Electromagnetic Induction and Alternating Currents

1Faraday's law and Lenz's law

A magnetic flux \Phi = BA\cos\theta threading a coil produces nothing at all while it sits still. Change it — move the magnet, spin the coil, shrink the area, ramp the field — and the coil develops an emf \varepsilon = -N\,d\Phi/dt. Only the rate of change counts: a huge flux held constant gives zero emf, and a tiny flux killed in a microsecond gives a large one.

The minus sign is Lenz's law, and it is a statement about energy rather than about geometry. The induced current always flows so as to oppose the change that created it, so pushing a magnet towards a coil is always resisted. Were the sign the other way, the magnet would be sucked in, speed up, and drive a larger current still — energy out of nothing.

Figure. The ramps are deliberately straight. While the flux climbs at a steady rate the emf is a constant negative step; while the flux is held constant the emf is exactly zero; and the final stretch takes the flux down in half the time, so the emf there is twice as tall and of the opposite sign. Zero on this axis is the level the emf trace holds through the flat middle section.

Getting Lenz's sign right

1. Name the flux and its directionFix a positive normal to the coil, then decide whether the flux through it is growing or shrinking.
2. Oppose the change, not the fluxThe induced current makes a field that fights the change: it resists a growing flux and props up a dying one.
3. Read the current off that fieldCurl the right hand round the coil with the thumb along the field the induced current has to produce.

2Motional emf

Slide a rod of length L along rails at speed v through a field B perpendicular to their plane, and the free charges in the rod feel q\mathbf{v}\times\mathbf{B}. They pile up at the ends until the electrostatic field they build balances that push, leaving \varepsilon = BLv across the rod. It is Faraday's law wearing a different hat: the loop's area grows at Lv, so d\Phi/dt = BLv.

Close the loop and the emf drives I = BLv/R. That current, sitting in the same field, feels F = BIL directed against the motion. Nothing here is free: whoever keeps the rod moving at constant speed does work at precisely the rate the resistor turns into heat.

Figure. The rod is the source and the resistor is the load; the outer rectangle marks the region in which the field exists. With the field into the page and the rod moving right, the current is driven up the rod, so the force BIL on the rod points back along the rails — opposing v, never helping it. The rail separation is what L means in ε = BLv.

Where the energy goes

1. The rod is the sourceThe moving rod is the seat of the emf, BLv. Everything else in the loop is just resistance.
2. Current, then dragI = BLv/R flows, and the retarding force on the rod is F = BIL = B²L²v/R.
3. The books balanceThe agent supplies Fv = B²L²v²/R, which is exactly ε²/R. Stop pushing and the rod coasts to a halt.

3Self-inductance and the energy it stores

A coil's own current threads its own turns, and N\Phi = LI defines the self-inductance. Change the current and the coil induces an emf against itself, \varepsilon = -L\,dI/dt — the electrical analogue of inertia. For a long solenoid the geometry is the whole story, L = \mu_0 n^2 A l, so L is fixed by the turns per metre, the cross-section and the length, and not at all by the current.

Building the current up costs work, and that work sits in the field as U = \tfrac12 LI^2. Put a resistance in series and the current cannot jump; it climbs as I = (V/R)(1 - e^{-Rt/L}), reaching 63% of its final value after one time constant \tau = L/R. At the very first instant an inductor is an open circuit; long afterwards it is a plain piece of wire.

Figure. Two runs of the same circuit, the second with the resistance doubled. Doubling R halves the final current V/R and halves the time constant L/R at once, so the dashed curve settles twice as low but gets there twice as fast, reaching 63% of its own final value in half the time. Both curves leave the origin on the same slope V/L, which the inductor alone decides.

How it works

1. Flux linkage defines LNΦ = LI. Work out the flux the coil's own current makes, divide by that current, and L drops out of the geometry alone.
2. Changing I costs a back-emfε = −L dI/dt opposes the change, so a coil resists being switched on and resists being switched off just as stubbornly.
3. The work done is storedIntegrating εI dt gives U = ½LI², held in the field inside the coil and handed back when the current dies.

