Electromagnetic Induction and Alternating Currents

A coil of $N$ turns and resistance $R$ sits in a magnetic field. You pull it out of the field — first gently over ten seconds, then with a violent jerk lasting a hundredth of a second. How much charge flows round the circuit each time?

Exactly the same amount. The speed of the pull is irrelevant:

$q=\int idt=\int \frac{N}{R}\frac{d\Phi }{dt}dt=\frac{N\Delta \Phi }{R}$

The time cancels out of the integral entirely. A faster pull produces a larger current for a shorter time, and the two effects compensate exactly.

This is why a ballistic galvanometer can measure a magnetic field by snatching a search coil out of it, and why the timing of the snatch never has to be recorded.

The pattern repeats through the chapter. Main gives $\epsilon =BvL$, one inductor and a formula for reactance. Advanced gives a rod whose speed varies along its length, a circuit caught mid-transient, and an AC network where the only sane approach is a phasor triangle.

1. Motional emf by integration

The general statement is

$\epsilon =\int \left(\vec{v}\times \vec{B}\right)\cdot d\vec{l}$

and $\epsilon =BvL$ is merely the case where $v$, $B$ and $L$ are mutually perpendicular and constant. When $v$ varies along the conductor, the integral is unavoidable.

A rod rotating about one end at angular speed $\omega$ in a perpendicular field has $v=\omega x$ at distance $x$, so

$\epsilon ={\int }_0^{L}B\left(\omega x\right)dx=\frac{1}{2}B\omega L^2$

not $B\omega L^2$, which is what using the tip speed would give. The correct value uses the speed of the midpoint, which is the average.

A rod sliding on rails through a closed circuit of resistance $R$ obeys

$i=\frac{BLv}{R},F_{retarding}=\frac{B^2{L}^{2}v}{R}$

so pushing it with a constant force gives a terminal velocity $v_T=FR/B^2{L}^2$, and releasing it gives an exponential decay with time constant $mR/B^2{L}^2$.

Illustration 1

A rod of length $0.5$ m rotates at $20$ rad s${}^{-1}$ about one end in a field of $0.4$ T perpendicular to its plane of rotation. Find the emf between its ends.

$\epsilon =\frac{1}{2}B\omega L^2=\frac{1}{2}\left(0.4\right)\left(20\right)\left(0.25\right)$

$\epsilon =1.0$ V

Using the tip speed of $10$ m s${}^{-1}$ would give $2.0$ V, double the truth. The factor of two is the difference between the maximum and the average of a quantity that grows linearly.

Illustration 2

A rod of mass $m$ slides without friction on horizontal rails of separation $L$ in a vertical field $B$, closed through resistance $R$. It is given an initial speed $v_0$. Find how far it travels before stopping.

$m\frac{dv}{dt}=-\frac{B^2{L}^{2}v}{R}$, and writing $\frac{dv}{dt}=v\frac{dv}{dx}$:

$mdv=-\frac{B^2{L}^2}{R}dx\Rightarrow x=\frac{m{v}_{0}R}{B^2{L}^2}$

The rod never formally stops, yet travels a finite distance. The velocity decays exponentially in time but the displacement converges, which is the same structure as a body in a viscous fluid.

2. Inductance from flux linkage

Self-inductance is defined by $N\Phi =LI$, and computing it means finding the flux the current produces through its own turns:

$L_{solenoid}=\frac{{\mu }_0{N}^{2}A}{l}={\mu }_0{n}^{2}Al,L_{coax}=\frac{{\mu }_{0}l}{2\pi }\ln \frac{b}{a}$

For two coils, $M=k\sqrt{L_1{L}_2}$ where the coupling coefficient $k$ lies between $0$ and $1$. Placed in series they give

$L=L_1+L_2\pm 2M$

with the plus sign when their fluxes reinforce and the minus sign when they oppose. Winding two identical coils in opposition on the same core gives $L\approx 0$ — the principle of a non-inductive resistor.

Illustration 3

A solenoid of $500$ turns, length $0.4$ m and cross-section $4\times 10^{-4}$ m${}^2$ carries $2$ A. Find its inductance and stored energy.

$L=\frac{{\mu }_0{N}^{2}A}{l}=\frac{\left(4\pi \times 10^{-7}\right)\left(2.5\times 10^5\right)\left(4\times 10^{-4}\right)}{0.4}$

$L=3.14\times 10^{-4}$ H

$U=\frac{1}{2}L{I}^2=\frac{1}{2}\left(3.14\times 10^{-4}\right)(4)=6.3\times 10^{-4}$ J

Inductance goes as $N^2$, not $N$, because doubling the turns both doubles the field and doubles the number of turns that field threads.

