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$\nabla \times \vec{E} = -\partial\vec{B}/\partial t$ and $\mathcal{E} = -d\Phi/dt$: induced electric fields, Lenz's law, mutual and self-inductance, RL circuits and magnetic energy $B^2/2\mu_0$.
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By the end of this lesson you will be able to compute induced EMFs and fields, find inductances, analyze RL circuits, and compute the energy stored in magnetic fields.
You know motional EMF and the flux rule from the last lesson, Faraday's law and Lenz's law from Physics C, and Stokes's theorem. In electrostatics the electric field had no curl. This lesson finds that changing magnetic fields give it one, and follows the consequences through inductance and magnetic energy.
| Term | What it means |
|---|---|
| Faraday's law | $\nabla \times \vec{E} = -\partial\vec{B}/\partial t$, or $\oint\vec{E} \cdot d\vec{l} = -d\Phi/dt$. |
| Induced electric field | The curling field produced by a changing magnetic field. |
| Lenz's law | Induced currents flow so as to oppose the change in flux that causes them. |
| Mutual inductance | $M$, with $\Phi_2 = MI_1$: flux through one circuit per unit current in another. |
| Self-inductance | $L$, with $\Phi = LI$: a circuit's own flux linkage per unit current. |
| Back EMF | $-L\,dI/dt$, the EMF a coil produces opposing changes in its own current. |
| Magnetic energy density | $u = B^2/2\mu_0$. |
Faraday found in 1831 that a changing magnetic flux through a circuit drives a current even when nothing moves. The EMF is
$$\mathcal{E} = \oint\vec{E} \cdot d\vec{l} = -\frac{d\Phi}{dt},$$
and since this holds for every loop, Stokes's theorem gives the differential form
$$\nabla \times \vec{E} = -\frac{\partial\vec{B}}{\partial t}.$$
A changing magnetic field produces an induced electric field that curls around it, unlike any electrostatic field. The minus sign is Lenz's law: the induced current's own field opposes the change.
When circuits share flux, a changing current in one induces an EMF in the other: $\Phi_2 = MI_1$ defines the mutual inductance $M$, which is the same in both directions. A single circuit links its own flux, $\Phi = LI$, and changing its current induces a back EMF $-L\,dI/dt$. For a long solenoid, $L = \mu_0N^2A/l$. Building up current against that back EMF stores energy
$$U = \tfrac{1}{2}LI^2 = \frac{1}{2\mu_0}\int B^2\,d\tau,$$
energy in the magnetic field with density $B^2/2\mu_0$.
Another way: picture
Picture the field inside a solenoid growing stronger. Around it, like a whirlpool around a rising column of water, an electric field curls in circles. Put a wire loop there and charges are pushed around it. The loop's own current makes a field that pushes back against the growth. A coil is sluggish for the same reason: every change in its current is resisted by the field it has already built.
Another way: steps
Checks. A steady flux induces nothing. Units: $d\Phi/dt$ in Wb/s is volts, and $L$ in henries is V s/A. Lenz's law must never let an induced current reinforce its cause. And for a solenoid, $\tfrac{1}{2}LI^2$ must equal $\tfrac{B^2}{2\mu_0}$ times the volume.
Motional EMF, from the last lesson, came from the magnetic force on moving charges. Faraday's law also covers circuits at rest in changing fields, where no magnetic force acts on stationary charges. The only way to push them is an electric field, and that field must have a curl: $\oint\vec{E} \cdot d\vec{l} \ne 0$. It is a genuinely different kind of electric field from the electrostatic one, which is always curl-free.
Einstein was struck that the same EMF appears whether a magnet moves past a coil or the coil moves past the magnet, even though the explanations differ — magnetic force in one frame, induced electric field in the other. His 1905 paper on special relativity opens with exactly this asymmetry, and relativity resolves it: electric and magnetic fields are two aspects of one field, and which you see depends on your motion.
