Electromagnetic Induction and AC Circuits

Big Picture

A changing magnetic flux through a loop induces an electromotive force (EMF) in the loop — this is Faraday's law. Lenz's law gives the direction: the induced current always opposes the change that created it. Inductors store energy in magnetic fields, analogous to capacitors storing energy in electric fields. In AC circuits, voltage and current oscillate sinusoidally; reactance, impedance, and phase describe the behavior.

ELI-10: Explain It Like I'm 10

Move a magnet near a coil of wire and electricity flows — without any battery. This is how generators work: spin a magnet near coils, and you get electricity. The faster you spin or the stronger the magnet, the more electricity. Transformers change voltage using the same idea: one coil creates a changing magnetic field, which induces voltage in a second coil. AC power is electricity that wiggles back and forth instead of flowing steadily — nearly all the power in your home is AC.

13.1 Magnetic Flux

Core Idea

Magnetic flux is the "amount of magnetic field" passing through a surface.

\[ \Phi_B = \int \mathbf{B} \cdot d\mathbf{A} = BA\cos\theta \]

Where \(\theta\) is the angle between \(\mathbf{B}\) and the normal to the surface. SI unit: weber (Wb). \(1\ \text{Wb} = 1\ \text{T·m}^2\).

Key insight: Only the component of the magnetic field perpendicular to the surface contributes to flux. When the field is parallel to the surface (\(\theta = 90^\circ\)), flux is zero. When the field is perpendicular (\(\theta = 0^\circ\)), flux is maximum: \(\Phi_B = BA\).

Worked Example — Computing Magnetic Flux:

A rectangular loop of dimensions 0.10 m × 0.15 m is placed in a uniform magnetic field of 0.80 T. The field makes a 30° angle with the normal to the loop's plane. Find the magnetic flux through the loop.

Solution:

1. Area: \(A = (0.10)(0.15) = 0.015\ \text{m}^2\)
2. Flux: \(\Phi_B = BA\cos\theta = (0.80)(0.015)\cos 30^\circ\)
3. \(\cos 30^\circ = \sqrt{3}/2 \approx 0.866\)
4. \(\Phi_B = (0.80)(0.015)(0.866) = 0.0104\ \text{Wb}\)

If the loop is rotated so \(\theta = 60^\circ\), the flux drops to \((0.80)(0.015)\cos 60^\circ = 0.0060\ \text{Wb}\). This change in flux is what drives induction.

ELI-10: Explain It Like I'm 10

Imagine holding a hoop in a steady wind. More wind passes through the hoop when you hold it face-on to the wind. Tilt it sideways and less wind goes through. Magnetic flux is how much magnetic "wind" passes through a loop. Change the amount — by moving the magnet, changing its strength, or rotating the loop — and interesting things happen.

13.2 Faraday's Law

Core Idea

A changing magnetic flux through a loop induces an EMF in the loop.

\[ \mathcal{E} = -N\frac{d\Phi_B}{dt} \]

The induced EMF \(\mathcal{E}\) (volts) is proportional to the rate of change of flux. The negative sign embodies Lenz's law.

Three ways to change flux (and induce EMF):

1. Change the magnetic field strength \(B\) (e.g., move a magnet closer or farther)
2. Change the area \(A\) of the loop (e.g., deform a flexible loop in a field)
3. Change the orientation \(\theta\) (e.g., rotate the loop — this is how generators work)

Lenz's Law

The direction of the induced current is such that its magnetic field opposes the change in flux that produced it. This is a consequence of energy conservation — if the induced current added to the change, you would get energy for free. Nature prevents this.

Conceptual Example — Magnet Falling Through a Copper Tube:

Drop a strong neodymium magnet through a vertical copper pipe. Instead of falling at \(g = 9.8\ \text{m/s}^2\), it drifts down slowly, taking several seconds to emerge. Why?

