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Wyatt's Notes — A-Level

Electromagnetism Unification -- Diagnostic Tests

Electromagnetism Unification — Diagnostic Tests

Unit Tests

UT-1: Faraday’s Law vs Lenz’s Law — Conceptual Distinction

Question:

A coil of N=500N = 500 turns and radius r=0.05 mr = 0.05\,\text{m} is placed in a uniform magnetic field B=0.40 TB = 0.40\,\text{T} directed along the axis of the coil. The field is reduced uniformly to zero in a time Δt=0.10 s\Delta t = 0.10\,\text{s}.

(a) Calculate the magnitude of the average EMF induced in the coil.

(b) Determine the direction of the induced current (clockwise or anticlockwise when viewed from the field direction).

(c) If the field were reversed (pointing in the opposite direction) and then reduced to zero over the same time, how would the induced EMF compare? Explain using both Faraday’s and Lenz’s laws.

Solution:

(a) By Faraday’s law: ε=−N◆LB◆ΔΦ◆RB◆◆LB◆Δt◆RB◆\varepsilon = -N\frac◆LB◆\Delta\Phi◆RB◆◆LB◆\Delta t◆RB◆

ΔΦ=Φfinal−Φinitial=0−BA=−Bπr2=−0.40×π×(0.05)2=−3.14×10−3 Wb\Delta\Phi = \Phi_{\text{final}} - \Phi_{\text{initial}} = 0 - BA = -B\pi r^2 = -0.40 \times \pi \times (0.05)^2 = -3.14 \times 10^{-3}\,\text{Wb}

∣ε∣=N◆LB◆∣ΔΦ∣◆RB◆◆LB◆Δt◆RB◆=500×◆LB◆3.14×10−3◆RB◆◆LB◆0.10◆RB◆=15.7 V|\varepsilon| = N\frac◆LB◆|\Delta\Phi|◆RB◆◆LB◆\Delta t◆RB◆ = 500 \times \frac◆LB◆3.14 \times 10^{-3}◆RB◆◆LB◆0.10◆RB◆ = 15.7\,\text{V}

(b) By Lenz’s law, the induced current opposes the change in flux. Since the flux (into the page, say) is decreasing, the induced current flows to maintain the flux by creating its own field in the same direction (into the page). By the right-hand grip rule, the current flows clockwise when viewed from the direction of the original field.

(c) If the field points in the opposite direction (say out of the page) and is reduced to zero:

ΔΦ=0−(−BA)=BA=+3.14×10−3 Wb\Delta\Phi = 0 - (-BA) = BA = +3.14 \times 10^{-3}\,\text{Wb}

The magnitude of the average EMF is the same: ∣ε∣=500×3.14×10−3/0.10=15.7 V|\varepsilon| = 500 \times 3.14 \times 10^{-3}/0.10 = 15.7\,\text{V}.

Faraday’s law gives the magnitude of the EMF from the rate of change of flux: same rate of change, same magnitude.

Lenz’s law gives the direction: the flux (out of the page) is decreasing, so the induced current creates a field out of the page to oppose the decrease. The current flows anticlockwise when viewed from the original field direction — the opposite direction to part (b).

The key distinction: Faraday’s law determines how much EMF is induced; Lenz’s law determines in which direction the current flows (it always opposes the change that causes it).

UT-2: EM Wave — Speed, Wavelength, and Frequency

Question:

An electromagnetic wave in vacuum has an electric field given by E=E0sin⁡(kx−ωt) j^E = E_0\sin(kx - \omega t)\,\hat{j}Where E0=30 V m−1E_0 = 30\,\text{V}\,\text{m}^{-1}, k=125.7 rad m−1k = 125.7\,\text{rad}\,\text{m}^{-1}And ω=3.77×1010 rad s−1\omega = 3.77 \times 10^{10}\,\text{rad}\,\text{s}^{-1}.

(a) Calculate the wavelength, frequency, and speed of the wave. Verify that the speed equals cc.

(b) Write the equation for the magnetic field component of this wave.

(c) Calculate the time-averaged intensity of the wave.

Take c = 3.00 \times 10^8\,\text{m}\,\text{s}^{-1}$$\varepsilon_0 = 8.85 \times 10^{-12}\,\text{F}\,\text{m}^{-1}$$\mu_0 = 4\pi \times 10^{-7}\,\text{H}\,\text{m}^{-1}.

