Electromagnetic induction: from circuits to generators

01 Start with a wire and a battery

Look at the battery, switch and resistor on the bench below. The leads join them into a route that can be closed or broken. The wire already contains charged particles: a metal contains electrons, and some can move through it. The battery does not manufacture a new batch of electrons or wait for one electron to travel all the way to the resistor before it works. Closing the route establishes an electrical influence along the circuit, called an electric field; the mobile charges respond to it.

Charge is a property that determines how a particle participates in electrical interactions. There are positive and negative charges. An electron has negative charge. We measure an amount of charge with the coulomb, abbreviated C, and use the symbol $Q$ for that amount. A coulomb contains the charge magnitude of a very large number of electrons; the markers in our drawings represent portions of charge, not individual electrons drawn to scale.

Choose a cross-section of the wire, as if we had drawn a counting line across it. If an amount $Q$ passes in a time interval $\Delta t$, the average current is the amount per unit time. This is the definition of average current, not a new experimental law. The symbol $\Delta$ means “change” or “interval”. One ampere, A, means one coulomb passing per second.

We conventionally point current in the direction positive charge would move. Mobile electrons in a metal drift in the opposite direction because they are negative. We will use conventional current consistently, including every direction arrow below.

Current tells us how much charge passes. We also need to know how much energy can be transferred per unit charge. A battery providing $2\,\text{J}$ for each $1\,\text{C}$ provides $2\,\text{V}$: voltage is measured in volts, V. A joule, J, is a unit of energy. Write transferred energy as $W$ and charge as $Q$. Energy delivered per charge and charge delivered per second answer different questions.

A resistor transfers electrical energy into thermal energy. For the simple resistor we use here, measurements give a current proportional to its voltage when temperature stays fixed. Its resistance $R$ is the ratio $V/I$ and is measured in ohms, $\Omega$. Rearranging $R=V/I$ gives $RI=V$, then $I=V/R$. This behaviour is called Ohm’s law; not every electrical device follows it.

We have now separated the charge already present, the energy supplied per charge and the current through a complete path. Keep that distinction. Next we ask whether a battery is the only way to supply this electrical driving action. We will need to describe a magnet before we can test that question.

02 How we describe a magnetic field

A magnet can affect another magnet without touching it. To describe the influence at different locations, we assign a magnetic field to each location. A field is a way of recording a local physical condition: which direction it points and how strong it is there. The arrows are our representation of that condition; they are not little material rods filling the space.

A small compass helps assign direction. Its north-seeking end points along the local field. Outside a bar magnet, field arrows leave its north pole and enter its south pole; the pattern continues through the magnet as closed field lines. We use $B$ for magnetic field magnitude and tesla, T, for its unit. When we say a field is uniform, we mean it has the same direction and magnitude throughout the region under discussion.

A field line is a drawing aid. Drawing twice as many lines on the same diagram does not physically double the field. We show field strength through an explicit numerical value; line density or darkness is only a visual guide. This matters because we are about to measure how a field passes through a chosen surface, and literal counting of a few drawn lines would give a misleading answer.

Electric current also produces a magnetic field. A current around a loop makes a field through its interior. A coil is a wire wound into several loops, called turns. Therefore a coil can respond to an external magnetic field and also create a field of its own. We will keep those two fields distinct when we determine the direction of induction.

03 One surface, a chosen direction, and flux

The copper loop below has an open middle. Turn on “Opening surface”: a translucent sheet fills exactly the opening bounded by that same wire. This is a drawing aid we have added, not a membrane in the apparatus. Turn it off and the wire remains unchanged. Why add it? We want to compare how a magnetic field passes through the whole opening, so we need to identify the area we are comparing.

Start with a uniform field passing straight through the sheet. For this perpendicular crossing, we define magnetic flux $\Phi$ as field strength multiplied by area $A$. Doubling the area doubles the amount of surface exposed to the same field. Doubling field strength doubles the flux through the same area. Flux is measured in webers, Wb.

Now rotate the copper frame. Its translucent opening surface turns with it. The blue field arrows keep their direction. The green arrow stays perpendicular to the surface; we call this a normal. It provides a definite direction to compare with the field, even when the face turns edge-on. There are two opposite normals; we choose one and keep tracking it. The angle between this chosen normal and the field is called theta, written $\theta$.