1. As the magnet falls, its magnetic field moves through the copper tube, which acts as a continuous conducting loop.
2. The moving magnet creates a changing magnetic flux through each cross-section of the tube.
3. By Faraday's law, this changing flux induces circulating currents (eddy currents) in the copper.
4. By Lenz's law, these eddy currents produce their own magnetic field that opposes the motion of the falling magnet — pushing upward against it.
5. The magnet reaches a low terminal velocity where the magnetic braking force balances gravity.

This is a dramatic demonstration of Lenz's law in action. The same principle is used in eddy-current braking systems on roller coasters and high-speed trains — no friction, no wear, purely electromagnetic braking. Note that a plastic tube shows no braking effect because plastic is an insulator and cannot support induced currents.

Worked Example 1 — Changing Field Strength

Problem: A 100-turn coil of area 0.020 m² is in a magnetic field that decreases from 0.50 T to 0 in 0.10 s, perpendicular to the coil. Find the induced EMF.

Solution: \(\Delta\Phi_B = \Delta(BA) = (0.50)(0.020) = 0.010\ \text{Wb}\). \(|\mathcal{E}| = N|\Delta\Phi_B/\Delta t| = 100(0.010/0.10) = 10\ \text{V}\).

Worked Example 2 — Rotating Loop (Generator Principle)

Problem: A single rectangular loop of area 0.050 m² rotates at 60 revolutions per second in a uniform magnetic field of 0.30 T. The rotation axis is perpendicular to the field. Find the maximum induced EMF.

Solution:

1. As the loop rotates, \(\theta = \omega t\) where \(\omega = 2\pi f = 2\pi(60) = 377\ \text{rad/s}\).
2. Flux: \(\Phi_B = BA\cos(\omega t)\).
3. By Faraday's law: \(\mathcal{E} = -\frac{d\Phi_B}{dt} = -BA\frac{d}{dt}[\cos(\omega t)] = BA\omega\sin(\omega t)\).
4. Maximum EMF occurs when \(\sin(\omega t) = 1\): \(\mathcal{E}_{\text{max}} = BA\omega\).
5. \(\mathcal{E}_{\text{max}} = (0.30)(0.050)(377) = 5.66\ \text{V}\).

This sinusoidal output is the fundamental principle behind all AC generators. With \(N\) turns, the peak EMF scales as \(N\), and the frequency of the output matches the rotational frequency.

ELI-10: Explain It Like I'm 10

Faraday discovered that a changing magnetic field creates electricity. Shove a magnet into a coil of wire and a current flows — but only while the magnet is moving. Pull it out and current flows the opposite direction. The current fights back against the motion (Lenz's law) — push a magnet into a coil and the coil's magnetic field pushes back against you. This is nature's way of enforcing energy conservation. Generators at power plants spin huge magnets near coils to produce the electricity that powers civilization.

13.3 Inductance

Core Idea

Self-inductance \(L\): a changing current in a coil induces an EMF in the same coil that opposes the change.

\[ \mathcal{E} = -L\frac{dI}{dt} \]

SI unit: henry (H). \(1\ \text{H} = 1\ \text{V·s/A}\).

Energy stored in an inductor: \(U = \frac{1}{2}LI^2\) (analogous to capacitor energy \(U = \frac{1}{2}CV^2\)).

RL circuits: current grows/decays exponentially with time constant \(\tau = L/R\).

• Current growth (switch closed): \(I(t) = I{\text{max}}(1 - e^{-t/\tau})\) where \(I{\text{max}} = \mathcal{E}/R\)
• Current decay (switch opened, with discharge path): \(I(t) = I_0 e^{-t/\tau}\)

At \(t = \tau\), current reaches ~63% of maximum during growth, and decays to ~37% of initial during decay. After \(5\tau\), the transient is effectively complete (>99%).

Worked Example — RL Circuit Current Growth

Problem: A 12 V battery is connected in series with a 200 Ω resistor and a 0.50 H inductor. Find (a) the time constant, (b) the maximum current, (c) the current after 2.5 ms, and (d) the time to reach 90% of maximum current.