Illustration 4

Two coils of $8$ mH and $2$ mH are connected in series, and the combination measures $14$ mH one way round and $6$ mH the other. Find $M$ and the coupling coefficient.

$L_1+L_2=10$ mH, so $2M=14-10=4$ mH and $M=2$ mH. The other connection gives $10-4=6$ mH, as observed.

$k=\frac{M}{\sqrt{L_1{L}_2}}=\frac{2}{\sqrt{16}}=0.5$

Half the flux of each coil reaches the other. Perfect coupling, $k=1$, needs a closed magnetic circuit, which is exactly what a transformer's iron core provides.

3. LR transients: the mirror image of a capacitor

An inductor resists changes in current, so its transient behaviour is the exact opposite of a capacitor's:

$i=\frac{\epsilon }{R}\left(1-e^{-Rt/L}\right),\tau =\frac{L}{R},U=\frac{1}{2}L{I}^2$

At $t=0$ the current cannot jump, so an inductor carrying no current behaves as an open circuit. At $t\to \infty$ the current is steady, so there is no emf across it and it behaves as a short circuit. A capacitor does precisely the reverse, and remembering the pairing is worth several marks.

Illustration 5

A coil of inductance $0.2$ H and resistance $5\Omega$ is connected to a $10$ V battery. Find the time constant, the initial rate of rise of current, and the final energy stored.

$\tau =\frac{L}{R}=\frac{0.2}{5}=0.04$ s

At $t=0$ the whole emf appears across the inductor: $\frac{di}{dt}=\frac{\epsilon }{L}=\frac{10}{0.2}=50$ A s${}^{-1}$

Final current $2$ A, so $U=\frac{1}{2}\left(0.2\right)(4)=0.4$ J.

The energy is stored in the magnetic field, at a density $B^2/2{\mu }_0$ — the exact magnetic counterpart of $\frac{1}{2}{\epsilon }_0{E}^2$.

4. Lenz's law is energy conservation

Lenz's law is not an extra rule about signs. It is the statement that induction cannot create energy, and it always resolves into the same instruction: the induced effect opposes the change that produced it.

Three standard consequences follow.

A magnet dropped down a copper pipe falls slowly. The changing flux drives circular currents in the pipe wall, which oppose the magnet's approach below and its departure above. The magnet reaches a terminal speed at which the gravitational power input exactly equals the heat dissipated in the pipe. Slit the pipe lengthwise and the circulating path is broken, so it falls freely.

A ring placed over the core of an AC solenoid jumps off. The induced current opposes the growing flux, so ring and coil momentarily behave as antiparallel currents and repel. A slit ring simply sits there.

Opening an inductive circuit produces a spark. The current cannot fall instantly, so as the contacts separate the inductor generates a back emf $Ldi/dt$ large enough to break down the air gap. That is why inductive loads need a diode or a snubber across them.

Illustration 6

A rod of mass $m$ and length $L$ slides down frictionless rails inclined at $\theta$, in a vertical field $B$, with the circuit closed by resistance $R$. Find the terminal velocity.

At terminal velocity the component of gravity along the incline balances the magnetic retarding force. The flux-cutting component of the field is $B\cos \theta$:

$mg\sin \theta =\frac{{\left(B\cos \theta \right)}^2{L}^2{v}_T}{R}$

$v_T=\frac{mgR\sin \theta }{B^2{L}^2{\cos }^2\theta }$

Check the energetics: at that speed the rate of loss of gravitational potential energy, $mg{v}_T\sin \theta$, equals $i^{2}R$ exactly. Nothing is left over to accelerate the rod.

Illustration 7

A $0.5$ H inductor carrying $3$ A is disconnected by a switch that interrupts the current in $1$ ms. Estimate the back emf across the contacts.

$\epsilon =L\frac{di}{dt}=0.5\times \frac{3}{10^{-3}}$

$\epsilon =1500$ V

From a supply that may have been only $12$ V. The inductor's stored energy of $\frac{1}{2}L{I}^2=2.25$ J has nowhere to go but the arc, which is why the contacts of relays and motor switches burn.

5. LC oscillation: an electrical pendulum

Connect a charged capacitor across an inductor and the charge oscillates:

$q=q_0\cos \omega t,i=-q_0\omega \sin \omega t,\omega =\frac{1}{\sqrt{LC}}$

The correspondence with mechanics is exact: $q$ plays the part of displacement, $i$ of velocity, $1/C$ of the spring constant and $L$ of the mass. Energy sloshes between $\frac{q^2}{2C}$ in the capacitor and $\frac{1}{2}L{i}^2$ in the inductor, with the total constant. Adding resistance damps it exactly as friction damps a pendulum.

