Induction and Inductance
Chapter 29 showed that a current produces a magnetic field. The reverse effect is more startling and, in its applications, more far-reaching: a changing magnetic field produces an electric field that can drive a current. This is Faraday's law of induction, and from it flow the electric generator, the transformer, the induction furnace, and the pickups of an electric guitar. This chapter defines magnetic flux, states Faraday's and Lenz's laws, shows how induction transfers energy, recasts the law as a statement about induced electric fields \(\oint\vec{E}\cdot d\vec{s} = -\,d\Phi_B/dt\), and then introduces the inductor — leading to RL circuits, the energy stored in a magnetic field, and mutual induction.
- Magnetic flux \(\Phi_B = \displaystyle\int \vec{B}\cdot d\vec{A}\) (reducing to \(\Phi_B = BA\) for a uniform perpendicular field), measured in webers, and Faraday's law \(\varepsilon = -\dfrac{d\Phi_B}{dt}\), with \(\varepsilon = -N\dfrac{d\Phi_B}{dt}\) for a coil of \(N\) turns.
- Lenz's law: the induced current flows so its own field opposes the change in flux that produced it — and how this conserves energy.
- Induction as energy transfer: the motional emf \(\varepsilon = BLv\), the opposing force \(F = \dfrac{B^{2}L^{2}v}{R}\), and eddy currents.
- The induced electric field \(\oint \vec{E}\cdot d\vec{s} = -\dfrac{d\Phi_B}{dt}\), which exists even with no wire present, and why electric potential loses meaning for it.
- Inductance \(L = \dfrac{N\Phi_B}{i}\) (the henry), the solenoid \(L/l = \mu_0 n^{2}A\), self-induction \(\varepsilon_L = -L\dfrac{di}{dt}\), RL circuits with \(\tau_L = L/R\), the stored energy \(U_B = \tfrac{1}{2}Li^{2}\), the density \(u_B = \dfrac{B^{2}}{2\mu_0}\), and mutual induction \(\varepsilon = -M\dfrac{di}{dt}\).
What Is Physics?
In the previous chapter we found that a current produces a magnetic field — a fact that surprised the scientists who discovered it. More surprising still was the reverse effect: a magnetic field can produce an electric field that drives a current. The link between a changing magnetic field and the electric field it induces is now called Faraday's law of induction.
What began as basic science in Michael Faraday's hands is now nearly everywhere. Induction is the basis of the electric guitars that drive rock and metal, of the electric generators that power cities and railways, and of the induction furnaces that melt metal in foundries. Before reaching such applications, though, we must look at two simple experiments.
Two Experiments
First experiment. A conducting loop is wired to a sensitive ammeter, with no battery in the circuit. Move a bar magnet toward the loop and a current suddenly appears; it vanishes the instant the magnet stops. Pull the magnet away and a current appears again, now in the opposite direction. Three lessons emerge: a current flows only while there is relative motion between loop and magnet; faster motion gives a larger current; and reversing either the motion or the pole reverses the current.
Second experiment. Place two loops near each other but not touching. Close a switch to turn on a current in the first loop and the meter on the second loop twitches briefly; open the switch and it twitches the other way. A current is induced in the second loop only while the current in the first loop is changing — never when it is steady, however large. The induced current is called an induced current, the work per unit charge driving it an induced emf, and the whole process induction. In both experiments, something is changing — and Faraday saw what.
Faraday's Law of Induction
Faraday realized that the "something" is the number of magnetic field lines passing through the loop. To make this quantitative we define the magnetic flux, exactly as we defined electric flux in Chapter 23 — the field integrated over the area of the loop.
Faraday's law then says that the induced emf equals the rate at which the flux changes — not the flux itself. The minus sign (justified in the next section by Lenz's law) records that the induced emf opposes the change.
Change the field magnitude \(B\); change the area \(A\) that lies in the field (expand the coil, or slide it in or out); or change the angle \(\theta\) between the field and the coil's normal (rotate the coil — the principle of the generator). Every induction problem in this chapter is one of these three in disguise.
