Faraday's Law and Lenz's Law

⚡ Faraday’s Law of Induction

Part 1 of 7 — Changing Flux Creates EMF

Faraday’s Law

$\mathcal{E} = -\frac{d\Phi_B}{dt}$

For $N$ loops: $\mathcal{E} = -N\frac{d\Phi_B}{dt}$

Lenz’s Law

The induced current flows in a direction that opposes the change in flux that caused it.

🔑 The negative sign in Faraday's law encodes Lenz's law. Nature resists changes in magnetic flux.

Ways to Change Flux

$\Phi_B = BA\cos\theta$ can change by changing:

1. $B$ — changing the field strength
2. $A$ — changing the area of the loop
3. $\theta$ — rotating the loop

Applying Lenz's Law — A Reliable Procedure

The minus sign in Faraday's law is bookkeeping; in practice you find the direction of the induced current with this three-step method:

1. Determine the existing flux through the loop (which way does $\vec{B}$ point through it, and is the flux into or out of the page?).
2. Decide whether that flux is increasing or decreasing.
3. The induced current opposes the change: if flux is increasing, the induced current creates field opposing it inside the loop; if decreasing, the induced current reinforces it. Use the right-hand rule to convert "field direction inside loop" into a current direction.

Energy interpretation. Lenz's law is conservation of energy in disguise. If the induced current aided the change, the flux would grow without bound and generate energy from nothing. The opposition guarantees you must do work against the induced effects — that mechanical work becomes the electrical energy dissipated as $I^2R$.

Common pitfall: a constant large flux induces nothing. Only $\frac{d\Phi_B}{dt} \neq 0$ produces an EMF. Always look for what is changing.

Worked Example — Differentiating the Flux

A square loop of side $a = 0.20\text{ m}$ lies flat (its plane perpendicular to $\vec{B}$) in a region where the field grows in time as $B(t) = B_0 + kt^2$, with $B_0 = 0.10\text{ T}$ and $k = 0.50\text{ T/s}^2$. Find the induced EMF magnitude at $t = 3.0\text{ s}$.

Step 1 — Write the flux. With $\theta = 0$, $\Phi_B = B(t)\,A = (B_0 + kt^2)a^2$.

Step 2 — Differentiate. Since $a^2$ is constant, $\frac{d\Phi_B}{dt} = a^2\frac{dB}{dt} = a^2(2kt)$.

Step 3 — Apply Faraday's law. $|\mathcal{E}| = \left|\frac{d\Phi_B}{dt}\right| = a^2(2kt)$.

Step 4 — Substitute. $|\mathcal{E}| = (0.20)^2 (2)(0.50)(3.0) = (0.04)(3.0) = 0.12\text{ V}$.

Notice the EMF grows linearly in time even though the field grows quadratically — differentiation drops the power by one. The minus sign in $\mathcal{E} = -\frac{d\Phi_B}{dt}$ tells us the induced current opposes the increase in $B$.

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🚂 Motional EMF

Part 2 of 7 — Moving Conductors in Fields

EMF in a Moving Rod

A rod of length $L$ moving at velocity $v$ perpendicular to $\vec{B}$:

$\mathcal{E} = BLv$

Derivation from Faraday’s Law

As the rod moves, the area of the circuit changes: $\frac{d\Phi}{dt} = B\frac{dA}{dt} = BL\frac{dx}{dt} = BLv$

Motional EMF and Force

The current in the circuit: $I = BLv/R$

The force on the rod: $F = BIL = B^2L^2v/R$

🔑 The magnetic braking force opposes the motion — this is the principle behind magnetic braking.

Two Views of Motional EMF

Flux view (Faraday). The moving rod changes the circuit area, so $\mathcal{E} = -\frac{d\Phi_B}{dt} = -B\frac{dA}{dt} = -BLv$. This is the bookkeeping picture you saw above.

Force view (microscopic). Inside the moving rod, each free charge $q$ feels a magnetic force $\vec{F} = q\vec{v}\times\vec{B}$ of magnitude $qvB$, pushing charges along the rod. This acts like a battery: the motional EMF is the work per unit charge,

$\mathcal{E} = \frac{W}{q} = \frac{(qvB)L}{q} = BLv.$

Both views give the same $BLv$ — a reassuring consistency check.