Solution:

(a) Wavelength: λ=2π/k=2π/125.7=0.050 m=50 mm\lambda = 2\pi/k = 2\pi/125.7 = 0.050\,\text{m} = 50\,\text{mm}

Frequency: f=ω/(2π)=3.77×1010/(2π)=6.00×109 Hz=6.0 GHzf = \omega/(2\pi) = 3.77 \times 10^{10}/(2\pi) = 6.00 \times 10^9\,\text{Hz} = 6.0\,\text{GHz}

Speed: v=ω/k=3.77×1010/125.7=3.00×108 m s−1=cv = \omega/k = 3.77 \times 10^{10}/125.7 = 3.00 \times 10^8\,\text{m}\,\text{s}^{-1} = c

Verified. This wave is in the microwave region.

(b) The magnetic field is perpendicular to both EE and the direction of propagation (+x+x). Since EE is in the j^\hat{j} direction and the wave propagates in +x$$B is in the k^\hat{k} direction.

B0=E0/c=30/(3.00×108)=1.00×10−7 TB_0 = E_0/c = 30/(3.00 \times 10^8) = 1.00 \times 10^{-7}\,\text{T}

B=B0sin⁡(kx−ωt) k^=1.00×10−7sin⁡(125.7x−3.77×1010t) k^\mathbf{B} = B_0\sin(kx - \omega t)\,\hat{k} = 1.00 \times 10^{-7}\sin(125.7x - 3.77 \times 10^{10}t)\,\hat{k}

(c) Time-averaged intensity: I=12ε0cE02I = \frac{1}{2}\varepsilon_0 c E_0^2

=0.5×8.85×10−12×3.00×108×900=0.5×8.85×10−12×2.7×1011=1.195 W m−2= 0.5 \times 8.85 \times 10^{-12} \times 3.00 \times 10^8 \times 900 = 0.5 \times 8.85 \times 10^{-12} \times 2.7 \times 10^{11} = 1.195\,\text{W}\,\text{m}^{-2}

UT-3: Displacement Current and Maxwell’s Insight

Question:

A parallel plate capacitor with circular plates of radius 0.10 m0.10\,\text{m} is being charged by a current I=2.0 AI = 2.0\,\text{A}. The plate separation is 5.0 mm5.0\,\text{mm}.

(a) Calculate the rate of change of electric field between the plates.

(b) Calculate the displacement current between the plates and verify it equals the conduction current.

(c) Explain why the concept of displacement current was necessary for Maxwell to predict electromagnetic waves.

Take ε0=8.85×10−12 F m−1\varepsilon_0 = 8.85 \times 10^{-12}\,\text{F}\,\text{m}^{-1}.

Solution:

(a) Area of plates: A=πr2=π(0.10)2=0.0314 m2A = \pi r^2 = \pi(0.10)^2 = 0.0314\,\text{m}^2

The conduction current equals the rate of charge flow onto the plates: I=dQ/dtI = dQ/dt

Since Q=CV=(ε0A/d)×Ed=ε0AEQ = CV = (\varepsilon_0 A/d) \times Ed = \varepsilon_0 AE:

I=dQdt=ε0AdEdtI = \frac{dQ}{dt} = \varepsilon_0 A \frac{dE}{dt}

dEdt=◆LB◆I◆RB◆◆LB◆ε0A◆RB◆=◆LB◆2.0◆RB◆◆LB◆8.85×10−12×0.0314◆RB◆=◆LB◆2.0◆RB◆◆LB◆2.779×10−13◆RB◆=7.20×1012 V m−1 s−1\frac{dE}{dt} = \frac◆LB◆I◆RB◆◆LB◆\varepsilon_0 A◆RB◆ = \frac◆LB◆2.0◆RB◆◆LB◆8.85 \times 10^{-12} \times 0.0314◆RB◆ = \frac◆LB◆2.0◆RB◆◆LB◆2.779 \times 10^{-13}◆RB◆ = 7.20 \times 10^{12}\,\text{V}\,\text{m}^{-1}\,\text{s}^{-1}

(b) Displacement current: Id=ε0◆LB◆dΦE◆RB◆◆LB◆dt◆RB◆=ε0AdEdtI_d = \varepsilon_0 \frac◆LB◆d\Phi_E◆RB◆◆LB◆dt◆RB◆ = \varepsilon_0 A \frac{dE}{dt}

Id=8.85×10−12×0.0314×7.20×1012=2.0 AI_d = 8.85 \times 10^{-12} \times 0.0314 \times 7.20 \times 10^{12} = 2.0\,\text{A}

Id=I=2.0 AI_d = I = 2.0\,\text{A}. The displacement current exactly equals the conduction current, as required by the continuity of current in Maxwell’s equations.

(c) Before Maxwell, Ampere’s law applied only to conduction currents in closed loops. For a charging capacitor, a conduction current flows in the wires but not between the plates. Maxwell introduced the displacement current to make Ampere’s law consistent: the changing electric field between the plates acts like a current (Id=ε0dΦE/dtI_d = \varepsilon_0 d\Phi_E/dt). This completed the symmetry of Maxwell’s equations: a changing electric field produces a magnetic field (just as Faraday’s law says a changing magnetic field produces an electric field). This mutual generation of EE and BB fields is what allows electromagnetic waves to propagate through space even in the absence of charges and currents.