4Why AC is quoted as rms

Over a full cycle the mean of i = i_0\sin\omega t is exactly zero, so the average current tells you nothing — and yet the wire still gets hot. Heating goes as i^2, and i^2 is positive whichever way the charge happens to be moving. Averaging the square and taking the root gives the one number that behaves like a DC current: i_{rms} = i_0/\sqrt2, and likewise V_{rms} = V_0/\sqrt2.

The factor \sqrt2 belongs to the sine wave and to nothing else — a square wave has i_{rms} = i_0. Every ammeter, every voltmeter and every mains rating you will meet is an rms value, so a 220 V supply peaks at 311 V and swings between +311 V and -311 V a hundred times a second.

Figure. One and three-quarter cycles of the current, with its square drawn over it. The current spends as long negative as positive and averages to zero; the square is never negative and averages to exactly half its peak. The root of that half is the 0.707 that turns a peak into an rms value.

Taking the rms of any waveform

1. Square itSquaring throws away the sign, which is the entire point: heat does not care which way the current went.
2. Average over a whole cycleThe mean of sin²ωt over a complete cycle is exactly ½, whatever the frequency happens to be.
3. Take the root√(i₀²/2) = i₀/√2 ≈ 0.707 i₀. For a square wave the mean of the square is i₀², so the rms is i₀ itself.

5Reactance: how L and C read the frequency

A resistor does not care how fast the current alternates; an inductor and a capacitor do almost nothing else. The inductor's opposition is X_L = \omega L, which grows without limit as the frequency rises, and the capacitor's is X_C = 1/(\omega C), which collapses. Both are in ohms and both are used exactly like a resistance: V = IX, with rms paired to rms or peak to peak.

Take the two extremes. At \omega \to 0, which is DC, X_L \to 0 and the inductor is a plain wire, while X_C \to \infty and the capacitor is a gap in the circuit. At very high frequency the roles swap exactly. That is the whole of filtering: an inductor in series blocks the high frequencies, a capacitor in series blocks DC.

Figure. Drawn for L = 2 mH and C = 8 μF. The inductor's reactance is a straight line through the origin, the capacitor's a hyperbola, and they cross at ω₀ = 7.9 × 10³ rad s⁻¹ where each is 15.8 Ω. Left of the crossing the capacitor dominates the pair and right of it the inductor does — which is precisely what makes a series LCR circuit capacitive below resonance and inductive above it.

6The ninety degrees, and wattless current

Reactance says how big the current is; it says nothing about when. Across a pure inductor v = L\,di/dt, so the voltage is largest exactly where the current is passing through zero, and the current reaches its own peak a quarter of a cycle after the voltage does — the current lags by 90^\circ. In a pure capacitor i = C\,dv/dt and the same argument runs the other way, so the current leads by 90^\circ.

That quarter-cycle offset earns more marks than the offset itself. The instantaneous power vi is positive for a quarter cycle and negative for the next in equal measure, so the average power in a pure inductor or a pure capacitor is exactly zero. Energy shuttles into the field and back out; none of it is dissipated. A current like that is called wattless.

Figure. A pure inductor across a sinusoidal supply, over two full periods. The voltage crest comes first and the current crest follows a quarter of a period behind it — compare the two peaks rather than the zero crossings and the lag is unmistakable. Wherever one curve is positive while the other is negative, the inductor is returning energy to the supply, and over a whole cycle it returns exactly as much as it took.

How it works

1. Voltage follows the slopeIn an inductor v = L di/dt, so the voltage peaks where the current curve is steepest — at the current's zero crossings.
2. Peaks a quarter cycle apartA sine and its derivative are 90° apart, and 90° of phase is a quarter of a period along the time axis.
3. The power averages awayvi is a sine times a cosine, that is a sine of twice the frequency, and its mean over a whole cycle is zero.

7Series LCR: impedance, phase and power

In series the same current passes through all three elements, so the current is the natural reference and the three voltages differ only in phase. V_R is in step with it, V_L is a quarter cycle ahead and V_C a quarter cycle behind, so V_L and V_C are 180^\circ apart and partly cancel. Combining the surviving reactance with R at right angles gives Z = \sqrt{R^2 + (X_L - X_C)^2} and \tan\phi = (X_L - X_C)/R, where \phi is the angle by which the supply voltage leads the current.

