Back to the on-screen lesson ·
Self-inductance $L = N\Phi/I$ and $\mu_0n^2A\ell$, the back emf $-L\,dI/dt$, RL transients with $\tau = L/R$, magnetic energy $\tfrac{1}{2}LI^2$, and LC oscillations.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to find inductances, solve RL and LC circuits, and compute magnetic energy.
From the last lesson you know that a changing flux induces an emf. From the Ampère's law lesson you know the field inside a solenoid. From the RC lesson you can solve first-order circuit equations, and from the oscillation lesson, $\ddot{x} = -\omega^2x$. This lesson puts them together: a coil induces an emf in itself, which makes a new circuit element.
| Term | What it means |
|---|---|
| Self-inductance | $L = N\Phi/I$, a coil's flux linkage per ampere of its own current. |
| Henry | The unit of inductance: $1$ H $= 1$ V·s/A. |
| Back emf | The emf $-L\,dI/dt$ an inductor develops, opposing changes in current. |
| RL circuit | A resistor and inductor, whose current changes with time constant $L/R$. |
| Magnetic energy density | $u = B^2/2\mu_0$, energy per unit volume of a magnetic field. |
| LC circuit | An inductor and capacitor, whose charge oscillates at $\omega = 1/\sqrt{LC}$. |
| Mutual inductance | $M$, the flux linkage in one coil per ampere in another. |
Current in a coil makes a field that threads the coil's own turns. The flux linkage is proportional to the current, and the constant is the inductance: $N\Phi = LI$. For a long solenoid, $L = \mu_0n^2A\ell$. When the current changes, Faraday's law gives a self-induced emf
$$\mathcal{E}_L = -L\frac{dI}{dt}.$$
It opposes the change: an inductor resists sudden increases and sudden decreases of current alike, but not steady current. With a resistor, switching on a battery gives
$$I = \frac{\mathcal{E}}{R}\left(1 - e^{-t/\tau}\right), \qquad \tau = \frac{L}{R}.$$
An inductor stores energy $U = \tfrac{1}{2}LI^2$ in its magnetic field. Connected to a capacitor, the energy sloshes between the two at $\omega = 1/\sqrt{LC}$.
Another way: picture
Picture an inductor as a heavy flywheel in a water pipe. Starting the flow means spinning up the flywheel, which takes time and effort; once it spins, water flows freely. Try to stop the flow suddenly and the flywheel keeps pushing. Current in an inductor has this inertia: it can change only gradually.
Another way: steps
A long solenoid with $n$ turns per meter carrying current $I$ has field $\mu_0nI$ inside. Each turn links flux $\mu_0nIA$, and there are $n\ell$ turns, so the flux linkage is $\mu_0n^2A\ell I$ and $L = \mu_0n^2A\ell$. Doubling the turns in the same length quadruples the inductance: twice the field, linked by twice the turns.
A solenoid $10$ cm long, $2$ cm across, with $1000$ turns has about $4$ mH. An iron core multiplies this by hundreds, which is how the inductors in power supplies reach henries. Like capacitance, inductance depends only on geometry and material, not on the current.
Switch a battery across an inductor and resistor in series. The loop rule gives $\mathcal{E} - IR - L\,dI/dt = 0$. At the first instant $I = 0$, so the inductor takes the whole emf: $dI/dt = \mathcal{E}/L$. As the current grows, the resistor takes more of the voltage and the current grows more slowly, approaching $\mathcal{E}/R$ exponentially with $\tau = L/R$.
The mathematics is the same as RC charging with the roles swapped: in an RC circuit the capacitor's voltage cannot jump, in an RL circuit the inductor's current cannot. Remove the battery and short the circuit, and the current decays as $I_0e^{-t/\tau}$, kept flowing for a while by the energy in the field.
Building up the current in an inductor takes work against its back emf: $dW = L I\,dI$, and integrating from zero to $I$ gives $U = \tfrac{1}{2}LI^2$. For a solenoid, $\tfrac{1}{2}(\mu_0n^2A\ell)I^2 = \dfrac{(\mu_0nI)^2}{2\mu_0}A\ell$, and $A\ell$ is the volume inside. So the energy lives in the field, with density
$$u_B = \frac{B^2}{2\mu_0},$$
the magnetic partner of $\tfrac{1}{2}\varepsilon_0E^2$. A $1$ T field stores about $400$ kJ per cubic meter, far more than any practical electric field, which is why large energies are stored magnetically in research magnets and some utility systems.