Integration Tests

IT-1: Generating EM Waves — LC Circuit Analogy (with Oscillations)

Question:

An LC circuit has L=1.0 μHL = 1.0\,\mu\text{H} and C=1.0 pFC = 1.0\,\text{pF}.

(a) Calculate the resonant frequency of the circuit.

(b) The electric field between the capacitor plates has maximum amplitude E0=100 V m−1E_0 = 100\,\text{V}\,\text{m}^{-1} and the plate area is 0.01 m20.01\,\text{m}^2. Calculate the maximum energy stored in the electric field and the maximum energy stored in the magnetic field.

(c) Explain how this circuit is analogous to an electromagnetic wave and state the wavelength of the radiation it would emit.

Solution:

(a) f=◆LB◆1◆RB◆◆LB◆2πLC◆RB◆=◆LB◆1◆RB◆◆LB◆2π1.0×10−6×1.0×10−12◆RB◆=◆LB◆1◆RB◆◆LB◆2π10−18◆RB◆=◆LB◆1◆RB◆◆LB◆2π×10−9◆RB◆=1.59×108 Hz=159 MHzf = \frac◆LB◆1◆RB◆◆LB◆2\pi\sqrt{LC}◆RB◆ = \frac◆LB◆1◆RB◆◆LB◆2\pi\sqrt{1.0 \times 10^{-6} \times 1.0 \times 10^{-12}}◆RB◆ = \frac◆LB◆1◆RB◆◆LB◆2\pi\sqrt{10^{-18}}◆RB◆ = \frac◆LB◆1◆RB◆◆LB◆2\pi \times 10^{-9}◆RB◆ = 1.59 \times 10^8\,\text{Hz} = 159\,\text{MHz}

This is in the VHF radio band.

(b) Maximum energy in electric field (when capacitor is fully charged):

UE=12ε0E02×volumeU_E = \frac{1}{2}\varepsilon_0 E_0^2 \times \text{volume}

We need the plate separation: C=ε0A/dC = \varepsilon_0 A/dSo d=ε0A/C=8.85×10−12×0.01/(1.0×10−12)=0.0885 md = \varepsilon_0 A/C = 8.85 \times 10^{-12} \times 0.01/(1.0 \times 10^{-12}) = 0.0885\,\text{m}

UE=12ε0E02Ad=12CV2U_E = \frac{1}{2}\varepsilon_0 E_0^2 Ad = \frac{1}{2}CV^2

Using UE=12CV2U_E = \frac{1}{2}CV^2: V0=E0d=100×0.0885=8.85 VV_0 = E_0 d = 100 \times 0.0885 = 8.85\,\text{V}

UE=0.5×10−12×78.32=3.92×10−11 JU_E = 0.5 \times 10^{-12} \times 78.32 = 3.92 \times 10^{-11}\,\text{J}

At maximum, all energy is in the electric field, so UB=UE=3.92×10−11 JU_B = U_E = 3.92 \times 10^{-11}\,\text{J}.

(c) In an LC circuit, energy oscillates between the electric field (capacitor) and the magnetic field (inductor), analogous to how in an EM wave, energy oscillates between the EE and BB fields. The frequency of the circuit determines the frequency of the emitted radiation.

Wavelength: λ=c/f=3.00×108/(1.59×108)=1.89 m\lambda = c/f = 3.00 \times 10^8/(1.59 \times 10^8) = 1.89\,\text{m}

This is a radio wave with wavelength approximately 1.9 m1.9\,\text{m}.

IT-2: EM Wave Intensity and Radiation Pressure (with Wave Properties)

Question:

A laser produces a beam of wavelength 532 nm532\,\text{nm} (green) with power 5.0 mW5.0\,\text{mW} and beam diameter 2.0 mm2.0\,\text{mm}. The beam is incident normally on a perfectly reflecting mirror.

(a) Calculate the intensity of the beam.

(b) Calculate the radiation pressure on the mirror.

(c) Calculate the number of photons per second striking the mirror and the momentum transferred per second.

Take c = 3.00 \times 10^8\,\text{m}\,\text{s}^{-1}$$h = 6.63 \times 10^{-34}\,\text{J}\,\text{s}.

