Skip to content
CalibratePhysics

Chapter 8

Electromagnetic Waves

CBSE · 3 subtopics

Formula SheetEvery formula this chapter can be examined on, in one table.

Constants and conversions CBSE expects

  • c=3×108c = 3\times10^{8} m s−1^{-1} — speed of every electromagnetic wave in vacuum, the same for all wavelengths.
  • μ0=4π×10−7\mu_0 = 4\pi\times10^{-7} T m A−1^{-1} — permeability of free space.
  • ε0=8.85×10−12\varepsilon_0 = 8.85\times10^{-12} C2^2 N−1^{-1} m−2^{-2} — permittivity of free space.
  • 14πε0=9×109\dfrac{1}{4\pi\varepsilon_0} = 9\times10^{9} N m2^2 C−2^{-2} — useful when ε0\varepsilon_0 is not quoted.
  • h=6.63×10−34h = 6.63\times10^{-34} J s =4.14×10−15= 4.14\times10^{-15} eV s; 11 eV =1.6×10−19= 1.6\times10^{-19} J — only for the photon-energy part of the spectrum question.
  • 1 A˚=10−101\ \text{\text{Å}} = 10^{-10} m, 11 nm =10−9= 10^{-9} m, 1 μ1\ \mum =10−6= 10^{-6} m.

Electric flux and displacement current

FormulaSymbols, units and conditions
ϕE=∮E⃗⋅dA⃗=EA(E⃗ uniform and perpendicular to the area A)\phi_E = \oint \vec E\cdot d\vec A = EA \quad (\vec E \text{ uniform and perpendicular to the area } A)
  • ϕE\phi_E = electric flux in N m2^2 C−1^{-1} (or V m), EE = field (N C−1^{-1} or V m−1^{-1}), AA = area (m2^2). This is the flux through the flat surface drawn between the capacitor plates.
ID=ε0 dϕEdtI_D = \varepsilon_0\,\dfrac{d\phi_E}{dt}
  • IDI_D = displacement current (A). Defining equation. It is non-zero only while the electric field (hence the flux) is changing with time; in any steady state ID=0I_D = 0.
E=σε0=QAε0ϕE=EA=Qε0ID=ε0ddt ⁣(Qε0)=dQdt=ICE = \dfrac{\sigma}{\varepsilon_0} = \dfrac{Q}{A\varepsilon_0} \qquad \phi_E = EA = \dfrac{Q}{\varepsilon_0} \qquad I_D = \varepsilon_0\dfrac{d}{dt}\!\left(\dfrac{Q}{\varepsilon_0}\right) = \dfrac{dQ}{dt} = I_C
  • σ\sigma = surface charge density (C m−2^{-2}), QQ = instantaneous charge on a plate (C). The standard two/three-mark result: inside a charging parallel-plate capacitor the displacement current equals the conduction (charging) current in the wire.
I=IC+IDI = I_C + I_D
  • Total current is continuous everywhere. Outside the plates IC=II_C = I, ID=0I_D = 0; between the plates IC=0I_C = 0, ID=II_D = I.
JD=ε0 dEdtJ_D = \varepsilon_0\,\dfrac{dE}{dt}
  • Displacement current density (A m−2^{-2}), for a uniform field over area AA, with ID=JDAI_D = J_D A.

Ampere-Maxwell law and Maxwell's equations

FormulaSymbols, units and conditions
∮B⃗⋅dl⃗=μ0 ⁣(IC+ε0dϕEdt)=μ0(IC+ID)\oint \vec B\cdot d\vec l = \mu_0\!\left(I_C + \varepsilon_0\dfrac{d\phi_E}{dt}\right) = \mu_0(I_C + I_D)
  • Ampere-Maxwell law (Maxwell's generalisation of Ampere's circuital law). B⃗\vec B in tesla, dl⃗d\vec l along the Amperian loop. Quote this whenever the question says "generalised form".
∮E⃗⋅dA⃗=Qε0∮B⃗⋅dA⃗=0\oint \vec E\cdot d\vec A = \dfrac{Q}{\varepsilon_0} \qquad \oint \vec B\cdot d\vec A = 0
  • Gauss's law for electricity (charge is a source of E⃗\vec E) and Gauss's law for magnetism (no magnetic monopoles).
∮E⃗⋅dl⃗=−dϕBdt\oint \vec E\cdot d\vec l = -\dfrac{d\phi_B}{dt}
  • Faraday's law: a time-varying magnetic flux ϕB\phi_B (weber) produces an electric field. These four, with the Lorentz force F⃗=q(E⃗+v⃗×B⃗)\vec F = q(\vec E + \vec v\times\vec B), are Maxwell's equations in vacuum.