Because an inductor's current cannot change instantly, opening a switch in a circuit with an inductor is dangerous. The current must fall to zero in the brief time the switch contacts part, so $dI/dt$ is enormous and $L\,dI/dt$ can be thousands of volts, enough to arc across the switch.
That is exactly how a car's ignition coil makes the spark for each cylinder: current is built up in a coil and then interrupted, and the collapsing field induces $20{,}000$ V or more. In other circuits, a diode placed across the inductor gives the current a safe path to decay, protecting transistors and relays.
Connect a charged capacitor to an inductor. The capacitor discharges through the inductor, but the current, once started, cannot stop suddenly, so it overshoots and charges the capacitor the other way. The loop rule gives $L\,d^2q/dt^2 + q/C = 0$: simple harmonic motion with $\omega = 1/\sqrt{LC}$.
The analogy with a mass on a spring is exact: charge for position, current for velocity, $L$ for mass, $1/C$ for the spring constant. Energy moves between the capacitor's electric field, $q^2/2C$, and the inductor's magnetic field, $\tfrac{1}{2}LI^2$. With resistance, the oscillation dies away, like a damped spring. LC circuits tune every radio.
Checking an answer. Units: H/Ω is s; $\sqrt{\text{H·F}}$ is s. The current at $t = 0$ must match its value just before switching. Total energy in an LC circuit must stay constant. And the inductor's voltage must be largest when the current is changing fastest.
Writing the inductor's emf as $L\,dI/dt$ assumes the flux linkage is proportional to the current, true for air cores and for iron below saturation. It also assumes the current is the same through every turn at each instant, which holds unless the frequency is so high that the coil's length approaches a wavelength.
Saying the current cannot jump follows from the voltage: a jump in $I$ would need an infinite $dI/dt$ and so an infinite emf, which no real source can supply. In practice a very fast change produces a very large voltage, which is the arc at the switch.
When two coils share flux, a changing current in one induces an emf in the other: $\mathcal{E}_2 = -M\,dI_1/dt$, with $M$ the mutual inductance. Wound on a common iron core, nearly all the flux of one coil passes through the other, and the emfs per turn are equal. So $V_2/V_1 = N_2/N_1$.
That ratio is how transformers step voltage up for transmission and down for homes. The utility pole transformer outside an American house takes about $7{,}200$ V from the distribution line and delivers $240$ V, split into two $120$ V halves. Wireless chargers and induction cooktops rely on mutual inductance without a shared core.
An inductor's inductance can be found from its RL time constant: charge it through a known resistor, watch the current rise on an oscilloscope, and read off the time to reach 63 percent of the final value. Then $L = \tau R$, with $R$ the total resistance including the coil's own wire.
Alternatively, connect it to a known capacitor and measure the resonant frequency: $L = 1/(4\pi^2f^2C)$. Meters that read inductance directly do something similar internally, driving the coil with a small alternating current and comparing its voltage with the current's rate of change.
Inductors and capacitors are mirror images, and setting them side by side makes each easier to remember. A capacitor stores energy in an electric field, $\tfrac{1}{2}CV^2$; an inductor in a magnetic field, $\tfrac{1}{2}LI^2$. A capacitor's voltage cannot jump; an inductor's current cannot. At the first instant after a switch closes, an empty capacitor acts as a wire and a current-free inductor acts as a break; long afterward, in steady DC, the capacitor is a break and the inductor a wire.
Their time constants mirror each other too: $RC$ grows with resistance, because a large resistor slows the charging current, while $L/R$ shrinks with resistance, because a large resistor makes the current settle at a smaller value more quickly. Put the two together with no resistance and neither can win: the energy passes back and forth forever at $\omega = 1/\sqrt{LC}$.