Power Balance

When you pull the rod at constant speed, you supply mechanical power $P_{\text{mech}} = Fv = \frac{B^2L^2v^2}{R}$. The resistor dissipates $P_{\text{elec}} = I^2R = \left(\frac{BLv}{R}\right)^2 R = \frac{B^2L^2v^2}{R}$. They are equal — every joule of work you do reappears as heat. This is the operating principle of regenerative braking and eddy-current brakes.

Worked Example — Terminal Velocity of a Sliding Rod

A conducting rod of mass $m = 0.10\text{ kg}$ and length $L = 0.50\text{ m}$ slides on frictionless rails of resistance-equivalent $R = 2.0\,\Omega$ inside a vertical field $B = 0.80\text{ T}$. It is released from rest and falls under gravity while staying horizontal. Find its terminal velocity.

Step 1 — Equation of motion. As the rod falls at speed $v$, the motional EMF is $\mathcal{E} = BLv$, driving current $I = \frac{BLv}{R}$. The magnetic force on this current opposes motion (Lenz), with magnitude $F_B = BIL = \frac{B^2L^2v}{R}$. Newton's second law gives

$m\frac{dv}{dt} = mg - \frac{B^2L^2}{R}v.$

Step 2 — Terminal condition. At terminal velocity the acceleration $\frac{dv}{dt} = 0$, so $mg = \frac{B^2L^2}{R}v_t$.

Step 3 — Solve. $v_t = \frac{mgR}{B^2L^2} = \frac{(0.10)(9.8)(2.0)}{(0.80)^2(0.50)^2} = \frac{1.96}{0.16} = 12.25\text{ m/s}.$

Step 4 — The full solution (calculus). Separating variables in Step 1 yields $v(t) = v_t\left(1 - e^{-t/\tau}\right)$ with time constant $\tau = \frac{mR}{B^2L^2}$. The speed approaches $v_t$ exponentially — exactly like charging in an RC circuit.

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🔗 Inductance

Part 3 of 7 — Self and Mutual Inductance

Self-Inductance

$\mathcal{E} = -L\frac{dI}{dt}$

where $L$ is the inductance. Units: Henry (H)

For a solenoid: $L = \mu_0 n^2 Al$

Mutual Inductance

$\mathcal{E}_2 = -M\frac{dI_1}{dt}$

Two coils that share magnetic flux have mutual inductance $M$.

Energy in an Inductor

$U = \frac{1}{2}LI^2$

Energy density: $u = \frac{B^2}{2\mu_0}$

🔑 An inductor stores energy in its magnetic field, just as a capacitor stores energy in its electric field.

Where Does $L = \mu_0 n^2 A l$ Come From?

Inductance is defined by the flux linkage per unit current: $L = \frac{N\Phi_B}{I}$.

For a solenoid of $n$ turns per meter and length $l$, the total turns are $N = nl$. The interior field is $B = \mu_0 n I$, so the flux through one turn is $\Phi_B = BA = \mu_0 n I A$. Therefore

$L = \frac{N\Phi_B}{I} = \frac{(nl)(\mu_0 n I A)}{I} = \mu_0 n^2 A l.$

The $n^2$ appears because each of the $\sim n$ turns both produces flux and links it.

The Capacitor ↔ Inductor Dictionary

This duality is why RL and RC circuits share the same exponential mathematics — and why LC circuits oscillate.

Worked Example — Energy Stored via Integration

A solenoid has inductance $L = 4.0\text{ mH}$. The current is increased from zero according to $I(t) = (5.0\text{ A/s})\,t$. (a) Find the self-induced (back) EMF. (b) Find the energy stored at $t = 2.0\text{ s}$ by integrating the power delivered.

Part (a) — Back EMF. $\mathcal{E} = -L\frac{dI}{dt} = -(4.0\times10^{-3})(5.0) = -2.0\times10^{-2}\text{ V} = -20\text{ mV}$. It is constant because $\frac{dI}{dt}$ is constant.