Lenz's Law
Faraday's law gives the size of the emf; Lenz's law gives its direction:
Two equivalent readings make it concrete. Opposition to pole movement: as a magnet's north pole approaches a loop, the loop responds by becoming a magnetic dipole that presents its own north pole toward the intruder, repelling it; pull the magnet away and the loop instead presents a south pole, trying to hold it back. Opposition to flux change: the induced field \(\vec{B}_{\text{ind}}\) points opposite an increasing applied flux and along a decreasing one — always fighting the change, never the field itself. Note carefully that \(\vec{B}_{\text{ind}}\) opposes the change in \(\Phi_B\), which does not always mean it points opposite \(\vec{B}\).
If the induced current aided the change instead of opposing it, a tiny nudge of the magnet would breed an ever-growing current and unlimited free energy. Lenz's law forbids this: you must do positive work against the induced effects, and that work is exactly what becomes electrical (and then thermal) energy.
Induction and Energy Transfers
Consider a rectangular loop of width \(L\) being pulled at constant velocity \(v\) out of a region of uniform field \(\vec{B}\). As the loop leaves, the area in the field shrinks, the flux \(\Phi_B = BLx\) falls, and an emf is induced. Replacing \(dx/dt\) with \(v\) gives the motional emf.
Induced Electric Fields
Place a copper ring in a magnetic field whose strength is rising steadily. A current appears in the ring — so an electric field must be present to push its electrons. That field was produced not by charges but by the changing magnetic flux. Remarkably, the field is there even if the ring is removed: a changing magnetic field induces an electric field in empty space. By symmetry the induced \(\vec{E}\) forms closed concentric circles around the region of changing flux.
For the symmetric case of a uniform field confined to a cylinder of radius \(R\) and changing at rate \(dB/dt\), a circular path of radius \(r\) gives \(E(2\pi r) = (\text{enclosed area})\,dB/dt\), and the field has a tidy two-part profile.
The field lines of an induced \(\vec{E}\) form closed loops, so a charge carried once around returns to its start having gained energy — yet it is at the same point, which cannot hold two potentials. Formally, \(\oint\vec{E}\cdot d\vec{s} = 0\) for static fields, but \(\oint\vec{E}\cdot d\vec{s} = -d\Phi_B/dt \ne 0\) here. Induced fields simply are not conservative.
Inductors and Inductance
Just as a capacitor stores a desired electric field, an inductor stores a desired magnetic field; the long solenoid is our prototype. Drive a current \(i\) through its \(N\) windings and the flux linkage \(N\Phi_B\) measures how much field the device bottles up per unit current.
Self-Induction
A coil's own changing current changes its own flux, so it induces an emf in itself — self-induction. Combining the flux linkage \(N\Phi_B = Li\) with Faraday's law gives the self-induced emf directly.
RL Circuits
Put an emf \(\varepsilon\), a resistor \(R\), and an inductor \(L\) in series. The loop rule gives \(L\,di/dt + Ri = \varepsilon\) — the same form as the RC circuit with \(L\to R\) and \(R\to 1/C\). The current does not jump to \(\varepsilon/R\) instantly; the inductor opposes the change, so it rises exponentially.
Initially the current cannot change instantly, so it stays at its prior value (zero, if just switched on) — the inductor looks like an open circuit. Once the current is steady, \(di/dt = 0\) and \(\varepsilon_L = 0\) — the inductor looks like a plain connecting wire. These two limits solve most circuit questions without any calculus.
Energy Stored in a Magnetic Field, and Its Density
Multiply the RL loop equation by \(i\) and read the terms by energy conservation: \(\varepsilon i\) is the power the battery supplies, \(i^{2}R\) the thermal loss, and the remainder \(Li\,di/dt\) the rate at which energy is stored in the magnetic field. Integrating gives the stored energy.
Mutual Induction
Return to two nearby coils. A current \(i_1\) in coil 1 links flux through coil 2; change \(i_1\) and an emf appears in coil 2. The coupling is measured by the mutual inductance \(M = N_2\Phi_{21}/i_1\). A non-obvious but exact result is that the coupling is symmetric: \(M_{21} = M_{12} = M\).
