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$\mathcal{E} = -N\,d\Phi/dt$ and Lenz's law, motional emf $BLv$ and magnetic drag, induced electric fields, eddy currents, and generators.
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By the end of this lesson you will be able to find induced emfs and currents from changing flux, their directions, and the forces and power involved.
From the last lesson you can compute magnetic flux through loops and coils. From the circuits lessons you know emf and Ohm's law. This lesson connects them: a changing flux produces an emf, which drives current around a circuit. This single law generates nearly all the electric power in the world.
| Term | What it means |
|---|---|
| Induction | The production of an emf by a changing magnetic flux. |
| Faraday's law | $\mathcal{E} = -N\,d\Phi/dt$. |
| Lenz's law | An induced current flows so that its own field opposes the change in flux. |
| Motional emf | The emf $BLv$ in a conductor moving through a field. |
| Induced electric field | A field made by a changing magnetic flux, with $\oint \vec{E}\cdot d\vec{l} = -d\Phi/dt$. |
| Eddy currents | Swirling currents induced in solid conductors by changing flux. |
| Generator | A coil turned in a field to make an alternating emf $NBA\omega\sin\omega t$. |
Michael Faraday found in 1831 that a current appears in a circuit only while the magnetic flux through it is changing. Faraday's law makes this exact:
$$\mathcal{E} = -N\frac{d\Phi}{dt}.$$
The flux $\Phi = \int \vec{B}\cdot d\vec{A}$ can change because $B$ changes, the area changes, or the angle changes; any of them induces an emf. The minus sign is Lenz's law: the induced current flows in the direction whose own magnetic field opposes the change. Push a magnet's north pole toward a loop and the loop's current makes a north pole facing it, pushing back. For a rod of length $L$ sliding at speed $v$ across a field, the emf is
$$\mathcal{E} = BLv.$$
Another way: picture
Picture a loop that dislikes change in the field through it. Push a magnet in, and the loop sets up a current to push the flux back out. Pull the magnet away, and the loop's current tries to hold the flux in. It never stops the change, but it always resists it, and the faster the change, the harder it pushes back.
Another way: steps
In $\Phi = BA\cos\theta$, each factor can change. A transformer changes $B$: alternating current in one coil makes the field in the core rise and fall. A sliding rod changes $A$: the circuit's area grows at rate $Lv$, so $d\Phi/dt = BLv$. A generator changes $\theta$: the coil turns at $\omega$, so $\Phi = BA\cos\omega t$ and $\mathcal{E} = NBA\omega\sin\omega t$.
In every case the emf is the rate of change of flux linkage. A large, steady flux induces nothing; a small flux changing quickly can induce a large emf. The spark from a car's ignition coil comes from collapsing a modest field in a fraction of a millisecond.
The minus sign is required by energy conservation. Suppose a rod slides along rails and the induced current's force pushed it forward. It would speed up, inducing more current, pushing harder: a runaway that creates energy from nothing. Instead, the force on the induced current, $BIL$, always opposes the motion.
To keep the rod moving at speed $v$, you must push with $F = B^2L^2v/R$, doing work at the rate $Fv = B^2L^2v^2/R$. That is exactly the power $\mathcal{E}^2/R$ dissipated in the resistor. Mechanical work becomes electrical energy becomes heat, with nothing gained or lost, which is how every generator works.
A conductor moving through a field carries its charges with it, and each feels $q\vec{v} \times \vec{B}$. In a rod of length $L$ moving perpendicular to both itself and $\vec{B}$, this pushes charges along the rod, and the work per charge over its length is $vBL$. That is the motional emf, and it agrees exactly with $d\Phi/dt$ for the circuit the rod completes.
An airliner flying at $250$ m/s through the Earth's vertical field of about $40$ μT has a motional emf of about $0.4$ V across its $40$ m wingspan. No current flows, since there is no circuit, but the emf is there, and NASA's tethered satellite experiments used a $20$ km wire to generate thousands of volts this way.
When the flux changes because $B$ changes, with nothing moving, what pushes the charges? Faraday's law says a changing magnetic field creates an electric field that circles around it: $\oint \vec{E}\cdot d\vec{l} = -d\Phi/dt$. Unlike the fields of static charges, this field has closed loops and is not conservative.
Inside a long solenoid whose current is changing, the induced field circles the axis, with $E \cdot 2\pi r = \pi r^2\,dB/dt$, so $E = (r/2)\,dB/dt$ inside. A betatron accelerates electrons with exactly this field. The induced field exists in empty space, with or without a wire to carry current, and it is half of what makes light possible.
A changing flux through a solid piece of metal induces currents that swirl within it. By Lenz's law they oppose the change, and they dissipate energy as heat. Drop a strong magnet down a copper pipe and it falls slowly, braked by eddy currents in the pipe.
Eddy current brakes slow roller coasters and high-speed trains without contact. Induction cooktops heat iron pans with eddy currents driven by a coil beneath the glass. Transformer cores, where eddy currents would waste energy, are built of thin insulated layers of iron that cut the current paths short.
Checking an answer. Units: Wb/s is V. A steady flux must give zero emf. The induced current's force must oppose the motion that caused it. And the mechanical power put in must equal the electrical power dissipated.