We can obtain the crossing area by projecting the sheet onto a plane perpendicular to the field. For a rectangular sheet rotated about one edge, that edge keeps its length $a$. Its other edge, length $b$, projects to $b\cos\theta$. Here cosine is the signed projection ratio: in a right triangle it is adjacent side divided by hypotenuse. Thus the original area is $ab$, while the signed projected area is $a(b\cos\theta)=A\cos\theta$. We label this signed projected area $A_{\perp}$. Multiply that crossing area by the uniform field to obtain the flux.

When the normal points with the field, $\theta=0^\circ$ and $\cos\theta=1$: flux is $BA$. When the field lies in the surface, $\theta=90^\circ$ and the projected crossing area is zero. When the normal points against the field, $\theta=180^\circ$ and flux is $-BA$. We have not lost the field; the minus sign preserves its direction relative to our choice.

For a field that varies across a surface, one product is not enough. Divide the surface into small patches. On each patch use its local field, area and normal, then add the signed contributions. The integral symbol is compact notation for the limit of these finer sums. Our experiments use a flat loop and a uniform field where the simple product is exact.

The index $i$ labels a patch; $\Delta A_i$ is its area. In the integral, $S$ names the surface, $dA$ is a tiny area contribution and $\hat{\mathbf n}$ is its unit normal, an arrow of length one specifying direction. The dot between the vectors means take the field component along that normal. No new physical law has been added here: this extends our definition of flux.

04 From a changing flux to an emf

We can now describe the magnetic situation through the loop. But that still leaves the question that started the lesson: can this arrangement drive charge without a battery? Compare the instrument while the field is held steady and while the field is changing. Before calculating its response, let us recall what the battery supplied: energy for each amount of charge, rather than new charge.

Suppose a source can supply 0.08 joule for each coulomb taken around a circuit. Dividing energy by charge gives 0.08 joule per coulomb, or 0.08 volt. This energy-per-charge driving quantity is called electromotive force, shortened to emf and written $\varepsilon$. The historical name contains “force”, but this is not a mechanical force measured in newtons. With a battery, chemical processes supply the energy; with induction, a changing magnetic field can provide the electrical driving action. A conducting path is still needed for a sustained current.

For a stationary loop, induced emf is the work the induced electric field would do per unit positive charge taken once around the loop. This is a whole-loop quantity. In a changing field it cannot in general be replaced by one path-independent voltage between two points. An open circuit can therefore have induced emf without a sustained current.

Faraday’s experiments established the connection we need: changing the magnetic flux through a circuit produces an emf around it. This is an experimental physical law. We will calculate its consequences, but the definition of flux and a drawing of field arrows alone cannot prove the law.

Let $\Phi_1$ be the flux at the beginning and $\Phi_2$ the flux at the end. Their change is final minus initial, $\Delta\Phi=\Phi_2-\Phi_1$. Dividing by the elapsed time gives the average rate of change. For a coil with $N$ identical turns connected in series, each turn contributes the same emf, so the contributions add. Faraday’s law then gives the average emf shown below. We will interpret its minus sign in the next direction experiment.

Suppose one turn’s flux increases from $0.004\,\text{Wb}$ to $0.012\,\text{Wb}$ in $2\,\text{s}$. Subtract first: the change is $0.008\,\text{Wb}$. Divide by time: each second contributes an average increase of $0.004\,\text{Wb}$. Twenty series turns multiply the emf by twenty. Therefore its magnitude is $0.08\,\text{V}$. This is an average; if the change is uniform, it is also the emf throughout the changing interval.

If the flux changes unevenly, the whole-interval average hides when the change was fastest. Take a smaller time interval around the moment of interest and calculate the same quotient. When these quotients approach a definite value as the interval shrinks, that value is the instantaneous rate, written $d\Phi/dt$. We do not put zero in a denominator; we take a limit of nonzero intervals.

For example, suppose a prescribed changing field gives $\Phi(t)=kt^2$, with $k=0.001\,\text{Wb/s}^2$ and time measured in seconds. We can find its rate without skipping the algebra. Expanding the square gives $t^2+2t\Delta t+(\Delta t)^2$; subtracting the original $t^2$ cancels that term. Dividing by the nonzero interval leaves $2kt+k\Delta t$. The last term tends to zero, so the rate is $2kt$. The formula describes a chosen input, not a claim that every real magnet follows a square law.

At $t=2\,\text{s}$, the instantaneous rate is $2\times0.001\times2=0.004\,\text{Wb/s}$. With twenty turns, the instantaneous emf is $-0.08\,\text{V}$. Between two and three seconds, however, the average rate is $(0.009-0.004)/1=0.005\,\text{Wb/s}$, giving $-0.10\,\text{V}$. The two answers differ because the flux is changing faster as time increases.