Illustration 8

A $2\mu$F capacitor charged to $50$ V is connected across a $50$ mH inductor. Find the oscillation frequency and the maximum current.

$\omega =\frac{1}{\sqrt{\left(0.05\right)\left(2\times 10^{-6}\right)}}=\frac{1}{10^{-3.5}}=3162$ rad s${}^{-1}$, so $f=503$ Hz

Energy conservation gives $\frac{1}{2}L{i}_{max}^2=\frac{1}{2}C{V}^2$:

$i_{max}=V\sqrt{\frac{C}{L}}=50\sqrt{\frac{2\times 10^{-6}}{0.05}}=0.316$ A

The quantity $\sqrt{L/C}$ has units of resistance and is called the characteristic impedance. It is the ratio of peak voltage to peak current in the oscillation.

6. Alternating current as a phasor triangle

For a sinusoidal source, represent each voltage as a rotating vector. Because the current is common to a series circuit, the three voltages sit at fixed angles relative to it: $V_R$ in phase, $V_L$ ahead by $90^{\circ }$, $V_C$ behind by $90^{\circ }$. Adding them as vectors gives

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

with $X_L=\omega L$ and $X_C=1/\omega C$.

Because $V_L$ and $V_C$ are antiphase, they subtract. A consequence that startles most students is that either one can individually be far larger than the applied voltage, and at resonance both usually are.

Illustration 9

A series circuit has $R=30\Omega$, $L=0.1$ H and $C=50\mu$F across $200$ V at $50$ Hz. Find $Z$, the current, and the voltage across the inductor.

$X_L=2\pi \left(50\right)\left(0.1\right)=31.4\Omega$; $X_C=\frac{1}{2\pi \left(50\right)\left(5\times 10^{-5}\right)}=63.7\Omega$

$Z=\sqrt{900+{\left(-32.3\right)}^2}=\sqrt{900+1043}=44.1\Omega$

$I=\frac{200}{44.1}=4.54$ A, so $V_L=4.54\times 31.4=143$ V

The circuit is capacitive here, since $X_C>X_L$, so the current leads the applied voltage — and it would lag if the frequency were raised past resonance.

7. Resonance, quality factor and power

At ${\omega }_0=1/\sqrt{LC}$ the reactances cancel, $Z=R$ is minimum, the current is maximum and the circuit is purely resistive. The sharpness of that peak is measured by

$Q=\frac{{\omega }_{0}L}{R}=\frac{1}{R}\sqrt{\frac{L}{C}}=\frac{{\omega }_0}{\Delta \omega }$

where $\Delta \omega =R/L$ is the bandwidth between the half-power points. A high $Q$ means a narrow, tall resonance — what a radio tuner needs to separate stations.

Average power in an AC circuit is

$P=V_{rms}{I}_{rms}\cos \varphi$

The factor $\cos \varphi =R/Z$ is the power factor. A purely reactive circuit has $\cos \varphi =0$ and carries a wattless current — real current, real heating in the wires, but zero average power delivered to the load. Industrial installations correct their power factor for exactly this reason.

Illustration 10

A series circuit has $R=10\Omega$, $L=0.5$ H and $C=20\mu$F. Find the resonant frequency, the quality factor and the bandwidth.

${\omega }_0=\frac{1}{\sqrt{\left(0.5\right)\left(2\times 10^{-5}\right)}}=\frac{1}{10^{-2.5}}=316$ rad s${}^{-1}$, so $f_0=50.3$ Hz

$Q=\frac{{\omega }_{0}L}{R}=\frac{316\times 0.5}{10}=15.8$

$\Delta \omega =\frac{R}{L}=20$ rad s${}^{-1}$, and $\frac{{\omega }_0}{\Delta \omega }=15.8$, confirming $Q$.

At resonance $V_L=V_C=Q$ times the applied voltage. With $Q=15.8$, a $100$ V supply puts $1580$ V across the inductor — which is a genuine hazard in resonant circuits.

Illustration 11

A load draws $10$ A at $230$ V with a power factor of $0.6$ lagging. Find the real power, and the capacitance needed at $50$ Hz to correct the power factor to unity.

$P=230\times 10\times 0.6=1380$ W

Reactive power $=230\times 10\times 0.8=1840$ VAR, which the capacitor must cancel:

$C=\frac{Q_{reactive}}{\omega V^2}=\frac{1840}{2\pi \left(50\right){\left(230\right)}^2}=1.1\times 10^{-4}$ F

The real power is unchanged by the correction. What falls is the current drawn, and with it the heating loss in the supply cables — which is what the customer is charged for.