Solution:

(a) Beam area: A=πr2=π(1.0×10−3)2=3.14×10−6 m2A = \pi r^2 = \pi(1.0 \times 10^{-3})^2 = 3.14 \times 10^{-6}\,\text{m}^2

Intensity: I=P/A=5.0×10−3/(3.14×10−6)=1592 W m−2I = P/A = 5.0 \times 10^{-3}/(3.14 \times 10^{-6}) = 1592\,\text{W}\,\text{m}^{-2}

(b) For a perfectly reflecting surface: Prad=2I/c=2×1592/(3.00×108)=1.061×10−5 PaP_{\text{rad}} = 2I/c = 2 \times 1592/(3.00 \times 10^8) = 1.061 \times 10^{-5}\,\text{Pa}

Force on mirror: F=Prad×A=1.061×10−5×3.14×10−6=3.33×10−11 NF = P_{\text{rad}} \times A = 1.061 \times 10^{-5} \times 3.14 \times 10^{-6} = 3.33 \times 10^{-11}\,\text{N}

(c) Energy per photon: Ephoton=hc/λ=6.63×10−34×3.00×108/(532×10−9)=3.74×10−19 JE_{\text{photon}} = hc/\lambda = 6.63 \times 10^{-34} \times 3.00 \times 10^8/(532 \times 10^{-9}) = 3.74 \times 10^{-19}\,\text{J}

Photons per second: n=P/Ephoton=5.0×10−3/(3.74×10−19)=1.34×1016 s−1n = P/E_{\text{photon}} = 5.0 \times 10^{-3}/(3.74 \times 10^{-19}) = 1.34 \times 10^{16}\,\text{s}^{-1}

Momentum per photon: p=E/c=3.74×10−19/(3.00×108)=1.25×10−27 kg m s−1p = E/c = 3.74 \times 10^{-19}/(3.00 \times 10^8) = 1.25 \times 10^{-27}\,\text{kg}\,\text{m}\,\text{s}^{-1}

For perfect reflection, momentum transfer per photon =2p=2.50×10−27 kg m s−1= 2p = 2.50 \times 10^{-27}\,\text{kg}\,\text{m}\,\text{s}^{-1}

Total momentum transfer per second =1.34×1016×2.50×10−27=3.34×10−11 N= 1.34 \times 10^{16} \times 2.50 \times 10^{-27} = 3.34 \times 10^{-11}\,\text{N}

This matches the force calculated in part (b) (Newton’s second law: force = rate of change of momentum).

IT-3: EM Wave in a Medium (with Refraction)

Question:

An EM wave of frequency 5.0×1014 Hz5.0 \times 10^{14}\,\text{Hz} travels from air into glass of refractive index n=1.50n = 1.50.

(a) Calculate the speed, wavelength, and frequency of the wave in the glass.

(b) Calculate the impedance of the glass and the reflection coefficient at normal incidence.

(c) Calculate the depth of penetration at which the intensity drops to 1/e1/e of its value at the surface, assuming the glass has an absorption coefficient α=0.50 m−1\alpha = 0.50\,\text{m}^{-1}.

Take c = 3.00 \times 10^8\,\text{m}\,\text{s}^{-1}$$Z_0 = 377\,\Omega (impedance of free space).

Solution:

(a) Frequency is unchanged: f=5.0×1014 Hzf = 5.0 \times 10^{14}\,\text{Hz}

Speed in glass: v=c/n=3.00×108/1.50=2.00×108 m s−1v = c/n = 3.00 \times 10^8/1.50 = 2.00 \times 10^8\,\text{m}\,\text{s}^{-1}

Wavelength in glass: λ=v/f=2.00×108/(5.0×1014)=4.0×10−7 m=400 nm\lambda = v/f = 2.00 \times 10^8/(5.0 \times 10^{14}) = 4.0 \times 10^{-7}\,\text{m} = 400\,\text{nm}

The wavelength decreases by the factor nn while the frequency remains constant.

(b) Impedance of glass: Z=Z0/n=377/1.50=251 ΩZ = Z_0/n = 377/1.50 = 251\,\Omega

Reflection coefficient (amplitude) at normal incidence: r=Z2−Z1Z2+Z1=251−377251+377=−126628=−0.201r = \frac{Z_2 - Z_1}{Z_2 + Z_1} = \frac{251 - 377}{251 + 377} = \frac{-126}{628} = -0.201

Reflectance (intensity): R=r2=0.0404=4.0%R = r^2 = 0.0404 = 4.0\%

About 4%4\% of the light intensity is reflected at the air-glass interface.

(c) Intensity in an absorbing medium: I=I0e−αxI = I_0 e^{-\alpha x}

At I=I0/eI = I_0/e: αx=1⇒x=1/α=1/0.50=2.0 m\alpha x = 1 \Rightarrow x = 1/\alpha = 1/0.50 = 2.0\,\text{m}

The penetration depth is 2.0 m2.0\,\text{m}. For glass with this absorption coefficient, the light penetrates deeply (typical window glass has much lower absorption in the visible range).