When the changing field has symmetry, Faraday's law finds the induced field just as Ampère's law found $\vec{B}$. Inside a long solenoid with field $B(t)$, a circle of radius $s$ has flux $\pi s^2B$, so $E \cdot 2\pi s = -\pi s^2\,dB/dt$ and $E = -\tfrac{s}{2}\,dB/dt$. Outside, the flux is fixed at $\pi R^2B$, giving $E = -\tfrac{R^2}{2s}\,dB/dt$.
The induced field outside is remarkable: it exists where $\vec{B}$ is zero. That is how a betatron works. Electrons circling in a doughnut-shaped tube are accelerated by the induced electric field of a changing magnetic flux through the middle; Donald Kerst built the first one at the University of Illinois in 1940, and betatrons were used for decades to produce x-rays for radiation therapy.
Two coils sharing flux form a transformer. An alternating current in the primary makes a changing flux, which induces an EMF in the secondary, $\mathcal{E}_2 = -M\,dI_1/dt$. When an iron core links all the flux through both coils, the voltages are in the ratio of their turns, $V_2/V_1 = N_2/N_1$, and nearly all the power passes from one to the other.
The mutual inductance is symmetric, $M_{12} = M_{21}$, a result that follows from writing the flux in terms of the vector potential and is not obvious from the geometry. That symmetry means a coil pair works equally well in either direction, which is why the same transformer design can step voltage up at a power plant and down at a substation.
A coil opposes changes in its own current. Connect a battery to an inductor and resistor, and the current rises as $I = \frac{V}{R}(1 - e^{-t/\tau})$ with $\tau = L/R$: at first the back EMF equals the battery's voltage and no current flows; later the current levels off. Break the circuit suddenly and the collapsing current induces a large back EMF, which is why switches on inductive loads spark.
That spark is useful in a car's ignition coil, which interrupts current in a primary winding to induce tens of kilovolts across a spark plug. It is harmful elsewhere, and circuits that switch motors or relays include a diode across the coil to give the current a safe path when the switch opens.
Increasing the current in an inductor takes work against the back EMF: power $LI\,dI/dt$, which integrates to $\tfrac{1}{2}LI^2$. For a solenoid, substituting $L = \mu_0n^2Al$ and $B = \mu_0nI$ gives $U = \frac{B^2}{2\mu_0}Al$: energy density $B^2/2\mu_0$ throughout the field's volume, the magnetic counterpart of $\tfrac{1}{2}\varepsilon_0E^2$.
Magnetic energy densities are large. At $1$ T, $B^2/2\mu_0 = 400$ kJ/m³, ten thousand times the energy density of an electric field at air's breakdown strength. An MRI magnet stores several megajoules, which must be released safely if the magnet quenches, that is, suddenly loses superconductivity; the stored energy boils off hundreds of litres of liquid helium in seconds.
Changing flux through a solid conductor induces currents that swirl within it, eddy currents. By Lenz's law they oppose the change: a magnet dropped through a copper pipe falls slowly, braked by eddy currents in the pipe walls, and a metal plate swinging into a magnet's gap stops abruptly.
Eddy currents are both tool and nuisance. Induction cooktops use them to heat iron pans directly; metal detectors at airports sense the eddy currents induced in hidden metal; trains and roller coasters use them as frictionless brakes. In transformer and motor cores they waste energy, which is why cores are built from thin laminations or ferrite powders that break up the current paths.
With Faraday's law, three of Maxwell's four equations are in hand: Gauss's law, $\nabla \cdot \vec{B} = 0$, and $\nabla \times \vec{E} = -\partial\vec{B}/\partial t$. The fourth, Ampère's law, still lacks the correction for changing electric fields. Maxwell noticed that just as a changing magnetic field makes a curling electric field, a changing electric field should make a curling magnetic field.
Put the two together and each changing field sustains the other: an electromagnetic wave, able to travel through empty space. The next lesson completes Maxwell's equations and the one after follows the energy they carry; the unit after that derives the waves. Faraday's law, discovered with coils and a galvanometer, turns out to be half the explanation of light.