Part (b) — Energy from the power integral. The instantaneous power the source delivers to the inductor is $P = \mathcal{E}_{\text{ext}} I = \left(L\frac{dI}{dt}\right)I$. The stored energy is

$U = \int_0^t P\,dt = \int_0^I L\frac{dI}{dt}\,I\,dt = \int_0^I L\,I\,dI = \tfrac{1}{2}LI^2.$

This derivation is why $U = \frac{1}{2}LI^2$. At $t = 2.0\text{ s}$, $I = (5.0)(2.0) = 10\text{ A}$, so

$U = \tfrac{1}{2}(4.0\times10^{-3})(10)^2 = \tfrac{1}{2}(4.0\times10^{-3})(100) = 0.20\text{ J}.$

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⏱️ RL Circuits

Part 4 of 7 — Inductors in DC Circuits

Current Growth (RL circuit with battery)

$I(t) = \frac{\mathcal{E}}{R}(1 - e^{-t/\tau})$

where $\tau = L/R$

Current Decay

$I(t) = I_0 e^{-t/\tau}$

Comparison with RC Circuits

🔑 Inductors resist changes in current, just as capacitors resist changes in voltage.

Reading the RL Curve

The growth solution $I(t) = \frac{\mathcal{E}}{R}\left(1 - e^{-t/\tau}\right)$ has three regimes worth memorizing:

Why $\tau = L/R$? Larger $L$ stores more magnetic energy and fights changes harder, slowing the response; larger $R$ dissipates energy faster, letting the current settle sooner. The product carries units of seconds: $[\text{H}/\Omega] = [\text{V}\cdot\text{s}/\text{A}]/[\text{V}/\text{A}] = \text{s}$.

Energy Accounting During Charging

As current builds, the battery delivers energy that splits between two destinations: heat in the resistor ($\int I^2 R\,dt$) and magnetic energy stored in the inductor ($\frac{1}{2}LI^2$). At steady state the inductor holds $\frac{1}{2}L I_{\max}^2$ while the resistor continues to dissipate $I_{\max}^2 R$ for as long as the circuit runs.

Worked Example — Solving the RL Loop Equation

A battery of EMF $\mathcal{E} = 12\text{ V}$ is connected in series with $R = 6.0\,\Omega$ and $L = 3.0\text{ H}$. The switch closes at $t = 0$. (a) Derive $I(t)$. (b) Find the current at $t = 0.50\text{ s}$. (c) Find $\frac{dI}{dt}$ at that instant.

Part (a) — Kirchhoff's voltage law. Going around the loop, $\mathcal{E} - IR - L\frac{dI}{dt} = 0$. Rearranging,

$\frac{dI}{dt} = \frac{\mathcal{E} - IR}{L}.$

Separating variables and integrating from $I=0$ gives $I(t) = \frac{\mathcal{E}}{R}\left(1 - e^{-t/\tau}\right)$ with $\tau = \frac{L}{R} = \frac{3.0}{6.0} = 0.50\text{ s}$.

Part (b) — Current at $t = \tau$. $I_{\max} = \frac{\mathcal{E}}{R} = \frac{12}{6.0} = 2.0\text{ A}$. At $t = \tau$, $I = 2.0(1 - e^{-1}) = 2.0(0.632) = 1.26\text{ A}$.

Part (c) — Slope by differentiating. $\frac{dI}{dt} = \frac{\mathcal{E}}{R}\cdot\frac{1}{\tau}e^{-t/\tau} = \frac{\mathcal{E}}{L}e^{-t/\tau}$. At $t = \tau$: $\frac{dI}{dt} = \frac{12}{3.0}e^{-1} = 4.0(0.368) = 1.47\text{ A/s}$. The inductor's back-EMF, $L\frac{dI}{dt} = 3.0(1.47) = 4.4\text{ V}$, is exactly what is left over after the resistor drop $IR = 1.26(6.0) = 7.6\text{ V}$.