Solution: (a) \(\tau = L/R = 0.50/200 = 0.0025\ \text{s} = 2.5\ \text{ms}\).

(b) Maximum (steady-state) current: \(I_{\text{max}} = \mathcal{E}/R = 12/200 = 0.060\ \text{A} = 60\ \text{mA}\). At steady state, the inductor behaves like a short circuit (zero resistance, ideal case).

(c) At \(t = 2.5\ \text{ms} = \tau\): \(I = I_{\text{max}}(1 - e^{-1}) = 0.060(1 - 0.368) = 0.060 \times 0.632 = 0.0379\ \text{A} \approx 38\ \text{mA}\).

(d) For 90%: \(0.90 = 1 - e^{-t/\tau}\) → \(e^{-t/\tau} = 0.10\) → \(-t/\tau = \ln(0.10) = -2.303\) → \(t = 2.303\tau = 2.303(2.5\ \text{ms}) = 5.76\ \text{ms}\).

ELI-10: Explain It Like I'm 10

An inductor is a coil that resists changes in current — like a flywheel resists changes in rotation speed. When you first turn on the current, the inductor fights the increase (it takes time to "spin up"). When you try to turn it off, the inductor fights the decrease (it wants to keep "spinning"). The energy is stored in the magnetic field around the coil, just as a capacitor stores energy in an electric field between its plates.

13.4 AC Circuits

Core Ideas

In AC circuits, voltage and current vary sinusoidally: \(v(t) = V_{\text{max}}\sin(\omega t)\).

RMS Values — Detailed Explanation

AC voltage and current are constantly changing, so how do we assign a single meaningful number? The root mean square (RMS) value answers this: it is the equivalent DC value that would deliver the same average power to a resistor.

For a sinusoidal signal \(v(t) = V_{\text{max}}\sin(\omega t)\):

\[ V_{\text{rms}} = \sqrt{\frac{1}{T}\int0^T v^2(t)\,dt} = \frac{V{\text{max}}}{\sqrt{2}} \approx 0.707\,V_{\text{max}} \]

Similarly, \(I{\text{rms}} = I{\text{max}}/\sqrt{2}\).

Why this matters: The "120 V" in US wall outlets means \(V{\text{rms}} = 120\ \text{V}\). The actual peak voltage is \(V{\text{max}} = 120\sqrt{2} \approx 170\ \text{V}\). The voltage swings from +170 V to −170 V, 60 times per second.

Worked Example — RMS and Peak Values:

A US wall outlet supplies \(V_{\text{rms}} = 120\ \text{V}\) at 60 Hz to a 240 Ω resistive heater. Find (a) the peak voltage, (b) the RMS current, (c) the peak current, and (d) the average power delivered.

Solution: (a) \(V{\text{max}} = V{\text{rms}}\sqrt{2} = 120 \times 1.414 = 170\ \text{V}\).

(b) \(I{\text{rms}} = V{\text{rms}}/R = 120/240 = 0.500\ \text{A}\).

(c) \(I{\text{max}} = I{\text{rms}}\sqrt{2} = 0.500 \times 1.414 = 0.707\ \text{A}\).

(d) Average power: \(P{\text{avg}} = I{\text{rms}}V{\text{rms}} = (0.500)(120) = 60\ \text{W}\). Equivalently, \(P{\text{avg}} = I_{\text{rms}}^2 R = (0.500)^2(240) = 60\ \text{W}\).

Note: for a purely resistive load, \(P{\text{avg}} = V{\text{rms}}I_{\text{rms}}\). For circuits with reactance, a power factor correction is needed.

Reactance

Reactance (opposition to AC, measured in Ω):

• Capacitive reactance: \(X_C = \frac{1}{\omega C}\). Capacitors pass high frequencies, block low frequencies/DC.
• Inductive reactance: \(X_L = \omega L\). Inductors pass low frequencies, block high frequencies.