With a resistor added to the LC loop, the charge obeys $L\ddot{q} + R\dot{q} + q/C = 0$, the damped oscillator of the mechanics course with $R$ in the role of friction. Small resistance gives a slowly decaying oscillation; large resistance gives a sluggish decay with no oscillation at all, and the dividing line, critical damping, comes at $R = 2\sqrt{L/C}$.
A hospital MRI magnet is a superconducting inductor of about $10$ H carrying hundreds of amperes, storing over a megajoule. Engineers at hospitals such as those in Houston's Texas Medical Center ramp the current up or down over hours, because a fast change would induce huge voltages across the windings.
If part of the coil stops superconducting, a quench, its resistance turns the stored $\tfrac{1}{2}LI^2$ into heat within seconds, boiling off hundreds of liters of liquid helium through a vent pipe on the roof. Protection circuits spread the heat through the whole coil so no single spot is damaged. A magnet storing $1.2$ MJ and dumping it in about $20$ s releases heat at around $60$ kW, as much as a dozen household furnaces running at once.
Every radio tuner contains an LC circuit whose resonant frequency, $1/(2\pi\sqrt{LC})$, is set to the station's broadcast frequency. The circuit responds strongly to signals at that frequency and weakly to others, picking one station out of the many that reach the antenna at once.
Old AM radios had a variable capacitor turned by the tuning knob; an FM station at $100$ MHz needs a much smaller $LC$, with inductances of a fraction of a microhenry. Modern phones tune digitally, but LC filters still pick out the bands for cellular, Wi-Fi and GPS signals.
Because an inductor resists current when a circuit is switched on, it can seem to act like a resistor. It does not: in a steady DC circuit an ideal inductor is just a wire, with no voltage across it. It pushes back only while the current is changing, in proportion to how fast.
A related error is to think an inductor's current drops to zero the instant a switch opens. The current cannot change instantly; if it has nowhere to go, the voltage rises until it finds a path, often as a spark across the switch.
A $50$ mH inductor and $10$ Ω resistor are switched across $20$ V. Find the time constant.
$\tau = \dfrac{0.050}{10} = 5.0\ \text{ms}$
$L/R$.
Find the initial rate of rise of current.
$\dfrac{dI}{dt} = \dfrac{\mathcal{E}}{L} = \dfrac{20}{0.050} = 400\ \text{A/s}$
The inductor takes the whole emf at first.
Find the final current.
$I = \dfrac{20}{10} = 2.0\ \text{A}$
The inductor acts as a wire.
Find the current after $5.0$ ms.
$I = 2.0(1 - e^{-1}) = 1.26\ \text{A}$
One time constant.
Find the final stored energy.
$U = \tfrac{1}{2} \times 0.050 \times 2.0^2 = 0.10\ \text{J}$
$\tfrac{1}{2}LI^2$.
A solenoid $20$ cm long, area $5.0$ cm², has $800$ turns. Find its turns per meter.
$n = \dfrac{800}{0.20} = 4000\ \text{m}^{-1}$
Turns over length.
Find its inductance.
$L = 4\pi \times 10^{-7} \times 4000^2 \times 5.0 \times 10^{-4} \times 0.20 = 2.0\ \text{mH}$
$\mu_0n^2A\ell$.
Find the field at $3.0$ A.
$B = 4\pi \times 10^{-7} \times 4000 \times 3.0 = 15\ \text{mT}$
$\mu_0nI$.
Find the stored energy.
$U = \tfrac{1}{2} \times 2.0 \times 10^{-3} \times 9.0 = 9.0\ \text{mJ}$
$\tfrac{1}{2}LI^2$.
Check with the energy density.
$\dfrac{B^2}{2\mu_0} \times A\ell = \dfrac{(0.015)^2}{2.51 \times 10^{-6}} \times 10^{-4} = 9.0\ \text{mJ}$
The two agree.
Find the emf if the current falls to zero in $1.0$ ms.
$\mathcal{E} = 2.0 \times 10^{-3} \times \dfrac{3.0}{0.0010} = 6.0\ \text{V}$
$L\,dI/dt$.
A $20$ μF capacitor charged to $50$ V is connected to an $80$ mH inductor. Find $\omega$.