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🔁 LC Circuits & Electromagnetic Oscillations

Part 5 of 7 — Energy Oscillations

LC Circuit

Energy oscillates between the capacitor (electric field) and inductor (magnetic field):

$q(t) = Q_0 \cos(\omega t + \phi)$

$\omega = \frac{1}{\sqrt{LC}}$

$T = 2\pi\sqrt{LC}$

Energy Exchange

$U_C = \frac{q^2}{2C}, \quad U_L = \frac{1}{2}LI^2$

$U_{\text{total}} = \frac{Q_0^2}{2C} = \text{constant}$

🔑 LC oscillation is the electromagnetic analog of SHM in mechanics. Charge ↔ position, current ↔ velocity, $L$ ↔ mass, $1/C$ ↔ spring constant.

The Mechanical Analogy in Detail

The LC loop equation $L\frac{d^2q}{dt^2} + \frac{q}{C} = 0$ is identical in form to the mass–spring equation $m\frac{d^2x}{dt^2} + kx = 0$. Match the terms:

Energy Timing

The energy sloshes between capacitor and inductor at twice the charge frequency (because energy $\propto q^2$ and $\propto I^2$). When $q$ is maximum, all energy is electric and $I = 0$; a quarter-period later $q = 0$, $I$ is maximum, and all energy is magnetic. With no resistance the total $U = \frac{Q_0^2}{2C}$ never changes — a real circuit's resistance slowly damps the oscillation (an RLC circuit).

Worked Example — Deriving the LC Differential Equation

An LC circuit has $L = 2.0\text{ mH}$ and $C = 8.0\,\mu\text{F}$. The capacitor starts fully charged with $Q_0 = 5.0\,\mu\text{C}$. (a) Show the charge obeys SHM. (b) Find the oscillation period. (c) Find the maximum current.

Part (a) — Kirchhoff's loop rule. The capacitor voltage equals the inductor back-EMF: $\frac{q}{C} = -L\frac{dI}{dt}$. With $I = \frac{dq}{dt}$, this becomes

$L\frac{d^2q}{dt^2} + \frac{q}{C} = 0 \quad\Longrightarrow\quad \frac{d^2q}{dt^2} = -\frac{1}{LC}\,q.$

This is the simple-harmonic equation $\frac{d^2q}{dt^2} = -\omega^2 q$ with $\omega = \frac{1}{\sqrt{LC}}$, so $q(t) = Q_0\cos(\omega t)$.

Part (b) — Period. $\omega = \frac{1}{\sqrt{(2.0\times10^{-3})(8.0\times10^{-6})}} = \frac{1}{\sqrt{1.6\times10^{-8}}} = \frac{1}{1.26\times10^{-4}} = 7.9\times10^{3}\text{ rad/s}$. Then $T = \frac{2\pi}{\omega} = 7.9\times10^{-4}\text{ s}$.

Part (c) — Maximum current. Differentiating, $I = \frac{dq}{dt} = -Q_0\omega\sin(\omega t)$, so $I_{\max} = Q_0\omega = (5.0\times10^{-6})(7.9\times10^{3}) = 4.0\times10^{-2}\text{ A} = 40\text{ mA}$. Check via energy: $\frac{1}{2}LI_{\max}^2 = \frac{Q_0^2}{2C}$ gives the same value.

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🛠️ EM Induction Workshop

Part 6 of 7 — Practice Strategies

Problem Types

A Decision Tree for Induction Problems

1. Is anything changing the flux? If $B$, $A$, and $\theta$ are all constant, $\mathcal{E} = 0$ — stop.
2. What is changing?
   • The field $B(t)$ → $\mathcal{E} = -N A\frac{dB}{dt}$ (differentiate the given $B(t)$).
   • The area (sliding rod) → $\mathcal{E} = BLv$.
   • The orientation (rotating coil) → $\mathcal{E} = NBA\omega\sin(\omega t)$.
3. Need the current? Divide by total resistance: $I = \mathcal{E}/R$.
4. Need a direction? Apply Lenz's law (oppose the change).
5. Need total charge? Use $q = \frac{|\Delta\Phi_B|}{R}$ — it depends only on the net flux change.