Putting It to Work
Problem. A long solenoid with \(n = 22\,000\ \mathrm{turns/m}\) carries \(i = 1.5\,\mathrm{A}\). At its center sits a closely packed 130-turn coil C of diameter \(d = 2.1\,\mathrm{cm}\). The solenoid current is reduced steadily to zero in 25 ms. Find the magnitude of the emf induced in coil C.
Solution. The field in the solenoid is \(B = \mu_0 n i\), so the initial flux through one turn of C (area \(A = \tfrac{1}{4}\pi d^{2} = 3.46\times10^{-4}\,\mathrm{m^{2}}\)) is \(\Phi_{B,i} = BA\). The final flux is zero.
The coil sits inside the solenoid's uniform field, so no integration is needed — just \(\Phi = BA\) through each of the 130 turns.
Problem. A square loop of side \(L = 0.10\,\mathrm{m}\) and resistance \(R = 0.40\,\Omega\) is pulled at \(v = 5.0\,\mathrm{m/s}\) out of a uniform field \(B = 1.2\,\mathrm{T}\). Find the induced emf, the current, the force needed, and the power dissipated.
Solution. Use the motional results in turn.
Check: \(i^{2}R = (1.5)^{2}(0.40) = 0.90\,\mathrm{W}\) — the mechanical power you supply equals the thermal power dissipated, as it must.
Problem. A circuit has three identical \(R = 9.0\,\Omega\) resistors, two identical inductors, and an ideal battery \(\varepsilon = 18\,\mathrm{V}\) (the inductors are arranged so all three resistors sit in parallel once the inductors act as wires). Find the battery current (a) just after the switch closes and (b) a long time after.
Solution (a). Just after closing, the inductors carry zero current and act as broken wires, leaving a single resistor in the loop.
Solution (b). Long after, currents are steady, so the inductors act as plain wires and the three resistors are in parallel, \(R_{\text{eq}} = R/3 = 3.0\,\Omega\).
The two limiting rules — open wire at \(t=0\), plain wire at \(t\to\infty\) — answer the question with no differential equation in sight.
Problem. A coil has inductance \(L = 53\,\mathrm{mH}\) and resistance \(R = 0.35\,\Omega\). A 12 V emf is applied. How much energy is stored in the magnetic field once the current reaches equilibrium?
Solution. The equilibrium current is \(i = \varepsilon/R\), then \(U_B = \tfrac{1}{2}Li^{2}\).
For perspective, a solenoid producing a field \(B = 1.0\,\mathrm{T}\) stores \(u_B = B^{2}/2\mu_0 \approx 4.0\times10^{5}\,\mathrm{J/m^{3}}\) in every cubic metre of its interior — the field is a real energy reservoir.
Problem. A small coil (radius \(R_2\), \(N_2\) turns) lies at the center of and coplanar with a large coil (radius \(R_1 \gg R_2\), \(N_1\) turns). Find \(M\), then evaluate it for \(N_1 = N_2 = 1200\), \(R_2 = 1.1\,\mathrm{cm}\), \(R_1 = 15\,\mathrm{cm}\).
Solution. The large coil makes an essentially uniform field \(B_1 = \mu_0 N_1 i_1/2R_1\) over the small coil. The flux linkage in the small coil is \(N_2\Phi_{21} = N_2 B_1(\pi R_2^{2})\), so
Computing \(M\) the other way — current in the small coil, awkward nonuniform flux through the large one — gives the identical 2.3 mH, a vivid illustration that \(M_{12} = M_{21}\) even though it is far from obvious.
Chapter Summary
\(\Phi_B = \int\vec{B}\cdot d\vec{A}\), or \(\Phi_B = BA\) for a uniform perpendicular field. Unit: weber, \(1\,\mathrm{Wb} = 1\,\mathrm{T\cdot m^{2}}\).
\(\varepsilon = -\dfrac{d\Phi_B}{dt}\); for \(N\) turns, \(\varepsilon = -N\dfrac{d\Phi_B}{dt}\).
Induced current opposes the flux change that makes it — energy conservation behind the minus sign.
Motional emf \(\varepsilon = BLv\), opposing force \(F = \dfrac{B^{2}L^{2}v}{R}\); eddy currents dissipate heat.
\(\oint\vec{E}\cdot d\vec{s} = -\dfrac{d\Phi_B}{dt}\); potential is undefined for it.
