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Electrodynamics

Electrodynamics is the regime of electromagnetism in which fields vary in time. Two coupled laws drive it: Faraday’s law makes a changing magnetic flux produce a circulating electric field, and the Maxwell–Ampère term (displacement current) makes a changing electric field produce a magnetic field. Together they let energy and information move through space, and they reduce to Electrostatics and Magnetostatics when the time derivatives vanish.

Electrodynamics is the engine behind transformers, generators, inductors, capacitors, induction heating, and the near field of every antenna. It is the bridge between the static laws you already know and the propagating waves of Electromagnetic Radiation.

Authoritative sources

Scope: near field, far field, and the quasi-static limit

Electrodynamics covers time-varying fields that have not yet detached from their source. Two boundaries keep its scope distinct from the other classes:

When the time derivatives are set to zero, Faraday’s law gives ∇×E=0\nabla\times\mathbf{E}=0 and Ampère’s law gives ∇×B=μ0J\nabla\times\mathbf{B}=\mu_0\mathbf{J}: the static limits of Electrostatics and Magnetostatics.

1. Faraday’s law of induction

Definition. A changing magnetic flux through a loop drives an electromotive force (EMF) around it:

E=∮E⋅dℓ=−dΦBdt,ΦB=∫SB⋅dA\mathcal{E} = \oint \mathbf{E}\cdot d\boldsymbol{\ell} = -\frac{d\Phi_B}{dt},\qquad \Phi_B = \int_S \mathbf{B}\cdot d\mathbf{A}

Differential form: ∇×E=−∂B/∂t\nabla\times\mathbf{E} = -\partial\mathbf{B}/\partial t. The minus sign is Lenz’s law: the induced current flows so as to oppose the change in flux that created it.

Details

The idea, step by step.

  1. Flux is the “amount of field” through a surface. The magnetic flux ΦB=∫B⋅dA\Phi_B = \int \mathbf{B}\cdot d\mathbf{A} counts how much B\mathbf{B} threads a surface. If the field strength, the area, or the tilt changes, the flux changes.

  2. A changing flux creates a circulating electric field. Faraday’s law says the line integral of E\mathbf{E} around any closed loop equals the negative rate of change of flux through that loop. Unlike the electrostatic field (which is conservative, ∇×E=0\nabla\times\mathbf{E}=0), this induced E\mathbf{E} is non-conservative: it forms closed loops with no source charge. This is the key new object in electrodynamics.

  3. Lenz’s law fixes the direction. The induced current’s own magnetic field opposes the change in flux. If flux into the page is increasing, the induced current circulates to make flux out of the page. This is energy conservation in disguise: you must do work pushing the magnet in, and that work becomes the electrical energy delivered.

B (out of page, increasing) induced E dΦ/dt < 0 ⇒ E circulates to oppose the change (Lenz)
  1. Worked example: a magnet entering a coil. Push a bar magnet into a coil of NN turns. The flux through the coil rises, so an EMF E=−N dΦB/dt\mathcal{E} = -N\,d\Phi_B/dt appears and a current flows while the magnet moves. Pull it out and the current reverses. Hold it still and the current stops — only the change matters.
  1. Authoritative source. Faraday’s law is the second of the four Maxwell equations in its modern form; see Maxwell (1865).
Example problem

A circular loop of radius r=5 cmr = 5\,\text{cm} is in a uniform magnetic field perpendicular to its plane. The field increases linearly from 00 to 0.4 T0.4\,\text{T} in 0.2 s0.2\,\text{s}. Find the induced EMF.

Solution

The flux is ΦB=B πr2\Phi_B = B\,\pi r^2 (field perpendicular, so the full area counts). The rate of change is constant:

dΦBdt=πr2dBdt=π(0.05)20.40.2=π(0.0025)(2)≈0.0157 T\cdotpm2/s\frac{d\Phi_B}{dt} = \pi r^2 \frac{dB}{dt} = \pi (0.05)^2 \frac{0.4}{0.2} = \pi (0.0025)(2) \approx 0.0157\ \text{T·m}^2/\text{s}

So ∣E∣=0.0157 V≈15.7 mV|\mathcal{E}| = 0.0157\ \text{V} \approx 15.7\ \text{mV}. By Lenz’s law the induced current circulates to oppose the increasing field.