Magnetic field produced by a displacement current

FormulaSymbols, units and conditions
B(2πr)=μ0I  ⇒  B=μ0I2πr(r>R, outside the plates)B(2\pi r) = \mu_0 I \;\Rightarrow\; B = \dfrac{\mu_0 I}{2\pi r} \quad (r > R,\ \text{outside the plates})
  • rr = distance from the axis, RR = plate radius. Same as the field of a long straight wire.
B=μ0r2πR2 ID(r<R, between the plates)B = \dfrac{\mu_0 r}{2\pi R^2}\, I_D \quad (r < R,\ \text{between the plates})
  • Only the fraction πr2/πR2\pi r^2/\pi R^2 of the displacement current is enclosed by a loop of radius rr. For the amplitude, put the peak current I0I_0 in place of IDI_D.
C=ε0Ad=ε0πR2dI=C dVdtC = \dfrac{\varepsilon_0 A}{d} = \dfrac{\varepsilon_0 \pi R^2}{d} \qquad I = C\,\dfrac{dV}{dt}
  • CC = capacitance (F), dd = plate separation (m). Gives the rate of change of potential difference when the charging current is constant.
XC=1ωCIrms=VrmsXC=Vrms ωCI0=2 IrmsX_C = \dfrac{1}{\omega C} \qquad I_{rms} = \dfrac{V_{rms}}{X_C} = V_{rms}\,\omega C \qquad I_0 = \sqrt2\, I_{rms}
  • For a capacitor across an ac source of angular frequency ω\omega (rad s−1^{-1}); needed for the "amplitude of BB between the plates" numerical.

Plane electromagnetic wave

FormulaSymbols, units and conditions
Ex=E0sin⁡(kz−ωt)By=B0sin⁡(kz−ωt)E_x = E_0\sin(kz - \omega t) \qquad B_y = B_0\sin(kz - \omega t)
  • Wave travelling along +z+z, E⃗\vec E along xx, B⃗\vec B along yy. E0E_0 in V m−1^{-1}, B0B_0 in T. The two fields are in the same phase.
k=2πλω=2πν=2πTv=ωkk = \dfrac{2\pi}{\lambda} \qquad \omega = 2\pi\nu = \dfrac{2\pi}{T} \qquad v = \dfrac{\omega}{k}
  • kk = propagation constant / wave number (rad m−1^{-1}), λ\lambda = wavelength (m), ν\nu = frequency (Hz), TT = period (s). Read kk and ω\omega straight off a given wave equation.
c=νλc = \nu\lambda
  • Connects any band of the spectrum to its wavelength; the only relation needed for "convert this frequency band to a wavelength band".
c=1μ0ε0=E0B0=EBc = \dfrac{1}{\sqrt{\mu_0\varepsilon_0}} = \dfrac{E_0}{B_0} = \dfrac{E}{B}
  • Two independent expressions for the speed in vacuum: from the constants of free space, and from the ratio of the field amplitudes (or of the instantaneous values). Both are examined.
v=1με=1μ0μr ε0εr=cμrεrv = \dfrac{1}{\sqrt{\mu\varepsilon}} = \dfrac{1}{\sqrt{\mu_0\mu_r\,\varepsilon_0\varepsilon_r}} = \dfrac{c}{\sqrt{\mu_r\varepsilon_r}}
  • μ,ε\mu, \varepsilon = permeability and permittivity of the medium; μr,εr\mu_r, \varepsilon_r = their relative (dimensionless) values. Speed in a material medium is always less than cc.
n=cv=μrεrn = \dfrac{c}{v} = \sqrt{\mu_r\varepsilon_r}
  • Refractive index of the medium. For most transparent media μr≈1\mu_r \approx 1, so n≈εrn \approx \sqrt{\varepsilon_r}.