05 Why a nearby magnet is not enough

Return to an ordinary magnet beside a coil. The magnet’s field at the coil becomes stronger as it approaches, so the flux changes. If we stop the magnet, a magnetic field is still present, but that field is no longer changing. The distinction is between having flux and changing flux. A large constant flux can give zero emf, while a smaller but rapidly changing flux can give a substantial emf.

Movement by itself is also insufficient. If a rigid closed loop translates entirely inside a uniform, steady field without rotating or changing area, its flux stays constant and its net induced emf is zero. Individual wire segments may have magnetic forces on their charges, but their contributions around the whole loop cancel. Moving into or out of a field region changes the area exposed to the field, so that is a different situation.

Likewise, we can leave every wire still and change the field using another coil whose current changes. Faraday’s law asks us to track the flux, not to search for a moving object at all costs. The three factors in $BA\cos\theta$ show three routes in the uniform-field case: change field strength, exposed area, or orientation.

06 Work out the direction with Lenz’s law

The magnitude calculation does not yet tell us which way current would flow. A current makes its own magnetic field, so its direction matters. Lenz’s law states that the induced emf acts in the direction whose current would oppose the change in flux. The word “change” carries the whole distinction: when the original flux is decreasing, the induced field helps maintain its original direction.

We can now read Faraday’s minus sign. Choose the normal toward yourself and pair it with counterclockwise positive circulation, using the right-hand rule. If outward flux increases, $\Delta\Phi$ is positive and circulation is clockwise. If outward flux decreases, $\Delta\Phi$ is negative, the emf becomes positive and circulation is counterclockwise. Changing our sign convention changes both labels, not the physical prediction.

Why is this opposition sensible? Suppose pushing a magnet into a closed resistive coil induced a current that helped the magnet move farther in. Motion would make more current; that current would accelerate the magnet; the resistor would also heat. This proposed self-amplifying process would create energy without a source. The observed opposing response instead makes the person pushing the magnet do work. Conservation of energy is a consistency check on the direction; the measured induction law supplies the quantitative relation.

Opposing a change does not mean perfectly cancelling it. Resistance, geometry and self-inductance determine how large the current becomes. A resistive coil usually cannot keep the original flux exactly unchanged. We will see the energy balance explicitly with a sliding rod.

07 What pushes the charge?

So far we have a reliable loop rule, but it is reasonable to ask what actually acts on the charge. There are two useful situations. A stationary wire in a changing magnetic field experiences an induced electric field. A wire moving through a magnetic field carries charges that experience magnetic forces. Both can contribute to the emf, and a real apparatus may contain both.

For the first situation, the electric field can circulate around a region of changing magnetic flux. It exists even if no wire is put there; a wire provides mobile charges with a conducting route. This is different from the electrostatic field of fixed charges, where the total work per charge around a closed path is zero. The changing-field case is precisely where that closed-path total can be nonzero.

For the moving-wire situation, use the experimentally established magnetic force law. A positive charge $q$ moving with speed $v$ perpendicular to a field $B$ experiences magnetic force magnitude $qvB$, perpendicular to both motion and field, given by the right-hand rule. The symbol $q$ refers to the charge we are following; it need not be a whole coulomb. An electric field $E$ exerts force $qE$ on the same positive charge. Below, $F_B$ labels magnetic force and $F_E$ labels electric force.

First imagine an isolated metal rod, with its length $\ell$, velocity and magnetic field mutually perpendicular. Magnetic forces redistribute the mobile electrons toward one end, leaving a positive charge imbalance at the other; the positive metal ions remain bound in the lattice. This charge separation builds an electric field in the rod. It grows until its electric force balances the magnetic separation force, giving the balancing field $E=vB$.

Why does this give a voltage? A constant electric force $qE$ acting over distance $\ell$ does work of magnitude $qE\ell$. Divide work by charge to obtain the potential difference magnitude $E\ell$. Therefore the open rod has an end-to-end potential difference $B\ell v$. In a rail circuit, the moving rod contributes the motional emf $B\ell v$; the complete circuit and its resistance then determine current.

Now verify the same magnitude from flux. During a short time $\Delta t$, a rod moving at constant speed advances a distance $v\Delta t$. The new rectangular strip has width $v\Delta t$ and height $\ell$, so its area is $\Delta A=\ell v\Delta t$. Multiply by the perpendicular uniform field to find $|\Delta\Phi|=B\ell v\Delta t$. Dividing by time cancels the nonzero $\Delta t$ and gives $B\ell v$ again. The local magnetic-force account and the loop-flux account agree.