Phones, electric toothbrushes and some electric buses charge without plugs by magnetic induction. A coil in the charging pad carries alternating current at $100$ to $200$ kHz, making an oscillating field that passes through a receiving coil in the device. Faraday's law gives a peak EMF of $NAB_0\omega$: with $10$ turns of $10$ cm² in a $0.1$ mT field at $100$ kHz, about $0.6$ V.
To reach the several volts a battery needs, both coils are tuned to resonance with capacitors, which multiplies the voltage, and the pad's field is shaped with ferrite sheets. The Qi standard used by most phones, and the magnetically aligned versions introduced later, transfer $15$ W or more at efficiencies around $70$ to $80$ percent. The same physics, at much higher power, charges electric vehicles through coils buried in parking spaces in pilot projects in several American cities.
An induction cooktop has a flat copper coil under its glass surface, driven at about $25$ kHz. The oscillating magnetic field passes through the glass, which does not conduct, and into the bottom of an iron or steel pan, where it induces eddy currents that heat the metal directly. The cooktop surface stays relatively cool, heated only by the pan.
Ferromagnetic pans work best because their high permeability concentrates the field in a thin layer near the surface, where the eddy currents flow through a small cross section with high resistance. Induction cooking transfers about $85$ percent of its energy to the food, compared with about $40$ percent for gas, and it is growing in American homes as cities and states encourage electric appliances.
A loop in a strong steady field has a large flux but no EMF. A loop in a weak field that changes quickly has a small flux but may have a large EMF. Only $d\Phi/dt$ matters. That is why transformers need alternating current: a steady direct current makes a steady flux and induces nothing in the secondary.
A related error is to think Lenz's law says the induced field opposes the field. It opposes the change. If the flux is decreasing, the induced current's field points the same way as the original field, trying to prop it up.
A $50$-turn coil of area $20$ cm² sits in a field that rises from $0$ to $0.40$ T in $0.10$ s. Find the flux change per turn.
$\Delta\Phi = \Delta B \cdot A = 0.40 \times 20 \times 10^{-4} = 8.0 \times 10^{-4}\ \text{Wb}$
The field is perpendicular to the coil.
Find the rate of change.
$\dfrac{\Delta\Phi}{\Delta t} = \dfrac{8.0 \times 10^{-4}}{0.10} = 8.0 \times 10^{-3}\ \text{Wb/s}$
Assuming a steady rise.
Multiply by the number of turns.
$|\mathcal{E}| = 50 \times 8.0 \times 10^{-3} = 0.40\ \text{V}$
Each turn contributes in series.
Find the direction.
$\text{induced field opposes } \vec{B}$
Lenz's law: the flux is increasing.
Find the current through a $4.0$ Ω coil.
$I = \dfrac{0.40}{4.0} = 0.10\ \text{A}$
While the field is changing.
A long solenoid of radius $R = 3.0$ cm has its field rising at $dB/dt = 5.0$ T/s. Choose a circle of radius $s$.
$\oint\vec{E} \cdot d\vec{l} = E \cdot 2\pi s$
By symmetry $\vec{E}$ circles the axis.
Find the flux inside for $s < R$.
$\Phi = \pi s^2B$
The uniform interior field.
Apply Faraday's law inside.
$E = \dfrac{s}{2}\dfrac{dB}{dt}$
Growing linearly with $s$.
Evaluate at $s = 2.0$ cm.
$E = 0.010 \times 5.0 = 0.050\ \text{V/m}$
Magnitude, circling opposite to the current's sense.
Apply Faraday's law outside.
$E = \dfrac{R^2}{2s}\dfrac{dB}{dt}$
The flux is limited to the solenoid's area.
Evaluate at $s = 6.0$ cm.
$E = \dfrac{(0.030)^2}{2 \times 0.060} \times 5.0 = 0.0375\ \text{V/m}$
Nonzero where $B = 0$.
A $12$ V battery connects at $t = 0$ to $L = 40$ mH in series with $R = 8.0$ Ω. Write the circuit equation.
