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Wyatt's Notes — University

Electrodynamics

4.1 Faraday”s Law of Induction

A changing magnetic field induces an electric field:

∇×E=−∂B∂t\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}

Lenz’s Law: The induced EMF opposes the change in flux that produced it.

Example. A circular loop of radius RR in a uniform magnetic field B(t)=B0cos⁡(ωt) z^\mathbf{B}(t) = B_0 \cos(\omega t)\,\hat{\mathbf{z}}.

The flux: ΦB=πR2B0cos⁡(ωt)\Phi_B = \pi R^2 B_0 \cos(\omega t).

The induced EMF: E=−dΦBdt=πR2B0ωsin⁡(ωt)\mathcal{E} = -\frac{d\Phi_B}{dt} = \pi R^2 B_0 \omega \sin(\omega t).

4.2 Displacement Current

Maxwell’s key insight: Ampere’s law ∇×B=μ0J\nabla \times \mathbf{B} = \mu_0 \mathbf{J} is inconsistent with The continuity equation. Adding the displacement current term μ0ε0∂E/∂t\mu_0 \varepsilon_0 \partial \mathbf{E}/\partial t Resolves this:

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

4.3 Worked Example

Problem. A parallel-plate capacitor with circular plates of radius RR is being charged by a Current II. Find the magnetic field between the plates at distance rr from the axis.

Solution. Between the plates, J=0\mathbf{J} = 0But there is a changing electric field. The Displacement current density is Jd=ε0∂E∂tJ_d = \varepsilon_0 \frac{\partial E}{\partial t}.

E=σε0=QπR2ε0E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}So ∂E∂t=IπR2ε0\frac{\partial E}{\partial t} = \frac{I}{\pi R^2 \varepsilon_0}.

By symmetry, use an Amperian loop of radius r<Rr \lt R:

∮B⋅dl=μ0ε0∂∂t∫E⋅dA\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 \varepsilon_0 \frac{\partial}{\partial t}\int \mathbf{E} \cdot d\mathbf{A}

B⋅2πr=μ0ε0⋅IπR2ε0⋅πr2=μ0Ir2R2B \cdot 2\pi r = \mu_0 \varepsilon_0 \cdot \frac{I}{\pi R^2 \varepsilon_0} \cdot \pi r^2 = \frac{\mu_0 I r^2}{R^2}

B=μ0Ir2πR2B = \frac{\mu_0 I r}{2\pi R^2}

■\blacksquare

4.4 Motional EMF

When a conductor moves through a magnetic field, the Lorentz force on the charges produces an EMF:

E=∮(v×B)⋅dl\mathcal{E} = \oint (\mathbf{v} \times \mathbf{B}) \cdot d\mathbf{l}

This is consistent with the flux rule E=−dΦB/dt\mathcal{E} = -d\Phi_B/dt since changing the Circuit’s geometry or position changes the flux.

Example: Rod sliding on rails

A conducting rod of length LL slides with velocity vv along two parallel rails connected by A resistor RRIn a uniform magnetic field B=B z^\mathbf{B} = B\,\hat{\mathbf{z}} perpendicular to The rail plane.

The motional EMF:

E=∫0L(v×B)⋅dl=vBL\mathcal{E} = \int_0^L (\mathbf{v} \times \mathbf{B}) \cdot d\mathbf{l} = vBL

The induced current: I=E/R=vBL/RI = \mathcal{E}/R = vBL/R.

The magnetic force on the rod: F=BIL=B2L2v/RF = BIL = B^2L^2v/R (opposing the motion, by Lenz’s law).

The power dissipated: P=I2R=v2B2L2/RP = I^2R = v^2B^2L^2/RWhich equals the mechanical power FvFv Supplied to the rod. ■\blacksquare

4.5 Derivation of Maxwell’s Correction

Problem with Ampere’s original law. The original Ampere’s law was ∇×B=μ0J\nabla \times \mathbf{B} = \mu_0 \mathbf{J}. Taking the divergence:

∇⋅(∇×B)=0=μ0∇⋅J\nabla \cdot (\nabla \times \mathbf{B}) = 0 = \mu_0 \nabla \cdot \mathbf{J}

This requires ∇⋅J=0\nabla \cdot \mathbf{J} = 0 at all times, which contradicts the continuity Equation ∇⋅J=−∂ρ/∂t\nabla \cdot \mathbf{J} = -\partial\rho/\partial t whenever charge density changes.

Resolution. Use Gauss’s law to rewrite the continuity equation:

∇⋅J=−∂ρ∂t=−∂∂t(ε0∇⋅E)=−∇⋅(ε0∂E∂t)\nabla \cdot \mathbf{J} = -\frac{\partial\rho}{\partial t} = -\frac{\partial}{\partial t}(\varepsilon_0 \nabla \cdot \mathbf{E}) = -\nabla \cdot \left(\varepsilon_0\frac{\partial \mathbf{E}}{\partial t}\right)

∇⋅(J+ε0∂E∂t)=0\nabla \cdot \left(\mathbf{J} + \varepsilon_0\frac{\partial \mathbf{E}}{\partial t}\right) = 0

This suggests modifying Ampere’s law to:

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

Now taking the divergence gives zero identically, consistent with charge conservation. The Term μ0ε0 ∂E/∂t\mu_0 \varepsilon_0\,\partial\mathbf{E}/\partial t is the displacement current.

Physical interpretation. The displacement current represents the time-varying electric field That produces a magnetic field just as a real current does. It is essential inside capacitors, Where J=0\mathbf{J} = 0 but ∂E/∂t≠0\partial\mathbf{E}/\partial t \neq 0.

4.6 Electromagnetic Induction: Worked Examples

Example: Loop falling through a magnetic field

A rectangular loop of width wwHeight ℓ\ellAnd resistance RR falls vertically under Gravity through a region of uniform magnetic field B=B x^\mathbf{B} = B\,\hat{\mathbf{x}} confined To a horizontal strip of height hh.

As the loop enters the field (top edge in, bottom edge out), the flux is ΦB=Bwx\Phi_B = Bwx where xx is the distance the top edge has penetrated.

The induced EMF: E=−Bw dx/dt=−Bwv\mathcal{E} = -Bw\,dx/dt = -Bwv.

The induced current: I=Bwv/RI = Bwv/RFlowing to oppose the change in flux (Lenz’s law).

The braking force: F=BwI=B2w2v/RF = BwI = B^2w^2v/R (upward).

Terminal velocity: mg=B2w2vterm/R  ⟹  vterm=mgR/(B2w2)mg = B^2w^2v_{\mathrm{term}/R \implies v_{\mathrm{term} = mgR/(B^2w^2)}}.

While entirely inside the field, ΦB\Phi_B is constant, so E=0\mathcal{E} = 0 and the loop Falls freely. As it exits, the braking force reappears. ■\blacksquare

Mutual inductance. When circuit 1 produces flux Φ21\Phi_{21} through circuit 2:

M=Φ21I1M = \frac{\Phi_{21}}{I_1}

The EMF induced in circuit 2 by a changing current in circuit 1:

E2=−MdI1dt\mathcal{E}_2 = -M\frac{dI_1}{dt}

Self-inductance. A circuit carrying current II produces flux Φ\Phi through itself:

L=NΦIL = \frac{N\Phi}{I}

The back-EMF:

E=−LdIdt\mathcal{E} = -L\frac{dI}{dt}

Energy stored in an inductor:

U=12LI2U = \frac{1}{2}LI^2

Example: Solenoid. A long solenoid of length ℓ\ell with NN turns, cross-sectional area AA:

L=μ0N2AℓL = \frac{\mu_0 N^2 A}{\ell}