Energy, intensity and momentum

FormulaSymbols, units and conditions
uE=12ε0E2uB=B22μ0u_E = \dfrac12\varepsilon_0 E^2 \qquad u_B = \dfrac{B^2}{2\mu_0}
  • Instantaneous energy densities of the electric and magnetic fields, in J m−3^{-3}.
uˉE=12ε0Erms2=14ε0E02uˉB=Brms22μ0=B024μ0uˉE=uˉB\bar u_E = \dfrac12\varepsilon_0 E_{rms}^2 = \dfrac14\varepsilon_0 E_0^2 \qquad \bar u_B = \dfrac{B_{rms}^2}{2\mu_0} = \dfrac{B_0^2}{4\mu_0} \qquad \bar u_E = \bar u_B
  • Averages over one cycle, using Erms=E0/2E_{rms} = E_0/\sqrt2 and Brms=B0/2B_{rms} = B_0/\sqrt2. The equality is the standard three-mark proof; energy is shared equally between the two fields.
uˉ=uˉE+uˉB=12ε0E02=B022μ0\bar u = \bar u_E + \bar u_B = \dfrac12\varepsilon_0 E_0^2 = \dfrac{B_0^2}{2\mu_0}
  • Total average energy density of the wave.
I=uˉ c=12ε0E02cI = \bar u\, c = \dfrac12\varepsilon_0 E_0^2 c
  • Intensity (average power per unit area) in W m−2^{-2}.
p=Uc(total absorption)p=2Uc(total reflection)p = \dfrac{U}{c} \quad (\text{total absorption}) \qquad p = \dfrac{2U}{c} \quad (\text{total reflection})
  • UU = energy delivered (J), pp = momentum transferred (kg m s−1^{-1}). This is what "em waves transport momentum" means.
Prad=Ic(absorbing surface)Prad=2Ic(reflecting surface)P_{rad} = \dfrac{I}{c} \quad (\text{absorbing surface}) \qquad P_{rad} = \dfrac{2I}{c} \quad (\text{reflecting surface})
  • Radiation pressure in N m−2^{-2} (Pa). Quote it to justify "an em wave exerts pressure on a surface".

Photon energy (for the spectrum question)

FormulaSymbols, units and conditions
E=hν=hcλE(in eV)=1240λ (in nm)E = h\nu = \dfrac{hc}{\lambda} \qquad E(\text{in eV}) = \dfrac{1240}{\lambda\ (\text{in nm})}
  • Photon energy rises steadily from radio waves (a tiny fraction of an eV) through visible light (a few eV) to gamma rays (of order MeV); the higher the photon energy, the more energetic the source — nuclear for gamma rays, inner-shell electrons for X-rays and UV, outer electrons for visible, molecular vibrations for infrared, circuit currents for radio.

The electromagnetic spectrum — bands, production, detection, uses

  • RadiationWavelength rangeFrequency rangeProductionDetectionMain uses
  • Radio wavesGreater than 0.10.1 m (up to ∼106\sim10^{6} m)Less than 3×1093\times10^{9} Hz (broadcast band 500500 kHz to 10001000 MHz)Rapid acceleration and deceleration of electrons in aerials, oscillating LC circuitsReceiver's aerialsRadio and TV broadcasting, AM (530530–17101710 kHz), FM (8888–108108 MHz), cellular phones (UHF), satellite communication
  • Microwaves0.10.1 m to 11 mm3×1093\times10^{9} Hz to 3×10113\times10^{11} Hz (often quoted 3×1083\times10^{8} to 3×10113\times10^{11} Hz)Special vacuum tubes — klystrons, magnetrons, Gunn diodesPoint contact diodesRadar and aircraft navigation, speed guns, microwave ovens, long-distance communication
  • Infrared11 mm to 700700 nm3×10113\times10^{11} Hz to 4×10144\times10^{14} HzVibration of atoms and molecules; hot bodiesThermopiles, bolometer, infrared photographic filmPhysical therapy and muscular strain, TV and appliance remote switches, night vision, photography through haze and fog, greenhouse warming, crop observation from satellites
  • Visible light700700 nm to 400400 nm4×10144\times10^{14} Hz to 7×10147\times10^{14} HzElectrons in atoms falling from a higher to a lower energy levelThe eye, photocells, photographic filmVision; gives us information about the world around us
  • Ultraviolet400400 nm to 11 nm8×10148\times10^{14} Hz to 3×10173\times10^{17} Hz (commonly quoted 8×10148\times10^{14} to 8×10168\times10^{16} Hz)Inner-shell electrons in atoms moving to a lower level; special UV lamps, very hot bodies, the Sun, welding arcsPhotocells, photographic filmKilling germs in water purifiers, LASIK eye surgery, sterilising and preserving foodstuff
  • X-rays11 nm to 10−310^{-3} nm3×10173\times10^{17} Hz to 3×10203\times10^{20} Hz (commonly quoted 101610^{16} to 102010^{20} Hz)X-ray tubes — bombarding a metal target of high atomic number with high-energy electrons; inner-shell electron transitionsPhotographic film, Geiger tubes, ionisation chamberMedical diagnosis such as detecting bone fractures, treatment of cancer, study of crystal structure by diffraction
  • Gamma raysLess than 10−310^{-3} nm (about 10−1010^{-10} m down to below 10−1410^{-14} m)Greater than about 3×10183\times10^{18} HzRadioactive decay of the nucleus; nuclear reactions; decay of subatomic particlesPhotographic film, Geiger tubes, ionisation chamberRadiotherapy to destroy cancer cells, sterilisation of medical instruments, nuclear studies
  • Order of increasing wavelength: gamma rays, X-rays, ultraviolet, visible, infrared, microwaves, radio waves. Order of increasing frequency and of increasing photon energy is exactly the reverse.
  • The boundaries are not sharp; X-rays and gamma rays overlap in wavelength and are distinguished by origin — gamma rays come from the nucleus, X-rays from outside it.
Short NotesThe chapter subtopic by subtopic, in the textbook's own order.3 subtopics
  1. 1Displacement Current9 board questions
  2. 2Electromagnetic Waves13 board questions
  3. 3Electromagnetic Spectrum29 board questions
The whole chapter in brief