2. Maxwell–Ampère law and the displacement current

Definition. Ampère’s law, completed by Maxwell, reads

∇×B=μ0J+μ0ε0∂E∂t\nabla\times\mathbf{B} = \mu_0\mathbf{J} + \mu_0\varepsilon_0\frac{\partial\mathbf{E}}{\partial t}

The extra term μ0ε0 ∂E/∂t\mu_0\varepsilon_0\,\partial\mathbf{E}/\partial t is the displacement current density. It is required so that the equation is consistent with charge conservation, and it is what lets a changing electric field generate a magnetic field.

Details

The idea, step by step.

  1. The original Ampère law has a hole. The steady-current form ∇×B=μ0J\nabla\times\mathbf{B}=\mu_0\mathbf{J} works for steady currents, where ∇⋅J=0\nabla\cdot\mathbf{J}=0. Take the divergence of both sides: ∇⋅(∇×B)=0\nabla\cdot(\nabla\times\mathbf{B})=0 (always), so the law requires ∇⋅J=0\nabla\cdot\mathbf{J}=0. But that fails for time-varying currents — charge piles up where current changes, so ∇⋅J=−∂ρ/∂t≠0\nabla\cdot\mathbf{J} = -\partial\rho/\partial t \neq 0.

  2. Charge conservation forces the fix. The continuity equation ∇⋅J=−∂ρ/∂t\nabla\cdot\mathbf{J} = -\partial\rho/\partial t is charge conservation. Combine it with Gauss’s law ∇⋅E=ρ/ε0\nabla\cdot\mathbf{E}=\rho/\varepsilon_0 to get ∇⋅J+ε0 ∂(∇⋅E)/∂t=0\nabla\cdot\mathbf{J} + \varepsilon_0\,\partial(\nabla\cdot\mathbf{E})/\partial t = 0. Adding ε0 ∂E/∂t\varepsilon_0\,\partial\mathbf{E}/\partial t to the current makes the divergence of the right-hand side vanish, restoring consistency. That added term is the displacement current.

  3. Concrete picture: a charging capacitor. Conduction current II flows in the wire and stops at the plate — no charge crosses the gap. Yet the current must be continuous around the loop. The growing electric field between the plates carries a displacement current Id=ε0 dΦE/dt=II_d = \varepsilon_0\,d\Phi_E/dt = I that exactly matches the conduction current, so B\mathbf{B} is continuous whether you thread the loop through the wire or through the gap.

I (conduction) I_d (displacement) E field between plates
  1. Why it matters beyond bookkeeping. The displacement term is not a bookkeeping trick: it is the term that, together with Faraday’s law, makes electromagnetic waves possible (Section 3). Without it, a changing field could not sustain itself in empty space.

  2. Authoritative source. Maxwell introduced the displacement current in A Dynamical Theory of the Electromagnetic Field (1865).

Example problem

A parallel-plate capacitor of plate area A=0.02 m2A = 0.02\,\text{m}^2 is charging with a conduction current I=4 AI = 4\,\text{A}. Find the displacement current and the rate of change of the electric field between the plates.

Solution

Continuity requires the displacement current to equal the conduction current: Id=I=4 AI_d = I = 4\,\text{A}. Since Id=ε0 A dE/dtI_d = \varepsilon_0\,A\,dE/dt,

dEdt=Idε0A=4(8.85×10−12)(0.02)≈2.26×1013 V/(m\cdotps)\frac{dE}{dt} = \frac{I_d}{\varepsilon_0 A} = \frac{4}{(8.85\times10^{-12})(0.02)} \approx 2.26\times10^{13}\ \text{V/(m·s)}

The huge rate reflects how little permittivity vacuum has: a modest current builds a very rapidly changing field.

3. The coupled Maxwell equations and wave generation

Definition. In vacuum the four Maxwell equations are

∇⋅E=ρε0,∇⋅B=0,∇×E=−∂B∂t,∇×B=μ0J+μ0ε0∂E∂t\nabla\cdot\mathbf{E}=\frac{\rho}{\varepsilon_0},\quad \nabla\cdot\mathbf{B}=0,\quad \nabla\times\mathbf{E}=-\frac{\partial\mathbf{B}}{\partial t},\quad \nabla\times\mathbf{B}=\mu_0\mathbf{J}+\mu_0\varepsilon_0\frac{\partial\mathbf{E}}{\partial t}

In a source-free region (ρ=0, J=0\rho=0,\ \mathbf{J}=0), taking the curl of Faraday’s law and substituting the Ampère law yields the wave equation for each field:

∇2E=μ0ε0 ∂2E∂t2,∇2B=μ0ε0 ∂2B∂t2\nabla^2\mathbf{E} = \mu_0\varepsilon_0\,\frac{\partial^2\mathbf{E}}{\partial t^2},\qquad \nabla^2\mathbf{B} = \mu_0\varepsilon_0\,\frac{\partial^2\mathbf{B}}{\partial t^2}

with wave speed c=1/μ0ε0≈3.0×108 m/sc = 1/\sqrt{\mu_0\varepsilon_0} \approx 3.0\times10^8\ \text{m/s}.