In the diagram, motion is rightward and the applied field points into the screen. The positive magnetic force in the rod is upward, so current goes up the rod and counterclockwise around the loop. That current creates an outward field, opposing the increasing inward flux. We obtain the same direction by both methods.

08 Where the electrical energy comes from

An emf does not tell us that energy is free. Close the rod circuit and it carries a current. The applied field then exerts a force on the current-carrying rod, opposite its motion. To keep the speed constant, an external pull must balance that magnetic drag. We will derive the amount rather than merely attach an energy label to the animation.

To find the force on the current, suppose the rod contains $n$ mobile positive charge carriers per unit volume, each of charge $q$, drifting along the rod at speed $u$. Let the wire’s cross-sectional area be $a$. During time $\Delta t$, carriers in a length $u\Delta t$ cross a counting plane. Their number is $na u\Delta t$, their charge is $naq u\Delta t$, and dividing by time gives current $I=naqu$. This is a convenient equivalent positive-carrier model; using negative electrons gives the same conventional-current result.

A length $\ell$ of rod contains $na\ell$ such carriers. Each has a magnetic force magnitude $quB$ associated with its drift along the rod. Add these forces: $F=(na\ell)quB$. Group the factors $naqu$ as the current just obtained, giving magnetic drag force $F=B\ell I$ opposing motion.

Mechanical work from a steady pull over a distance $x$ is $Fx$. Divide by the time taken and use $x/\Delta t=v$ to obtain mechanical power $P_{\text{in}}=Fv$. Electrical energy transferred to charge $Q$ is $\varepsilon Q$; divide by time and use $Q/\Delta t=I$ to obtain electrical power $P_{\text{electric}}=\varepsilon I$. Power means energy transferred per second and is measured in watts, W.

Finally, for our resistive load $\varepsilon=IR$, so its heating power is $\varepsilon I=(IR)I=I^2R$. This proves the equality of input mechanical power and output heating in the ideal steady model. The energy comes from the person or machine pulling the rod, while the magnetic field enables the transfer.

There is a subtle point: a magnetic force is perpendicular to a charge’s total instantaneous velocity, so that force alone does no work on a point charge. In the moving conductor, charge separation, electric fields and the external mechanical constraint act together. It would therefore be misleading to describe the static magnetic field as a fuel that is consumed to power the resistor.

09 Turn the motion into a generator

A sliding rod cannot keep moving along a finite rail forever. A rotating coil solves the practical problem: it can return to its starting position and repeat. Its area and the external field stay constant, but the angle changes continuously, so its flux changes. A generator converts the mechanical work needed to maintain that rotation into electrical energy.

Angular speed $\omega$ tells us how quickly an angle changes. To define a radian, take a circular arc of length $x$ on radius $r$ and set its angle to $x/r$. Therefore $x=r\theta$ when the angle is in radians. Divide by time for uniform rotation: the speed of a point at radius $r$ is $v=r\omega$. Starting face-on, the angle at time $t$ is $\theta=\omega t$, and the previously derived flux rule gives flux $\Phi=BA\cos(\omega t)$.

We can find the emf without assuming you know the derivative of cosine. Take a rectangular coil rotating about an axis through its centre, parallel to its two long sides. Call each long side’s length $\ell$ and its distance from the axis $r$. Their separation is $2r$, so the enclosed area is $A=2r\ell$. Each long side moves at speed $r\omega$.

When the normal is aligned with the field, each long side is moving parallel to the field at that instant, so no magnetic separation force acts along it. As the coil turns by $\theta$, the side’s velocity turns through that angle relative to the field. Its component perpendicular to the field has magnitude $r\omega|\sin\theta|$: sine is the perpendicular projection ratio, opposite side divided by hypotenuse in a right triangle. This is the component that enters the rod result.

Each long side contributes emf magnitude $B\ell r\omega|\sin\theta|$. The two sides move oppositely, but they are also traversed in opposite directions when we go around the loop, so their loop contributions add. The short sides contribute no emf along themselves for this geometry. Two long sides and $N$ series turns therefore give $2NB\ell r\omega|\sin\theta|$. Replace $2r\ell$ by $A$, and use the sign convention consistent with decreasing positive flux at the start: the signed result is $NBA\omega\sin\theta$.

The maximum magnitude of sine is one, so the peak emf $\varepsilon_0$ is $NBA\omega$. More turns, a larger area, a stronger field or faster rotation can each increase it within the assumptions of this model. A half-turn reverses the sign of the emf. With a resistive load this gives a current that reverses periodically: alternating current, AC. Maintaining the rotation under load requires mechanical work; the generator is not an energy source independent of its drive.