8.1 Introduction — what Maxwell did

  • Chapter 4: a current produces a magnetic field. Chapter 6: a changing magnetic field produces an electric field. Maxwell asked the converse — does a changing electric field produce a magnetic field? — and answered yes.
  • Applying Ampere's circuital law to a point outside a capacitor being charged by a time-varying current, Maxwell found an inconsistency. To remove it he introduced an extra current, the displacement current, due to the time-varying electric field.
  • Maxwell's equations, together with the Lorentz force formula, express all the basic laws of electromagnetism.
  • Their most important prediction is the existence of electromagnetic waves — coupled, time-varying electric and magnetic fields that propagate through space.
  • The speed the equations gave, c=1/μ0ε0≈3×108c = 1/\sqrt{\mu_0\varepsilon_0} \approx 3\times10^8 m s−1^{-1}, matched the optically measured speed of light, so light is an electromagnetic wave. Maxwell thus unified electricity, magnetism and optics.
  • Hertz demonstrated em waves experimentally in 1887 (NCERT also mentions 1885 in the introduction); J. C. Bose in Calcutta produced much shorter waves (2525 mm to 55 mm); Marconi transmitted them over many kilometres, starting radio communication.

8.2 Displacement current

  • Definition: the current that appears, in addition to conduction current, wherever the electric field (and hence the electric flux) in a region is changing with time. ID=ε0dϕEdtI_D = \varepsilon_0\dfrac{d\phi_E}{dt}
  • The inconsistency it removes: for a circular Amperian loop of radius rr round the wire feeding a capacitor, the flat surface pierced by the wire gives ∮B⃗⋅dl⃗=μ0I\oint\vec B\cdot d\vec l = \mu_0 I, but a pot-shaped or tiffin-shaped surface with the same rim, whose bottom lies between the plates, is pierced by no conduction current and gives zero. Two answers for the same loop integral — impossible.
  • Maxwell's fix: the flat surface between the plates does have something crossing it — the electric field. Adding ε0 dϕE/dt\varepsilon_0\, d\phi_E/dt to the enclosed current makes the answer the same for every surface bounded by the loop.
  • For a parallel-plate capacitor, E=Q/(Aε0)E = Q/(A\varepsilon_0), so ϕE=Q/ε0\phi_E = Q/\varepsilon_0 and ID=ε0 dϕE/dt=dQ/dt=ICI_D = \varepsilon_0\, d\phi_E/dt = dQ/dt = I_C. The displacement current between the plates exactly equals the conduction current in the wire, so total current is continuous round the whole circuit.
  • Outside the plates: only conduction current (IC=II_C = I, ID=0I_D = 0). Between the plates: only displacement current (IC=0I_C = 0, ID=II_D = I).
  • Displacement current has all the same physical effects as conduction current; in particular it produces a magnetic field. The field measured just inside the gap (point M) equals the field just outside (point P).
  • In a dc circuit at steady state the capacitor is fully charged, the field between the plates is constant, so ID=0I_D = 0, and the insulating gap carries no conduction current — hence no steady current flows in a capacitor branch. During charging or discharging, QQ changes, so a momentary current does flow.