8. Transformers and eddy currents

An ideal transformer conserves power, so

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

Real losses come from four sources: copper loss ($I^{2}R$ in the windings), flux leakage, hysteresis in the core, and eddy currents. The last is why cores are built from thin laminations: the power dissipated by eddy currents goes as the square of the lamination thickness, so slicing a core into ten sheets cuts the loss by a factor of a hundred.

Illustration 12

A step-down transformer converts $2200$ V to $220$ V with $95\%$ efficiency and delivers $10$ A. Find the primary current.

Output power $=220\times 10=2200$ W

Input power $=\frac{2200}{0.95}=2316$ W

$I_p=\frac{2316}{2200}=1.05$ A

A perfect transformer would draw exactly $1$ A, and the extra $5\%$ is the loss. Note that voltage steps down while current steps up, which is why transmission lines run at high voltage and low current.

Illustration 13

A solid iron core is replaced by one made of $20$ laminations of the same total cross-section. By what factor does the eddy-current loss fall?

Eddy loss $\propto t^2$, and each lamination is $\frac{1}{20}$ as thick.

Loss per lamination falls by $400$, but there are $20$ of them, so the total falls by

$\frac{400}{20}=20$ times

Laminating is not merely helpful, it is essential. Without it, a mains transformer would waste most of its input as heat in the core.

Illustration 14

A coil of $200$ turns and resistance $40\Omega$ is pulled out of a field of $0.5$ T. Its area is $2\times 10^{-3}$ m${}^2$. Find the charge that flows.

$\Delta \Phi =BA=0.5\times 2\times 10^{-3}=10^{-3}$ Wb

$q=\frac{N\Delta \Phi }{R}=\frac{200\times 10^{-3}}{40}=5\times 10^{-3}$ C

No time appears anywhere in the calculation, which is exactly why a ballistic galvanometer works: it integrates the current pulse and reports only the total charge.

Summary

• Charge flowed depends only on flux change: $q=\frac{N\Delta \Phi }{R}$, independent of how fast the change happened.
• General motional emf $\epsilon =\int \left(\vec{v}\times \vec{B}\right)\cdot d\vec{l}$; $BvL$ is only the constant-$v$ special case.
• Rod rotating about one end: $\epsilon =\frac{1}{2}B\omega L^2$ — the average speed, not the tip speed.
• Rod on rails: $i=\frac{BLv}{R}$, retarding force $\frac{B^2{L}^{2}v}{R}$, terminal velocity $\frac{FR}{B^2{L}^2}$, stopping distance $\frac{m{v}_{0}R}{B^2{L}^2}$.
• $L=\frac{{\mu }_0{N}^{2}A}{l}$ for a solenoid — quadratic in $N$; $M=k\sqrt{L_1{L}_2}$ with $0\le k\le 1$.
• Coils in series: $L=L_1+L_2\pm 2M$, the sign depending on whether the fluxes reinforce.
• Inductor transients are the mirror of a capacitor's: open at $t=0$, short at $t\to \infty$, with $\tau =L/R$ and $U=\frac{1}{2}L{I}^2$.
• Lenz's law is energy conservation: a magnet falls slowly down a copper pipe, and a ring on an AC solenoid jumps off.
• Opening an inductive circuit forces a huge $Ldi/dt$, so the back emf can be a hundred times the supply voltage.
• $LC$ oscillation is SHM with $\omega =1/\sqrt{LC}$: $q$ for displacement, $i$ for velocity, $L$ for mass, $1/C$ for stiffness.
• $Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2}$, $\tan \varphi =\frac{X_L-X_C}{R}$; $V_L$ and $V_C$ subtract and either can exceed the applied voltage.
• Resonance at ${\omega }_0=1/\sqrt{LC}$: $Z$ minimum, current maximum, circuit purely resistive.
• $Q=\frac{{\omega }_{0}L}{R}=\frac{1}{R}\sqrt{\frac{L}{C}}=\frac{{\omega }_0}{\Delta \omega }$, and at resonance $V_L=V_C=Q$ times the supply voltage.
• $P=V_{rms}{I}_{rms}\cos \varphi$; a purely reactive circuit carries a wattless current delivering zero average power.
• Transformer: $\frac{V_s}{V_p}=\frac{N_s}{N_p}=\frac{I_p}{I_s}$; eddy loss goes as the square of lamination thickness.