\(L = \dfrac{N\Phi_B}{i}\) (henry); solenoid \(\dfrac{L}{l} = \mu_0 n^{2}A\).
\(\varepsilon_L = -L\dfrac{di}{dt}\); rise \(i = \dfrac{\varepsilon}{R}(1 - e^{-t/\tau_L})\), \(\tau_L = \dfrac{L}{R}\).
\(U_B = \tfrac{1}{2}Li^{2}\), \(u_B = \dfrac{B^{2}}{2\mu_0}\); mutual \(\varepsilon = -M\dfrac{di}{dt}\).
Problems
Take \(\mu_0 = 4\pi\times10^{-7}\,\mathrm{T\cdot m/A}\). For each problem, first identify which of the three flux knobs is turning — \(B\), \(A\), or \(\theta\) — then apply Faraday's law for the magnitude and Lenz's law for the direction. In RL circuits, lean on the two limits: an inductor is an open wire just after switching and a plain wire long after.
- A circular loop of diameter 10 cm has its normal at \(30^\circ\) to a uniform 0.50 T field. The loop is rotated so that \(\vec{B}\) sweeps a cone about the field direction at 100 rev/min while the angle stays fixed. What emf is induced?
- An elastic conducting loop of radius 12.0 cm lies perpendicular to a uniform 0.800 T field. When released, its radius shrinks at an instantaneous rate of 75.0 cm/s. Find the emf induced at that instant.
- The flux through a loop varies as \(\Phi_B = 6.0t^{2} + 7.0t\) (milliwebers, seconds). (a) Find the magnitude of the induced emf at \(t = 2.0\,\mathrm{s}\). (b) Is the induced current through the load resistor to the left or the right (state your sign convention)?
- A 120-turn coil of radius 1.8 cm and resistance \(5.3\,\Omega\) is coaxial with a solenoid of 220 turns/cm and diameter 3.2 cm. The solenoid current drops from 1.5 A to zero in 25 ms. What current is induced in the coil?
- A rod of length \(L = 10\,\mathrm{cm}\) is pulled at \(v = 5.0\,\mathrm{m/s}\) along frictionless rails through a 1.2 T field (out of the page). The rod's resistance is \(0.40\,\Omega\); the rails are negligible. Find (a) the emf, (b) the current, (c) the thermal-energy rate, and (d) the external force needed.
- A long solenoid of diameter 12.0 cm has an interior field of 30.0 mT that is made to decrease at 6.50 mT/s. Find the magnitude of the induced electric field (a) 2.20 cm and (b) 8.20 cm from the axis.
- The inductance of a closely packed 400-turn coil is 8.0 mH. Find the magnetic flux through the coil when the current is 5.0 mA.
- A long cylindrical solenoid has 100 turns/cm and radius 1.6 cm. (a) Find its inductance per metre of length. (b) If the current changes at 13 A/s, what emf is induced per metre?
- A 12 H inductor carries 2.0 A. At what rate must the current change to produce a 60 V self-induced emf?
- A solenoid of inductance 6.30 mH is in series with a \(1.20\,\mathrm{k\Omega}\) resistor and a 14.0 V battery. (a) How long until the current reaches 80.0% of its final value? (b) What is the current at \(t = 1.0\tau_L\)?
- A coil with inductance 2.0 H and resistance \(10\,\Omega\) is connected to a 100 V battery. (a) Find the equilibrium current. (b) How much energy is stored in the field at equilibrium?
- A solenoid 85.0 cm long with cross-section \(17.0\,\mathrm{cm^{2}}\) has 950 turns carrying 6.60 A. Find (a) the magnetic energy density inside and (b) the total stored energy (neglect end effects).
- Two coils at fixed positions have mutual inductance \(M\). When coil 1 has no current and coil 2's current increases at 15.0 A/s, the emf in coil 1 is 25.0 mV. (a) Find \(M\). (b) When coil 2 carries no current and coil 1 carries 3.60 A, what is the flux linkage in coil 2?
- A multiloop circuit has identical batteries, inductors, and resistors. Rank the current drawn from the battery (a) just after the switch closes and (b) long after, by reasoning from the open-wire and plain-wire limits of an inductor.