$\omega = \dfrac{1}{\sqrt{0.080 \times 20 \times 10^{-6}}} = 790\ \text{rad/s}$
$1/\sqrt{LC}$.
Find the frequency.
$f = \dfrac{790}{2\pi} = 126\ \text{Hz}$
Cycles per second.
Find the initial charge.
$Q = 20 \times 10^{-6} \times 50 = 1.0\ \text{mC}$
$CV$.
Find the total energy.
$U = \tfrac{1}{2} \times 20 \times 10^{-6} \times 50^2 = 25\ \text{mJ}$
All in the capacitor at first.
Find the greatest current.
$\tfrac{1}{2}LI_{\max}^2 = 0.025 \Rightarrow I_{\max} = 0.79\ \text{A}$
All in the inductor a quarter period later.
Check with $\omega Q$.
$I_{\max} = \omega Q = 790 \times 0.0010 = 0.79\ \text{A}$
Like $v_{\max} = A\omega$ for a spring.
Find when the capacitor first has zero charge.
$t = \dfrac{T}{4} = \dfrac{1}{4 \times 126} = 2.0\ \text{ms}$
A quarter cycle.
Find the stored energy.
$U = \tfrac{1}{2} \times 0.20 \times 3.0^2 = 0.90\ \text{J}$
$\tfrac{1}{2}LI^2$.
Find the rate of change.
$\dfrac{\Delta I}{\Delta t} = \dfrac{3.0}{0.010} = 300\ \text{A/s}$
A steady fall.
Multiply by the inductance.
A long solenoid has inductance $5$ mH. It is rewound with twice as many turns in the same length and on the same form. What is its inductance now?
Complete the worked solution: the current in a $0.5$ H inductor is raised steadily from zero to $6$ A in $0.050$ s. Find the emf across it during the rise in V, the energy stored at the end in J, and the average power delivered to it in W.
Multiply the inductance by the rate of change.
$\mathcal{E} = L\dfrac{\Delta I}{\Delta t} =$ e
Steady rise, steady emf.
Find the stored energy.
$U = \tfrac{1}{2}LI^2 =$ u
At the final current.
Divide the energy by the time.
$\bar{P} = \dfrac{U}{\Delta t} =$ p
All of it went into the field.
Check with the instantaneous power.
$P = \mathcal{E}I, \text{ rising from } 0 \text{ to } \mathcal{E} \times 6$
Its average is half the final value.
Match each inductance idea to its expression.
| $\mu_0n^2A\ell$ | $-L\,dI/dt$ | $L/R$ | $\tfrac{1}{2}LI^2$ | |
|---|---|---|---|---|
| solenoid inductance | ||||
| self-induced emf | ||||
| RL time constant | ||||
| stored energy |
A $20$ mH inductor and a $4$ Ω resistor are switched across a $12$ V battery. Fill in the current long afterward in A, the time constant in ms, and the energy then stored in the inductor in mJ.
| value | |
|---|---|
| final current (A) | |
| time constant (ms) | |
| stored energy (mJ) |
A $6$ V battery is switched at $t = 0$ across a $30$ mH inductor in series with a $4$ Ω resistor. Write the voltage across the inductor, in V, as a formula in $t$ (ms).
Answer:
A charged $15$ μF capacitor is connected across a $60$ mH inductor with negligible resistance. At what frequency does the charge slosh back and forth, in Hz?
Answer: Hz oscillation
A service engineer at a hospital in Houston, Texas, needs to know how much energy a superconducting MRI magnet holds before it is ramped down. Its inductance is $20$ H and it carries $300$ A. How much energy is stored, in MJ?
Answer: MJ in the field
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A $6$ V battery is switched at $t = 0$ across a $50$ mH inductor in series with a $10$ Ω resistor. Write the voltage across the inductor, in V, as a formula in $t$ (ms).
Answer:
You can analyze circuits with inductors. Explain to someone why opening a switch on a coil can make a spark.
21. Your turn: a $0.20$ H inductor carries $3.0$ A. How much energy does it store, and what emf appears if the current stops in $0.010$ s?, step 3
$\mathcal{E} = 0.20 \times 300 = 60\ \text{V}$
Opposing the fall.