10 Two coils and a transformer

We can also transfer energy between two separate coils. Drive an alternating current through the first coil, called the primary. Its changing magnetic field links a second coil, called the secondary, and induces an emf there. The charges in the primary wire do not have to jump into the secondary wire. Energy is transferred through the electromagnetic field.

For an ideal transformer, assume every turn of both windings links the same changing core flux, and ignore winding resistance, magnetic losses and leakage flux. Let the primary have $N_p$ turns and the secondary $N_s$ turns. Each turn has the same flux rate. Faraday’s law therefore gives $\varepsilon_p=-N_p d\Phi/dt$ and $\varepsilon_s=-N_s d\Phi/dt$. Dividing their nonzero amplitudes cancels the shared rate, giving voltage ratio equal to turns ratio: $V_s/V_p=N_s/N_p$.

The symbols $V_p$ and $V_s$ here represent effective AC voltages. “Effective” means a constant voltage that produces the same average heating in a resistor. Since the entire secondary waveform is the primary waveform multiplied by the same turn ratio, its effective value is multiplied by that ratio too. Coil winding direction also determines relative polarity; the displayed ratio compares magnitudes.

Doubling secondary voltage does not double available power for free. In a lossless transformer with a resistive load, average input and output powers are equal: $V_pI_p=V_sI_s$. Divide by the nonzero product $V_pI_s$ to get $I_p/I_s=V_s/V_p=N_s/N_p$. Thus a voltage increase is accompanied by an inverse current relation. Changing the load changes how much power the source must provide.

A steady current produces a steady core flux, so it cannot maintain secondary induction after the switching transient. This is why an ordinary transformer needs a changing input. Electronic power supplies may begin with a DC source and switch it rapidly, creating the changing current the transformer requires. Real devices also lose energy as heat and can saturate their cores, so the ideal ratios describe a useful limit, not a promise of lossless hardware.

11 A coil can induce an emf in itself

Until now, we often treated the induced current as emf divided immediately by resistance. But a coil’s own current makes magnetic flux. If its current changes, that self-produced flux changes too, inducing another emf in the same winding. This is self-induction. It explains why a coil can resist a change in current without preventing current itself.

Add the flux linked by all turns and call the result flux linkage $\lambda$. For identical turns, $\lambda=N\Phi$. In a fixed coil whose materials respond linearly, linkage is proportional to its current. We define the proportionality constant as inductance $L$: $\lambda=LI$. This proportionality is a model for that coil and operating range, not a universal rule for every magnetic material. Applying Faraday to the linkage and holding $L$ constant gives the self-induced emf.

Inductance is measured in henries, H. Rearranging the emf relation shows $1\,\text{H}=1\,\text{V}\cdot\text{s/A}$: an inductance of one henry corresponds to an emf magnitude of one volt when current changes by one ampere per second. Rising positive current gives a negative self-induced emf; falling positive current gives a positive emf. It opposes the change, exactly as Lenz’s law required.

Where does the extra work done against self-induction go? In an ideal inductor it is stored in the magnetic field. We can calculate it using small increments of current. To increase current, the source must overcome a voltage $V_L=L\,dI/dt$. In a short interval $dt$, charge $dQ=I\,dt$ passes. The small energy supplied is voltage times charge, so $dU=(L\,dI/dt)(I\,dt)=LI\,dI$. The time factors cancel; the energy depends on how far the current is increased.

To add these contributions from zero current to a final current $I_f$, plot $LI$ vertically against $I$ horizontally. Because $L$ is constant, the plot is a straight line from the origin to height $LI_f$. Each small energy is a narrow area under that line. All the areas together form a right triangle: half the rectangle with base $I_f$ and height $LI_f$. Therefore the stored energy is $U=\tfrac12 LI_f^2$. The factor one-half comes from building the field gradually; using final voltage or final current for every earlier increment would overcount.

When current decreases, the coil can return this stored energy to the circuit. Trying to stop current very quickly makes $|\Delta I/\Delta t|$ large, so the required voltage can be large. This is why practical circuits include paths or components that manage switching transients. The lesson’s steady resistor approximation is useful when self-induction has a negligible effect on the timescale being studied; it is not the full response of every real coil.

12 Use the chain of reasoning yourself

For each problem, keep the same order: identify the loop and choose its normal; find which part of the flux changes; calculate final minus initial and divide by time for an average; determine direction with Lenz’s law; calculate current if a circuit model is specified; and identify the energy source.