Watch the Calculus

Most Physics C induction problems hand you a time-dependent quantity — $B(t)$, $\Phi_B(t)$, or a geometry that gives $A(t)$ — and ask for the EMF. The move is almost always differentiate, then evaluate at the requested instant. If instead they ask for accumulated charge or the area under an EMF-vs-time graph, you integrate. Identifying "differentiate vs. integrate" is half the battle.

Worked Example — The AC Generator

A flat coil of $N = 200$ turns and area $A = 0.015\text{ m}^2$ rotates at angular speed $\omega = 120\text{ rad/s}$ in a uniform field $B = 0.25\text{ T}$. Find (a) the EMF as a function of time and (b) its peak value.

Step 1 — Flux through the rotating coil. With the coil's normal making angle $\theta = \omega t$ with $\vec{B}$, $\Phi_B = BA\cos(\omega t)$ per turn.

Step 2 — Differentiate (Faraday's law for N turns).

$\mathcal{E} = -N\frac{d\Phi_B}{dt} = -NBA\frac{d}{dt}\cos(\omega t) = NBA\omega\sin(\omega t).$

Step 3 — Peak EMF. The sine factor maxes at 1, so

$\mathcal{E}_{\max} = NBA\omega = (200)(0.25)(0.015)(120) = 90\text{ V}.$

So $\mathcal{E}(t) = 90\sin(120t)\text{ V}$. This is exactly why power-grid generators output a sinusoidal AC voltage — the rotation turns a constant field into an oscillating flux, and the derivative of a cosine is a sine.

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📋 EM Induction Review

Part 7 of 7 — Summary

Key Formulas

Threads That Tie the Unit Together

Everything starts with flux. $\Phi_B = \int \vec{B}\cdot d\vec{A}$, and an EMF appears only when that flux changes in time. Faraday's law $\mathcal{E} = -\frac{d\Phi_B}{dt}$ is the master equation; motional EMF ($BLv$) and the generator EMF ($NBA\omega\sin\omega t$) are just special cases you get by computing $\frac{d\Phi_B}{dt}$ for a particular geometry.

Inductance packages self-flux. Defining $L = N\Phi_B/I$ lets us write the back-EMF as $\mathcal{E} = -L\frac{dI}{dt}$ and the stored energy as $U = \frac{1}{2}LI^2$, both obtained by calculus.

Circuits are differential equations. Apply Kirchhoff's voltage law with an inductor term $L\frac{dI}{dt}$:

• One inductor + resistor → first-order equation → exponential ($\tau = L/R$).
• Inductor + capacitor → second-order equation → sinusoidal ($\omega = 1/\sqrt{LC}$).

The recurring skill is translating a physical setup into $\frac{d\Phi_B}{dt}$ or a loop equation, then differentiating or integrating. Master that and the whole unit collapses into one idea.

Worked Example — Cumulative Free-Response Style

A single conducting loop of area $A = 0.040\text{ m}^2$ and resistance $R = 0.50\,\Omega$ lies in a field perpendicular to its plane that varies as $B(t) = (0.60\text{ T}) e^{-t/2.0}$ ($t$ in seconds). Find (a) the induced EMF, (b) the induced current, and (c) the charge that flows through the loop between $t = 0$ and $t = \infty$.

Part (a) — Differentiate the flux. $\Phi_B = AB(t) = (0.040)(0.60)e^{-t/2.0} = 0.024\,e^{-t/2.0}$. Then

$\mathcal{E} = -\frac{d\Phi_B}{dt} = -0.024\left(-\tfrac{1}{2.0}\right)e^{-t/2.0} = 0.012\,e^{-t/2.0}\text{ V}.$

Part (b) — Ohm's law. $I(t) = \frac{\mathcal{E}}{R} = \frac{0.012}{0.50}e^{-t/2.0} = 0.024\,e^{-t/2.0}\text{ A}$.

Part (c) — Charge by integration. $q = \int_0^{\infty} I\,dt = \frac{|\Delta\Phi_B|}{R}$. Since $\Phi_B$ drops from $0.024\text{ Wb}$ to $0$, $q = \frac{0.024}{0.50} = 0.048\text{ C}$. Note the total charge depends only on the net flux change, not on how fast it happens — a key Physics C result.

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