Details

The idea, step by step.

  1. Each law feeds the other. Faraday’s law: a changing B\mathbf{B} makes a curling E\mathbf{E}. The Maxwell–Ampère law: a changing E\mathbf{E} makes a curling B\mathbf{B}. Neither field needs a charge or current to keep going once it is moving — each regenerates the other.

  2. Chase the algebra. Take the curl of Faraday’s law: ∇×(∇×E)=−∂(∇×B)/∂t\nabla\times(\nabla\times\mathbf{E}) = -\partial(\nabla\times\mathbf{B})/\partial t. Use the vector identity ∇×(∇×E)=∇(∇⋅E)−∇2E\nabla\times(\nabla\times\mathbf{E}) = \nabla(\nabla\cdot\mathbf{E}) - \nabla^2\mathbf{E}; in a source-free region ∇⋅E=0\nabla\cdot\mathbf{E}=0, so the left side is −∇2E-\nabla^2\mathbf{E}. Substitute the Ampère law for ∇×B\nabla\times\mathbf{B} and the wave equation falls out.

  3. The speed is a surprise. The wave speed c=1/μ0ε0c = 1/\sqrt{\mu_0\varepsilon_0} depends only on the vacuum permittivity and permeability — the same constants that govern static electricity and magnetism. Maxwell recognized this equals the measured speed of light, unifying optics with electromagnetism.

  4. Transverse, in phase, perpendicular. A plane wave traveling in direction k^\hat{\mathbf{k}} has E\mathbf{E} and B\mathbf{B} perpendicular to each other and to the direction of travel, in phase, with B=E/cB = E/c. The field energy flows in the direction E×B\mathbf{E}\times\mathbf{B} (Section 6).

  5. Authoritative source. The wave prediction is the climax of Maxwell (1865); the propagating-wave treatment continues in Electromagnetic Radiation.

Example problem

A plane wave in vacuum has peak electric field E0=120 V/mE_0 = 120\ \text{V/m}. Find the peak magnetic field and the average intensity (power per unit area).

Solution

The peak magnetic field is B0=E0/c=120/(3.0×108)=4.0×10−7 TB_0 = E_0/c = 120/(3.0\times10^8) = 4.0\times10^{-7}\ \text{T}. The average intensity is the time-average of the Poynting vector:

⟨S⟩=12E02μ0c=E022 Z0=12022×377≈19.1 W/m2\langle S\rangle = \frac{1}{2}\frac{E_0^2}{\mu_0 c} = \frac{E_0^2}{2\,Z_0} = \frac{120^2}{2\times 377} \approx 19.1\ \text{W/m}^2

where Z0=μ0c≈377 ΩZ_0 = \mu_0 c \approx 377\ \Omega is the impedance of free space.

4. Applications of induction

Definition. Faraday’s law, applied to engineered geometries, gives the workhorses of electrical technology: transformers (change voltage via coupled coils), generators (convert mechanical rotation into EMF), eddy currents (induced circulating currents in bulk conductors), and induction heating (eddy currents and hysteresis that heat a conductor).

Details

The idea, step by step.

  1. Transformer: coupling two coils through shared flux. A primary coil driven by AC creates a time-varying flux in a common core; that flux links the secondary coil and induces an EMF there. The turns ratio sets the voltage ratio: V2/V1=N2/N1V_2/V_1 = N_2/N_1. Power is (ideally) conserved, so current scales inversely: I2/I1=N1/N2I_2/I_1 = N_1/N_2.
~ N₁ N₂ shared flux Φ
  1. Generator: Faraday in reverse of a motor. Rotate a coil in a magnetic field and the flux through it varies sinusoidally, ΦB=Φ0cos⁡(ωt)\Phi_B = \Phi_0\cos(\omega t), giving E=Φ0ωsin⁡(ωt)\mathcal{E} = \Phi_0\omega\sin(\omega t) — an AC voltage. This is how most of the world’s electricity is made.