$V = IR + L\dfrac{dI}{dt}$
Kirchhoff's loop rule with the back EMF.
Solve for the current.
$I(t) = \dfrac{V}{R}\left(1 - e^{-t/\tau}\right)$
Starting from zero.
Find the time constant.
$\tau = \dfrac{L}{R} = \dfrac{0.040}{8.0} = 5.0\ \text{ms}$
The response time.
Find the final current.
$I_\infty = \dfrac{12}{8.0} = 1.5\ \text{A}$
Reached after several time constants.
Find the initial rate of rise.
$\dfrac{dI}{dt}\Big|_0 = \dfrac{V}{L} = \dfrac{12}{0.040} = 300\ \text{A/s}$
At first the whole battery voltage drives the inductor.
Find the current after one time constant.
$I(\tau) = 1.5(1 - e^{-1}) = 0.95\ \text{A}$
Sixty-three percent of final.
Find the stored energy at the end.
$U = \tfrac{1}{2}LI_\infty^2 = \tfrac{1}{2} \times 0.040 \times 2.25 = 45\ \text{mJ}$
In the magnetic field.
Find what happens if the circuit is broken suddenly.
$\mathcal{E} = -L\dfrac{dI}{dt} \to \text{very large}$
The stored energy must go somewhere: often a spark.
Write the inductor energy.
$U = \tfrac{1}{2}LI^2$
Work done establishing the current.
Substitute the values.
$U = \tfrac{1}{2} \times 0.020 \times 3^2$
SI units.
Evaluate the energy.
A wire loop sits at rest in a steady, uniform field of $7$ T perpendicular to its plane. What EMF is induced around it?
Complete the worked solution: a $19$ V battery is connected at $t = 0$ to an $55$ mH inductor in series with $10$ Ω. Find the time constant in ms, the final current in A, and the energy finally stored in mJ.
Divide inductance by resistance.
$\tau = \dfrac{L}{R} =$ t
The current reaches $63$ percent of its final value after one $\tau$.
Find the final current.
$I_\infty = \dfrac{V}{R} =$ i
Once steady, the inductor has no back EMF.
Find the stored energy.
$U = \tfrac{1}{2}LI_\infty^2 =$ u
Held in the magnetic field.
Match each statement to its equation.
| $\nabla \times \vec{E} = -\partial\vec{B}/\partial t$ | $-d\Phi/dt$ | $\mu_0N^2A/l$ | $\tfrac{1}{2}LI^2$ | |
|---|---|---|---|---|
| Faraday's law, differential form | ||||
| the EMF around a loop | ||||
| a solenoid's inductance | ||||
| an inductor's energy |
A long solenoid has inductance $102$ mH. Fill in the inductance, in mH, if it had twice the turns, twice the length (same turns), or twice the cross-sectional area.
| $L$ (mH) | |
|---|---|
| twice the turns | |
| twice the length | |
| twice the area |
A coil of $40$ turns, each of area $20$ cm², sits in a uniform field perpendicular to its plane that is increasing at $16$ mT/s. What EMF is induced, in mV?
Answer: mV
A solenoid has $800$ turns, radius $1.5$ cm and length $30$ cm, and carries $3$ A. Treating it as long, how much energy does its magnetic field store, in mJ?
Answer: mJ
A phone's wireless-charging receiver coil has $12$ turns of area $12$ cm². The charging pad makes a field through it oscillating at $125$ kHz with amplitude $0.08$ mT. What is the peak EMF induced, in volts?
Answer: V
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A long solenoid has inductance $72$ mH. Fill in the inductance, in mH, if it had twice the turns, twice the length (same turns), or twice the cross-sectional area.
| $L$ (mH) | |
|---|---|
| twice the turns | |
| twice the length | |
| twice the area |
You can apply Faraday's law and inductance. Explain to someone why a transformer does not work with direct current.
19. Your turn: a solenoid has $L = 20$ mH and carries $3$ A. How much energy does it store?, step 3
$U = 0.090\ \text{J}$
Ninety millijoules in the field.