Faraday's law is an experimental law, confirmed with enormous precision, but its two halves have different explanations. For a moving conductor, the emf is the magnetic force $q\vec{v} \times \vec{B}$ on its charges. For a stationary circuit in a changing field, it is the induced electric field. Einstein noted that which explanation applies depends on the observer's frame, a puzzle that led him to special relativity.
The flux rule holds for any surface bounded by the circuit, since the flux through any two such surfaces differs by the flux through a closed surface, which is zero.
Almost all electricity is generated by Faraday's law. In a power plant, steam, water or wind turns a rotor carrying electromagnets inside stationary coils. The flux through each coil rises and falls as the magnets pass, inducing an alternating emf at exactly $60$ Hz across the United States grid.
The generators at Grand Coulee Dam in Washington, the largest in the country, each produce several hundred megawatts. The turning force they need grows with the current drawn from them, by Lenz's law, so when more homes switch on lights, the turbines must push harder, and the grid operator dispatches more water or steam.
A rod falling on vertical rails through a horizontal field meets a drag force $B^2L^2v/R$ that grows in proportion to its speed, just like linear air drag. Its motion obeys $m\,dv/dt = mg - (B^2L^2/R)v$, the same equation as the drag lesson, with terminal speed $mgR/B^2L^2$ and time constant $mR/B^2L^2$.
At terminal speed, gravity's work goes entirely into heat in the resistor, at the rate $mgv$. This is the principle of the eddy current brakes on drop towers at amusement parks, where rows of magnets slow a falling car smoothly, with no contact and no wear, and more strongly the faster it falls.
Lenz's law turns direction questions into a short routine. First, find which way the flux through the loop points and whether it is growing or shrinking. Second, decide which way the induced field must point to oppose that change: against the flux if it is growing, along it if it is shrinking. Third, use the right-hand grip rule to find the current that makes that induced field.
For a north pole approaching a loop from the left, the flux through the loop points to the right and is growing. The induced field must point left, so, seen from the magnet's side, the current circulates counterclockwise, and the loop presents its own north pole to the magnet, repelling it. Pull the magnet away and every step reverses: the induced current attracts the retreating magnet.
The same routine works for sliding rods. As a rod moves to enlarge a loop in a field pointing into the page, the flux into the page grows, so the induced current must make a field out of the page inside the loop, which means a counterclockwise current. The force $I\vec{L} \times \vec{B}$ on the rod then points back against its motion, as energy conservation requires.
The wind turbines across Iowa, which gets more than half its electricity from wind, turn a rotor about $15$ times a minute. A gearbox or a direct-drive generator with many magnetic poles converts that slow rotation into the rapidly changing flux needed to make useful emf, which power electronics then match to the grid's $60$ Hz.
Lenz's law sets how hard the wind must push. The more current the grid draws, the larger the opposing torque on the rotor, and the turbine's controls adjust the blade pitch to capture just enough wind. In a storm the blades feather and the generator brakes the rotor electromagnetically.
An induction cooktop has a coil under the glass carrying current that alternates tens of thousands of times a second. Its changing field induces eddy currents in the iron base of a pan, heating the pan directly while the glass stays relatively cool. Aluminum and copper pans do not work well, because their low resistance and lack of iron change how the currents heat.
Wireless phone chargers use the same idea with a coil in the phone: the charging pad's alternating field induces an emf in the phone's coil, which is rectified to charge the battery. Electric toothbrushes, sealed against water, have charged this way for decades.
A loop inside an MRI magnet sits in a field of several tesla, yet no current flows in it while it rests. A strong field does not induce an emf; only a change in flux does. The loop feels an emf only as it is moved in or out, turned, or as the field is ramped.
A second error is to read Lenz's law as saying the induced field opposes the field. It opposes the change in flux: if the flux is decreasing, the induced field points the same way as the original, trying to keep the flux from falling.
A $200$-turn coil of area $0.0050$ m² faces a field rising steadily from $0.10$ T to $0.50$ T in $0.20$ s. Find the flux change per turn.
$\Delta\Phi = 0.0050 \times 0.40 = 2.0\ \text{mWb}$
$A\,\Delta B$.
Find the rate of change.
$\dfrac{\Delta\Phi}{\Delta t} = \dfrac{2.0 \times 10^{-3}}{0.20} = 0.010\ \text{Wb/s}$
Steady rise.
Find the emf.
$\mathcal{E} = 200 \times 0.010 = 2.0\ \text{V}$
$N\,d\Phi/dt$.
Find the current through $5.0$ Ω.
$I = \dfrac{2.0}{5.0} = 0.40\ \text{A}$
Ohm's law.
Find the charge that flows.
$Q = I\Delta t = 0.40 \times 0.20 = 0.080\ \text{C}$
Also $N\Delta\Phi/R$, whatever the time taken.
A $0.40$ m rod slides at $5.0$ m/s on rails joined by $2.0$ Ω in a $0.60$ T field. Find the emf.
$\mathcal{E} = BLv = 0.60 \times 0.40 \times 5.0 = 1.2\ \text{V}$
Motional emf.