  2. Eddy currents: induction in bulk metal. A changing field induces circulating currents throughout a conductor, not just in a wire. These currents create their own opposing field (Lenz’s law), producing magnetic braking, damping, and losses.

  3. Induction heating: eddy currents as a heater. A high-frequency field drives strong eddy currents in a metal workpiece; the resistance of the metal converts that current to heat. Because the field is applied without contact, it heats the metal directly and rapidly — used in induction cooktops, melting, and hardening.

  4. Authoritative source. The transformer and generator relations follow directly from Faraday’s law in Maxwell (1865).

Example problem

A step-down transformer has N1=1000N_1 = 1000 primary turns and N2=250N_2 = 250 secondary turns. The primary is connected to a 240 V240\,\text{V} (rms) supply and the secondary feeds a 12 Ω12\,\Omega load. Find the secondary voltage and the primary current (ideal transformer).

Solution

The turns ratio is N2/N1=0.25N_2/N_1 = 0.25, so V2=0.25×240=60 VV_2 = 0.25 \times 240 = 60\,\text{V}. The secondary current is I2=V2/R=60/12=5 AI_2 = V_2/R = 60/12 = 5\,\text{A}. Ideal power conservation gives V1I1=V2I2V_1 I_1 = V_2 I_2, so

I1=V2I2V1=60×5240=1.25 AI_1 = \frac{V_2 I_2}{V_1} = \frac{60 \times 5}{240} = 1.25\ \text{A}

Equivalently I1=I2 N2/N1=5×0.25=1.25 AI_1 = I_2\,N_2/N_1 = 5 \times 0.25 = 1.25\,\text{A}.

5. Quasi-static limit and lumped circuit theory

Definition. When a circuit’s size LL is much smaller than the wavelength λ=c/f\lambda = c/f of its signals, the fields are quasi-static and the circuit reduces to ideal lumped elements: a resistor RR (dissipates), an inductor LL (stores magnetic energy, V=L dI/dtV = L\,dI/dt), and a capacitor CC (stores electric energy, I=C dV/dtI = C\,dV/dt), obeying Kirchhoff’s laws. A series RLC circuit is a damped harmonic oscillator with natural frequency ω0=1/LC\omega_0 = 1/\sqrt{LC}.

Details

The idea, step by step.

  1. Why lumping is an approximation. The full field solution has a phase that varies across the circuit by ∼2πL/λ\sim 2\pi L/\lambda. When L≪λL \ll \lambda, that phase variation is negligible, so the voltage and current are the same at every point of a given element — the premise of lumped circuit theory. This is the same L≪λL \ll \lambda criterion from the Scope section, and it is why a 50 MHz50\,\text{MHz} circuit (λ≈6 m\lambda \approx 6\,\text{m}) can be lumped on a 10 cm10\,\text{cm} board but a 3 GHz3\,\text{GHz} circuit (λ≈10 cm\lambda \approx 10\,\text{cm}) cannot.

  2. Inductor and capacitor are the two storage modes. An inductor’s voltage is proportional to the rate of change of current (V=L dI/dtV = L\,dI/dt): it resists changes in current and stores energy U=12LI2U = \tfrac{1}{2}LI^2 in its magnetic field. A capacitor’s current is proportional to the rate of change of voltage (I=C dV/dtI = C\,dV/dt): it resists changes in voltage and stores energy U=12CV2U = \tfrac{1}{2}CV^2 in its electric field. These are the circuit shadows of Faraday’s law and the displacement current.

  3. RLC as an oscillator. In a series RLC circuit, energy sloshes between the inductor’s magnetic field and the capacitor’s electric field, exactly like a mass–spring (see Harmonic oscillator). The undamped natural frequency is ω0=1/LC\omega_0 = 1/\sqrt{LC}; the resistor damps the oscillation with time constant τ=2L/R\tau = 2L/R.

V(t) R L C
  1. Authoritative source. The lumped-element model and its L≪λL \ll \lambda validity are standard; see Griffiths, Introduction to Electrodynamics, and any circuits text.
Example problem

A series RLC circuit has L=10 mHL = 10\,\text{mH} and C=250 nFC = 250\,\text{nF}. Find the resonant frequency. If R=20 ΩR = 20\,\Omega, find the quality factor QQ.

