\frac{V_s}{V_p}=\frac{N_s}{N_p}=\frac{I_p}{I_s}$

Voltage and current trade inversely, conserving power in the ideal case. A transformer creates no energy and cannot raise power. It works only on AC, since a steady current gives no changing flux and hence no secondary emf.

That single limitation decided the current wars. Transmission needs very high voltage to keep $I^{2}R$ losses small, then a step down to safe voltages — and only AC can be stepped efficiently, so the world grid is AC.

Illustration 12

A transformer steps 2200 V down to 220 V to run a 4.4 kW load, and is 90 per cent efficient. Find the turns ratio, the secondary current and the primary current.

$\frac{N_p}{N_s}=\frac{V_p}{V_s}=\frac{2200}{220}=10$

The load draws its power at the secondary:

$I_s=\frac{4400}{220}=20\text{A}$

The primary must supply more than the output, since a tenth of it is lost in the core and windings:

$P_p=\frac{4400}{0.9}=4889\text{W}\Rightarrow I_p=\frac{4889}{2200}=2.2\text{A}$

The ideal relation $I_p/I_s=N_s/N_p$ would have given exactly 2 A. Efficiency always pushes the primary current up, never down, and reaching for the ideal ratio when an efficiency has been quoted is the standard slip here.

Summary

• Induction responds to change, never to the field, so a huge steady flux induces nothing.
• $\epsilon =-Nd\Phi /dt$, and flux changes through $B$, $A$ or $\theta$.
• Charge is $N\Delta \Phi /R$, independent of how fast the change happened.
• Lenz's law is conservation of energy — an assisting current would give energy from nothing.
• Motional emf $Bvl$ follows from Faraday or from the Lorentz force, and the two always agree.
• A sliding rod's mechanical power exactly equals $I^{2}R$; a rod on vertical rails reaches $v_T=mgR/B^2{l}^2$.
• Eddy currents heat and brake deliberately, and are suppressed by lamination where unwanted.
• An inductor is electrical inertia: $\epsilon =-LdI/dt$, $L={\mu }_0{n}^{2}Al$, energy $\frac{1}{2}L{I}^2$.
• Breaking an inductive circuit fast gives an enormous $LdI/dt$ — the switch spark.
• $M=k\sqrt{L_1{L}_2}$ and is symmetric. Inductances combine like resistances, not like capacitances.
• RMS is the DC-equivalent value $I_0/\sqrt{2}$; 220 V mains peaks at 311 V. Superposed DC and AC: square-add $I_{DC}^2+I_0^2/2$.
• Current lags in an inductor, leads in a capacitor — CIVIL. $X_L=\omega L$ rises with frequency, $X_C=1/\omega C$ falls, neither dissipates.
• $Z=\sqrt{R^2+(X_L-X_C{)}^2}$; voltages add as phasors, which is why each can exceed the supply.
• Resonance at ${\omega }_0=1/\sqrt{LC}$: $Z$ minimum at $R$, current maximum, $\varphi =0$; $Q$ sets sharpness and voltage magnification.
• $P=V_{rms}{I}_{rms}\cos \varphi$; a pure reactance carries wattless current and a choke exploits it.
• A generator's emf peaks where the flux is zero, a quarter cycle from maximum flux.
• A transformer trades voltage against current, cannot raise power, and works only on AC — which is why the grid is AC.