Find the current.
$I = \dfrac{1.2}{2.0} = 0.60\ \text{A}$
Ohm's law.
Find the magnetic force on the rod.
$F = BIL = 0.60 \times 0.60 \times 0.40 = 0.144\ \text{N}$
Opposing the motion.
Find the power needed to keep it moving.
$P = Fv = 0.144 \times 5.0 = 0.72\ \text{W}$
Mechanical power in.
Check against the electrical power.
$I^2R = 0.60^2 \times 2.0 = 0.72\ \text{W}$
Energy is conserved.
Find the direction of the current.
$\text{its field opposes the growing flux}$
Lenz's law.
A $50$-turn coil of area $0.020$ m² spins at $30$ revolutions per second in a $0.40$ T field. Find $\omega$.
$\omega = 2\pi \times 30 = 188.5\ \text{rad/s}$
Radians per second.
Write the flux through one turn.
$\Phi = 0.40 \times 0.020\cos\omega t = 0.0080\cos\omega t\ \text{Wb}$
$BA\cos\omega t$.
Differentiate for the emf.
$\mathcal{E} = N\omega(0.0080)\sin\omega t$
$-N\,d\Phi/dt$.
Find the peak emf.
$\mathcal{E}_{\max} = 50 \times 188.5 \times 0.0080 = 75.4\ \text{V}$
$NBA\omega$.
Find when the emf peaks.
$\text{when the coil's plane is parallel to } \vec{B}$
The flux is zero but changing fastest.
Connect it to $30$ Ω and find the peak current.
$I_{\max} = \dfrac{75.4}{30} = 2.5\ \text{A}$
Ohm's law at the peak.
Find the average power.
$\bar{P} = \tfrac{1}{2}\mathcal{E}_{\max}I_{\max} = 94\ \text{W}$
The average of $\sin^2$ is one half.
Find the change in flux.
$\Delta\Phi = 3.0\ \text{mWb}$
In magnitude.
Apply Faraday's law.
$\mathcal{E} = 100 \times \dfrac{3.0 \times 10^{-3}}{0.050}$
$N\Delta\Phi/\Delta t$.
Evaluate the emf.
Pushing a bar magnet into a coil at a steady pace produces an average emf of $8$ mV. The same push, from the same start to the same finish, is repeated in half the time. What is the average emf now?
Complete the worked solution: a $34$-turn coil $10$ cm by $10$ cm faces a magnetic field that rises steadily by $0.7$ T in $0.10$ s. The coil's ends are joined through a total resistance of $2.0$ Ω. Find the change in flux through one turn in mWb, the induced emf in V, and the current in A.
Find the change in flux through one turn.
$\Delta\Phi = A\,\Delta B = 0.010 \times 0.7 =$ f
In mWb.
Apply Faraday's law.
$\mathcal{E} = N\dfrac{\Delta\Phi}{\Delta t} =$ e
A steady rise gives a steady emf.
Divide the emf by the resistance.
$I = \dfrac{\mathcal{E}}{R} =$ i
Ohm's law for the coil circuit.
Find the direction.
$\text{its field opposes the rising field}$
Lenz's law.
Match each induction idea to its expression.
| $-N\,d\Phi/dt$ | $BLv$ | $NBA\omega$ | opposes the change | |
|---|---|---|---|---|
| Faraday's law | ||||
| sliding rod | ||||
| spinning coil peak | ||||
| Lenz's law |
A metal rod $0.50$ m long slides at $7$ m/s along two rails joined by a $0.25$ Ω resistor, in a uniform $0.7$ T field perpendicular to the rails. Fill in the induced emf in V, the current in A, and the magnetic force opposing the rod's motion in N.
| value | |
|---|---|
| emf (V) | |
| current (A) | |
| magnetic force (N) |
The flux through each turn of a $4$-turn coil is $\Phi = 5t^3 - 3t$ mWb, with $t$ in seconds. Write the induced emf, in mV, as a formula in $t$, using $\mathcal{E} = -N\,d\Phi/dt$.
Answer:
A $90$ g metal rod $0.20$ m long slides without friction down two vertical rails joined at the top by a $0.4$ Ω resistor, in a horizontal $0.50$ T field perpendicular to the rails. With $g = 10$ m/s², what terminal speed does it reach, in m/s?
Answer: m/s terminal speed
An engineer at a wind farm in Iowa tests a simple generator model: a coil of $50$ turns, each of area $0.03$ m², spinning in a $0.5$ T field at $60$ revolutions per second. What is its peak emf, in V?
Answer: V peak
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
The flux through each turn of a $4$-turn coil is $\Phi = 5t^3 - 6t$ mWb, with $t$ in seconds. Write the induced emf, in mV, as a formula in $t$, using $\mathcal{E} = -N\,d\Phi/dt$.
Answer:
You can apply Faraday's law. Explain to someone why a magnet falls slowly through a copper pipe.
21. Your turn: a $100$-turn coil's flux per turn falls steadily from $3.0$ mWb to zero in $0.050$ s. Find the emf., step 3
$\mathcal{E} = 6.0\ \text{V}$
Six volts while the flux falls.