Solution

The resonant frequency is

f0=12πLC=12π(10×10−3)(250×10−9)=12π2.5×10−9≈3.18 kHzf_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{(10\times10^{-3})(250\times10^{-9})}} = \frac{1}{2\pi\sqrt{2.5\times10^{-9}}} \approx 3.18\ \text{kHz}

The quality factor is Q=ω0L/R=(1/LC) L/R=L/C/RQ = \omega_0 L / R = (1/\sqrt{LC})\,L/R = \sqrt{L/C}/R:

Q=10×10−3/(250×10−9)20=4000020=20020=10Q = \frac{\sqrt{10\times10^{-3}/(250\times10^{-9})}}{20} = \frac{\sqrt{40000}}{20} = \frac{200}{20} = 10

A Q=10Q = 10 means the oscillation decays over roughly Q/π≈3Q/\pi \approx 3 cycles — a lightly damped resonance.

6. Poynting theorem and energy flow

Definition. Electromagnetic energy density and flux are

u=12(ε0E2+B2μ0),S=1μ0 E×Bu = \tfrac{1}{2}\left(\varepsilon_0 E^2 + \frac{B^2}{\mu_0}\right),\qquad \mathbf{S} = \frac{1}{\mu_0}\,\mathbf{E}\times\mathbf{B}

Poynting’s theorem is the local energy-conservation law:

∂u∂t+∇⋅S=−J⋅E\frac{\partial u}{\partial t} + \nabla\cdot\mathbf{S} = -\mathbf{J}\cdot\mathbf{E}

The Poynting vector S\mathbf{S} points in the direction energy flows; J⋅E\mathbf{J}\cdot\mathbf{E} is the rate at which the field does work on charges.

Details

The idea, step by step.

  1. Energy lives in the field, not the wire. The density uu has two equal-and-symmetric parts: electric (12ε0E2\tfrac{1}{2}\varepsilon_0 E^2), stored in capacitors, and magnetic (B2/(2μ0)B^2/(2\mu_0)), stored in inductors. The field is the real carrier of energy.

  2. The Poynting vector is the energy current. S=E×B/μ0\mathbf{S} = \mathbf{E}\times\mathbf{B}/\mu_0 has units of W/m² and points where the energy is going. In a plane wave it points along the direction of propagation, with magnitude S=E2/(μ0c)S = E^2/(\mu_0 c).

  3. The surprising direction: energy flows through space. In a charging capacitor, the energy does not travel up the wire and into the plates. The Poynting vector points inward from the surrounding space into the gap between the plates — the field energy flows through the space around the wires and into the region where it is being stored. This is one of the most counterintuitive and important results in electrodynamics.

E S = E×B/μ₀ flows into the gap
  1. Authoritative source. Poynting’s theorem follows from the Maxwell equations; see Maxwell (1865) and Griffiths, Introduction to Electrodynamics.
Example problem

A long straight wire carries a steady current II and has a uniform electric field EE along its surface (due to its resistance) and a circular magnetic field B=μ0I/(2πr)B = \mu_0 I/(2\pi r) at radius rr. Show that the Poynting vector points radially into the wire, and find the power per unit length entering the wire at its surface r=ar = a.

Solution

At the surface, E\mathbf{E} is axial and B\mathbf{B} is azimuthal, so S=E×B/μ0\mathbf{S} = \mathbf{E}\times\mathbf{B}/\mu_0 points radially inward — energy flows from the surrounding field into the wire, where it is dissipated as heat. Its magnitude at r=ar=a is

S=Eμ0⋅μ0I2πa=EI2πaS = \frac{E}{\mu_0}\cdot\frac{\mu_0 I}{2\pi a} = \frac{E I}{2\pi a}

The power per unit length is the flux through the cylindrical surface of radius aa:

Pℓ=S⋅(2πa)=EI\frac{P}{\ell} = S \cdot (2\pi a) = E I

which is exactly the Joule heating per unit length, I2R/ℓ=IEI^2 R/\ell = I E (since E=Rℓ−1IE = R\ell^{-1} I). Energy conservation closes perfectly.