Impedance

Impedance \(Z\) combines resistance and reactance. For series RLC:

\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]

Ohm's law for AC: \(V{\text{rms}} = I{\text{rms}} Z\).

The phase angle \(\phi\) between voltage and current is given by:

\[ \tan\phi = \frac{X_L - X_C}{R} \]

• If \(X_L > X_C\): circuit is inductive, voltage leads current (\(\phi > 0\))
• If \(X_C > X_L\): circuit is capacitive, current leads voltage (\(\phi < 0\))
• If \(X_L = X_C\): resonance, voltage and current are in phase (\(\phi = 0\))

Worked Example — Series RLC Circuit

Problem: A series RLC circuit has \(R = 40\ \Omega\), \(L = 0.30\ \text{H}\), and \(C = 50\ \mu\text{F}\). It is connected to a 120 V (RMS), 60 Hz AC source. Find (a) the inductive reactance, (b) the capacitive reactance, (c) the impedance, (d) the RMS current, and (e) the phase angle. (f) Is the circuit more inductive or capacitive?

Solution: (a) \(\omega = 2\pi f = 2\pi(60) = 377\ \text{rad/s}\). \(X_L = \omega L = 377 \times 0.30 = 113\ \Omega\).

(b) \(X_C = 1/(\omega C) = 1/(377 \times 50 \times 10^{-6}) = 1/0.01885 = 53.1\ \Omega\).

(c) \(Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{40^2 + (113 - 53.1)^2} = \sqrt{1600 + 3588} = \sqrt{5188} = 72.0\ \Omega\).

(d) \(I{\text{rms}} = V{\text{rms}}/Z = 120/72.0 = 1.67\ \text{A}\).

(e) \(\tan\phi = (X_L - X_C)/R = (113 - 53.1)/40 = 59.9/40 = 1.498\) → \(\phi = \arctan(1.498) = 56.3^\circ\).

(f) Since \(X_L > X_C\) and \(\phi > 0\), the circuit is inductive overall — the voltage leads the current by 56.3°.

Phase Relationships (Series RLC)

• Resistor: current and voltage are in phase.
• Capacitor: current LEADS voltage by 90° ("ICE" — I before E in C).
• Inductor: voltage LEADS current by 90° ("ELI" — E before I in L).

RLC Resonance

At the resonant frequency, \(X_L = X_C\), so \(\omega_0 L = 1/(\omega_0 C)\), giving:

\[ \omega_0 = \frac{1}{\sqrt{LC}}, \qquad f_0 = \frac{1}{2\pi\sqrt{LC}} \]

At resonance:

• Impedance is minimum: \(Z = R\) (reactances cancel)
• Current is maximum: \(I{\text{rms}} = V{\text{rms}}/R\)
• Voltage and current are in phase (\(\phi = 0\))
• The voltages across L and C can be much larger than the source voltage (voltage magnification)

Worked Example — Resonance:

Problem: A series RLC circuit has \(L = 0.20\ \text{H}\) and \(C = 8.0\ \mu\text{F}\). (a) Find the resonant frequency. (b) If \(R = 10\ \Omega\) and \(V_{\text{rms}} = 50\ \text{V}\), find the current at resonance and the voltage across the capacitor at resonance.

Solution: (a) \(f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{(0.20)(8.0 \times 10^{-6})}} = \frac{1}{2\pi\sqrt{1.6 \times 10^{-6}}}\).

\(\sqrt{1.6 \times 10^{-6}} = 1.265 \times 10^{-3}\). \(f_0 = 1/(2\pi \times 1.265 \times 10^{-3}) = 1/0.00795 = 126\ \text{Hz}\).

(b) At resonance, \(Z = R = 10\ \Omega\). \(I{\text{rms}} = V{\text{rms}}/R = 50/10 = 5.0\ \text{A}\).

