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Electrodynamics | Physics - Wyatt's Notes

A changing magnetic field induces an electric field:

×E=Bt\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).

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 μ0ε0E/t\mu_0 \varepsilon_0 \partial \mathbf{E}/\partial t Resolves this:

×B=μ0J+μ0ε0Et\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}

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} = 0 But there is a changing electric field. The Displacement current density is Jd=ε0EtJ_d = \varepsilon_0 \frac{\partial E}{\partial t}.

E=σε0=QπR2ε0E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0} So Et=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:

Bdl=μ0ε0tEdA\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 \varepsilon_0 \frac{\partial}{\partial t}\int \mathbf{E} \cdot d\mathbf{A}

B2πr=μ0ε0IπR2ε0πr2=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

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=Bz^\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

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=μ0J\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=ρt=t(ε0E)=(ε0Et)\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+ε0Et)=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+μ0ε0Et\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 μ0ε0E/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/t0\partial\mathbf{E}/\partial t \neq 0.

4.6 Electromagnetic Induction: Worked Examples

Section titled “4.6 Electromagnetic Induction: Worked Examples”
Example: Loop falling through a magnetic field

A rectangular loop of width wwHeight \ell And resistance RR falls vertically under Gravity through a region of uniform magnetic field B=Bx^\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=Bwdx/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 \ell with NN turns, cross-sectional area AA:

L=μ0N2AL = \frac{\mu_0 N^2 A}{\ell}

flowchart TD
A[4_Electrodynamics] --> B[Key Concepts]
A --> C[Core Principles]
A --> D[Practical Applications]
B --> E[Fundamental definitions]
C --> F[Design patterns]
D --> G[Real-world usage]

Electrodynamics is the physics of how changing fields create each other. Faraday’s law says a changing magnetic field whips up an electric field like stirring tea creates a whirlpool. The displacement current is Maxwell’s brilliant addition: a changing electric field produces a magnetic field just as a real current does, which is what allows electromagnetic waves to propagate through empty space. Lenz’s law is nature’s stubbornness: when you try to change the magnetic flux through a loop, the loop fights back by inducing a current that opposes the change. This is why a magnet falling through a copper pipe slows down even though copper is not magnetic. Self-inductance is like electrical inertia: an inductor resists changes in current the way a massive object resists changes in velocity. The energy stored in an inductor’s magnetic field is like the kinetic energy of a moving mass.

Mistake 1: Forgetting the minus sign in Faraday’s law The induced EMF is E=dΦB/dt\mathcal{E} = -d\Phi_B/dt, not +dΦB/dt+d\Phi_B/dt. The minus sign encodes Lenz’s law: the induced current opposes the change in flux that produced it. Dropping the sign leads to predictions of perpetual motion machines where induced currents accelerate rather than resist motion.

Mistake 2: Treating displacement current as actual charge flow The displacement current μ0ε0E/t\mu_0\varepsilon_0\,\partial\mathbf{E}/\partial t is not a real current carried by moving charges. It is a term in Maxwell’s equations that accounts for the magnetic field produced by a time-varying electric field. Inside a charging capacitor, J=0\mathbf{J} = 0 but E/t0\partial\mathbf{E}/\partial t \neq 0, so the displacement current fills the gap.

Mistake 3: Applying Faraday’s law to non-conservative fields Faraday’s law ×E=B/t\nabla \times \mathbf{E} = -\partial\mathbf{B}/\partial t applies to induced electric fields, which are non-conservative. Students sometimes attempt to define a scalar potential for these fields, but EV\mathbf{E} \neq -\nabla V when B/t0\partial\mathbf{B}/\partial t \neq 0. The scalar potential formulation only works in the static limit.

This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.

Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.

Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.

This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.

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