7. Retarded potentials and causality

Definition. Fields at a point respond to sources at an earlier time, because changes propagate at cc. The scalar and vector potentials are retarded potentials:

ϕ(r,t)=14πε0∫ρ(r′, t−R/c)R d3r′,A(r,t)=μ04π∫J(r′, t−R/c)R d3r′\phi(\mathbf{r},t) = \frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}',\, t - R/c)}{R}\,d^3r',\qquad \mathbf{A}(\mathbf{r},t) = \frac{\mu_0}{4\pi}\int \frac{\mathbf{J}(\mathbf{r}',\, t - R/c)}{R}\,d^3r'

where R=∣r−r′∣R = |\mathbf{r}-\mathbf{r}'| is the distance from the source point. The fields are then E=−∇ϕ−∂A/∂t\mathbf{E} = -\nabla\phi - \partial\mathbf{A}/\partial t and B=∇×A\mathbf{B} = \nabla\times\mathbf{A}.

Details

The idea, step by step.

  1. No instantaneous action at a distance. In the static limit the potentials look instantaneous, but the full solution uses the source at the retarded time t−R/ct - R/c: the field at r\mathbf{r} today depends on what the source did R/cR/c ago. This is causality built into Maxwell’s equations — information and energy never travel faster than cc.

  2. The retarded time is the light-travel time. A source RR meters away influences you only after R/cR/c seconds. For a source 1 m1\,\text{m} away that delay is ∼3 ns\sim 3\,\text{ns}; for the Sun (1.5×1011 m1.5\times10^{11}\,\text{m}) it is ∼500 s\sim 500\,\text{s}. You see the Sun as it was eight minutes ago.

  3. Near field vs. far field from the same solution. Expanding the retarded potentials for a small source: the terms that fall off as 1/r21/r^2 and 1/r31/r^3 are the near (reactive) field — they store energy locally and belong to electrodynamics. The term that falls off as 1/r1/r is the radiation (far) field — it carries energy away and is the subject of Electromagnetic Radiation. The boundary between them is around r∼λ/(2π)r \sim \lambda/(2\pi).

  4. Authoritative source. The retarded-potential solution of the wave equation is standard; see Jackson, Classical Electrodynamics, and Griffiths, Introduction to Electrodynamics.

Example problem

A radio transmitter 300 km300\,\text{km} away switches on. How long after the switch does the signal reach a receiver, and what is the wavelength of a 100 MHz100\,\text{MHz} signal?

Solution

The travel time is t=d/c=300×103/(3.0×108)=1.0×10−3 s=1 mst = d/c = 300\times10^3 / (3.0\times10^8) = 1.0\times10^{-3}\,\text{s} = 1\,\text{ms}. The wavelength is

λ=cf=3.0×108100×106=3 m\lambda = \frac{c}{f} = \frac{3.0\times10^8}{100\times10^6} = 3\ \text{m}

Since the receiver is 10510^5 wavelengths away, it is deep in the far field, where the 1/r1/r radiation term dominates.

Part of Electromagnetism.

Quizz

Test yourself

Problem. A single loop of wire (radius r=10 cmr = 10\,\text{cm}) lies in a uniform magnetic field perpendicular to its plane. The field is switched on and rises linearly to B=0.5 TB = 0.5\,\text{T} over 0.1 s0.1\,\text{s}.

  1. What is the magnitude of the induced EMF?
  2. If the loop has resistance R=0.5 ΩR = 0.5\,\Omega, what is the induced current, and which way does it flow (viewed from the side the field is increasing toward)?
  3. Explain, using energy conservation, why the current flows that way.
Solution
  1. The flux is ΦB=B πr2\Phi_B = B\,\pi r^2, so
∣E∣=πr2dBdt=π(0.10)20.50.1=π(0.01)(5)≈0.157 V|\mathcal{E}| = \pi r^2 \frac{dB}{dt} = \pi (0.10)^2 \frac{0.5}{0.1} = \pi (0.01)(5) \approx 0.157\ \text{V}
  1. The current is I=E/R=0.157/0.5≈0.31 AI = \mathcal{E}/R = 0.157/0.5 \approx 0.31\,\text{A}. By Lenz’s law it flows so as to oppose the increasing field: viewed from the side the field points toward, it circulates counter-clockwise, producing a field back the other way.

  2. The induced current’s field opposes the change, so the growing external field and the induced field partially cancel inside the loop. Pushing the field up therefore requires external work; that work is exactly the electrical energy dissipated as I2RI^2 R heat in the loop. If the current flowed the other way it would aid the change, creating energy from nothing — violating conservation. This is Lenz’s law as a statement of energy conservation.