\(X_C\) at resonance: \(\omega_0 = 2\pi f_0 = 2\pi(126) = 792\ \text{rad/s}\). \(X_C = 1/(\omega_0 C) = 1/(792 \times 8.0 \times 10^{-6}) = 1/0.00634 = 158\ \Omega\).

Voltage across capacitor: \(VC = I{\text{rms}} X_C = 5.0 \times 158 = 790\ \text{V}\).

Notice: the capacitor voltage (790 V) is nearly 16× the source voltage (50 V). This voltage magnification is why resonance must be handled carefully in power systems — and why it is so useful in radio tuning.

Transformers

A transformer consists of two coils (primary and secondary) wound on a common iron core. An AC current in the primary creates a changing magnetic flux in the core, which induces an EMF in the secondary.

\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \]

• Step-up: \(N_s > N_p\) → higher voltage, lower current
• Step-down: \(N_s < N_p\) → lower voltage, higher current

For an ideal transformer (100% efficient): power in = power out:

\[ V_p I_p = V_s I_s \quad\Rightarrow\quad \frac{I_s}{I_p} = \frac{N_p}{N_s} \]

Why only AC? A transformer requires a changing magnetic flux to induce EMF in the secondary. DC produces a constant flux; no induction occurs after the initial turn-on transient. This is one reason AC won the "war of the currents" for power distribution.

Worked Example — Transformer:

Problem: A transformer has 500 turns on the primary and 50 turns on the secondary. The primary is connected to 120 V (RMS) AC. (a) Is this a step-up or step-down transformer? (b) Find the secondary voltage. (c) If a 24 Ω load is connected to the secondary, find the secondary current and the primary current (assume an ideal transformer).

Solution: (a) \(N_s < N_p\) (50 < 500), so this is a step-down transformer.

(b) \(V_s = V_p \times (N_s/N_p) = 120 \times (50/500) = 120 \times 0.10 = 12.0\ \text{V}\).

(c) Secondary current: \(I_s = V_s/R = 12.0/24 = 0.500\ \text{A}\).

For an ideal transformer, \(V_p I_p = V_s I_s\) → \(I_p = V_s I_s / V_p = (12.0 \times 0.500)/120 = 6.0/120 = 0.050\ \text{A} = 50\ \text{mA}\).

Check via turns ratio: \(I_p/I_s = N_s/N_p = 50/500 = 0.10\), so \(I_p = 0.10 \times I_s = 0.050\ \text{A}\). ✓

The voltage was stepped down by a factor of 10, and the current stepped up by a factor of 10. Power is conserved: \(P = V_p I_p = 120 \times 0.050 = 6.0\ \text{W}\) and \(P = V_s I_s = 12.0 \times 0.500 = 6.0\ \text{W}\).

ELI-10: Explain It Like I'm 10

AC power wiggles back and forth 60 times per second (in the US). The "120 volts" in your wall is a kind of average (RMS). Capacitors act like frequency filters: they let high frequencies through but block steady current. Inductors do the opposite: they let steady current through but resist rapid wiggling. An RLC circuit at its resonant frequency is like pushing a swing at exactly the right rhythm — the response gets very strong. A transformer changes voltage by trading volts for amps: high voltage in = more turns on the input coil; low voltage out = fewer turns on the output coil. Power stays roughly the same (minus some losses).

13.5 Common Misconceptions

Misconception 1: "A steady magnetic field induces current."

Reality: Only a changing magnetic flux induces EMF. A stationary magnet near a stationary coil produces no current. The flux must change — by motion, changing field strength, or changing orientation. This is the most common misunderstanding of Faraday's law.

Misconception 2: "Lenz's law means the induced current always points opposite to the applied field."

Reality: Lenz's law says the induced current opposes the change in flux, not the flux itself. If flux is decreasing, the induced current creates a field in the same direction as the applied field to oppose the decrease. If flux is increasing, the induced field opposes the applied field. The sign depends on whether flux is growing or shrinking.

Misconception 3: "Inductors block current entirely."