Power is where the phase pays. Only the resistor dissipates anything, so P = V_{rms}I_{rms}\cos\phi with the power factor \cos\phi = R/Z. When X_L > X_C the circuit is inductive and \phi is positive; when X_C wins, \phi is negative and the current leads instead.

Figure. The impedance triangle for the worked example, drawn to scale: 40 Ω of resistance along the reference direction, 30 Ω of net reactance at right angles to it, and the 50 Ω impedance closing the triangle at 37° — an angle that really does measure 37° on the page, once the box's width-to-height ratio has been allowed for. Multiply every side by the current, 4 A, and the same triangle becomes the voltage triangle: 160 V, 120 V and 200 V.

Solving any series LCR circuit

1. Reactances firstWork out X_L = ωL and X_C = 1/(ωC) at the frequency you were actually given, not at 50 Hz by habit.
2. Then the impedanceZ = √(R² + (X_L − X_C)²). Subtract the two reactances before squaring, never after.
3. Current, then phaseI = V/Z gives the size, tan φ = (X_L − X_C)/R the timing, and cos φ = R/Z the power.

8Series resonance

X_L climbs with frequency and X_C falls, so somewhere they must be equal. That point is resonance, at \omega_0 = 1/\sqrt{LC}. There the reactive parts cancel outright: the impedance collapses to its smallest possible value Z = R, the current peaks at V/R, and it runs exactly in phase with the supply, so the power factor is 1 and the circuit behaves as a bare resistor.

The voltages across L and C do not vanish at resonance. They are equal and opposite, each of size IX_L, and each can be several times the supply voltage. Their ratio to the supply is the quality factor Q = X_L/R = (1/R)\sqrt{L/C}, which is also how sharply the current peak is tuned: a small R gives a tall narrow peak, a large R a low broad one.

Figure. Both curves are the same L and C with only R changed, so both peak at exactly the same ω₀ = 7.9 × 10³ rad s⁻¹. Cutting R from 20 Ω to 5 Ω raises the peak current four-fold and narrows it at the same time — the height of the peak and the sharpness of the tuning are the same quantity, Q. Far from resonance the two curves converge, because out there the impedance is set by the reactance rather than by R.

How it works

1. Set X_L equal to X_CωL = 1/(ωC) gives ω₀ = 1/√(LC), an expression with no R anywhere in it.
2. Read off the consequencesZ = R is a minimum, I = V/R is a maximum, φ = 0 and the power factor is 1.
3. Then check the magnificationV_L = V_C = QV with Q = (1/R)√(L/C), which is often several times the supply voltage.

9Mutual inductance and the transformer

Set two coils near one another and a current in the first threads flux through the second: N_2\Phi_2 = MI_1 defines the mutual inductance, and changing the first current induces \varepsilon_2 = -M\,dI_1/dt in the second. M depends only on the geometry and on how well the two are coupled, and it comes out the same number whichever coil you choose to drive.

A transformer is that idea with the coupling pushed as close to perfect as it can be got, both windings sharing one iron core. Every turn then links the same flux, so the volts per turn are common to both windings and V_s/V_p = N_s/N_p. An ideal transformer creates no energy, so V_pI_p = V_sI_s and the current ratio is the turns ratio upside down: step the voltage up and you step the current down by the same factor.

Figure. Ideal transformer: V_p/V_s = N_p/N_s. Here a 4:1 turns ratio steps 240 V down to 60 V. Mutual inductance is why a changing primary current drives the secondary at all.

Why a transformer needs AC

1. Only a changing flux does anythingFeed a transformer with DC and, once the switch-on transient has died, dΦ/dt is zero and so is the output.
2. One flux, two windingsA closed iron core keeps essentially all the flux inside it, so every turn of both windings sees the same dΦ/dt.
3. Energy is conserved, not voltageV_pI_p = V_sI_s for an ideal transformer. Real ones lose a few per cent to copper heating, eddy currents and hysteresis.