Reality: Inductors oppose changes in current, not current itself. In a DC circuit at steady state, an ideal inductor behaves like a short circuit (zero resistance) — current flows freely. The inductor only "fights back" during transients when current is changing.

Misconception 4: "Higher voltage always means more power."

Reality: Power is the product of voltage and current (\(P = VI\) for DC, \(P = V{\text{rms}}I{\text{rms}}\cos\phi\) for AC). Transformers trade voltage for current at constant power (ideally). A step-up transformer increases voltage but decreases available current; a step-down transformer does the reverse. Power in ≈ power out.

Misconception 5: "\(V{\text{rms}} = V{\text{max}}/2\) for AC."

Reality: \(V{\text{rms}} = V{\text{max}}/\sqrt{2} \approx 0.707\,V_{\text{max}}\), not half. The RMS value comes from squaring, averaging, and then taking the square root — not a simple average. The average of a pure sinusoid over a full cycle is zero! RMS is the correct measure for power calculations.

Misconception 6: "At resonance, the circuit draws no power."

Reality: At resonance, the impedance is minimum (\(Z = R\)), so the current is maximum and the circuit draws maximum power from the source (\(P = I_{\text{rms}}^2 R\)). The reactive elements (L and C) exchange energy with each other, but the resistor still dissipates real power. The voltages across L and C cancel, but individually they can be very large.

Misconception 7: "Transformers work with DC if the voltage is high enough."

Reality: Transformers fundamentally require a changing magnetic flux. DC produces a static flux after the initial transient, so no continuous EMF is induced in the secondary. This is why power distribution uses AC — transformers enable efficient voltage conversion for long-distance transmission.

Topic Summary

• Magnetic flux \(\Phi_B = BA\cos\theta\).
• Faraday's law: \(\mathcal{E} = -N d\Phi_B/dt\). Changing flux induces EMF.
• Lenz's law: induced current opposes the change in flux.
• Inductance \(L\): \(\mathcal{E} = -L dI/dt\). Energy stored: \(U = \frac{1}{2}LI^2\).
• RL circuits: exponential growth/decay with \(\tau = L/R\).
• AC circuits: \(X_C = 1/(\omega C)\), \(X_L = \omega L\), impedance \(Z = \sqrt{R^2 + (X_L - X_C)^2}\).
• RMS values: \(V{\text{rms}} = V{\text{max}}/\sqrt{2}\), \(I{\text{rms}} = I{\text{max}}/\sqrt{2}\).
• Resonance at \(\omega_0 = 1/\sqrt{LC}\). Transformers: \(V_s/V_p = N_s/N_p\).
• Phase: ELI (voltage leads current in inductor), ICE (current leads voltage in capacitor).

Essential Equations

Concept Check

1. A magnet is dropped through a vertical copper pipe. It falls slower than in free fall. Why? (Hint: Lenz's law.)
2. In an AC circuit, why does a capacitor act like an open circuit at very low frequencies?
3. What happens to the resonant frequency if you double both \(L\) and \(C\) in a series RLC circuit?
4. Explain why a transformer only works with AC, not DC.
5. An inductor is connected to an AC source. Describe the phase relationship between voltage and current.
6. A series RLC circuit has \(R = 50\ \Omega\), \(L = 0.10\ \text{H}\), and \(C = 20\ \mu\text{F}\). Is the circuit inductive or capacitive at 400 Hz? (Compute \(X_L\) and \(X_C\) to decide.)
7. A step-up transformer has 100 primary turns and 1000 secondary turns, connected to 12 V AC. What is the secondary voltage? If the secondary delivers 0.20 A, what is the primary current (ideal transformer)?
8. The peak voltage of a European wall outlet is about 325 V. What is the RMS voltage? (European standard is 230 V — verify this matches.)

Open Educational References

• OpenStax, College Physics, Chapter 23: Electromagnetic Induction, AC Circuits
• OpenStax, University Physics, Volume 2, Chapters 13–15: Induction, AC Circuits