Maxwell's equations in vacuum

EquationNameWhat it says
∮E⃗⋅dA⃗=Qε0\oint\vec E\cdot d\vec A = \dfrac{Q}{\varepsilon_0}Gauss's law for electricityElectric charge is the source of the electric field
∮B⃗⋅dA⃗=0\oint\vec B\cdot d\vec A = 0Gauss's law for magnetismNo magnetic monopoles exist; magnetic field lines are closed
∮E⃗⋅dl⃗=−dϕBdt\oint\vec E\cdot d\vec l = -\dfrac{d\phi_B}{dt}Faraday's law of inductionA changing magnetic field produces an electric field
∮B⃗⋅dl⃗=μ0IC+μ0ε0dϕEdt\oint\vec B\cdot d\vec l = \mu_0 I_C + \mu_0\varepsilon_0\dfrac{d\phi_E}{dt}Ampere-Maxwell lawConduction current and a changing electric field both produce a magnetic field
  • The last two make the laws almost symmetric: a changing B⃗\vec B makes E⃗\vec E, and a changing E⃗\vec E makes B⃗\vec B. They are not perfectly symmetric because there are no magnetic monopoles.
  • That mutual regeneration is exactly what allows a self-sustaining electromagnetic wave.

8.3.1 Sources of electromagnetic waves

  • Stationary charges give only an electrostatic field. Charges in uniform motion (steady current) give a magnetic field that does not change with time. Neither radiates.
  • Accelerated charges radiate electromagnetic waves. An oscillating charge is an accelerating charge, so it radiates; an oscillating electric dipole is the basic source of em waves.
  • The wave frequency equals the frequency of oscillation of the source charge. The energy carried away comes from the energy of the oscillating charge.
  • Yellow light needs about 6×10146\times10^{14} Hz, but electronic circuits reach only about 101110^{11} Hz — which is why the first experimental demonstration (Hertz) was in the radio region.

8.3.2 Nature of electromagnetic waves

  • E⃗\vec E and B⃗\vec B are perpendicular to each other and both perpendicular to the direction of propagation — em waves are transverse.
  • E⃗\vec E and B⃗\vec B oscillate in the same phase: they reach their maxima, zeros and minima at the same instants.
  • E⃗×B⃗\vec E \times \vec B points along the direction of propagation; E⃗\vec E, B⃗\vec B and the propagation direction form a right-handed set.
  • Standard forms: Ex=E0sin⁡(kz−ωt)E_x = E_0\sin(kz-\omega t) and By=B0sin⁡(kz−ωt)B_y = B_0\sin(kz-\omega t) for a wave travelling along +z+z, with k=2π/λk = 2\pi/\lambda, ω=2πν\omega = 2\pi\nu and speed ω/k\omega/k.
  • Amplitudes are related by E0=cB0E_0 = cB_0, so B0B_0 is smaller than E0E_0 by a factor of 3×1083\times10^8.
  • They need no material medium; they are self-sustaining oscillations of the fields in vacuum. This is what makes them different from sound and from waves on a string.
  • In vacuum every em wave travels at the same speed cc, independent of wavelength — the ratio of the speeds of two light waves of different wavelengths in vacuum is 1:11:1.
  • The constancy of cc is so well established that it is used to define the standard of length.
  • In a medium the speed falls to v=1/με=c/μrεrv = 1/\sqrt{\mu\varepsilon} = c/\sqrt{\mu_r\varepsilon_r}; the refractive index is n=c/vn = c/v.
  • Em waves carry energy and momentum, and are electrically neutral, so they are not deflected by electric or magnetic fields.
  • Energy is shared equally between the electric and magnetic fields: uˉE=uˉB=14ε0E02\bar u_E = \bar u_B = \dfrac14\varepsilon_0E_0^2.

Standard diagram in words — linearly polarised plane wave

  • Draw the propagation direction (say the zz-axis) horizontally across the page.
  • Draw the electric field as a sine curve in the xx–zz plane, its arrows along ±x\pm x, drawn as a function of zz at a fixed instant.
  • Draw the magnetic field as a sine curve in the yy–zz plane, its arrows along ±y\pm y, with the same wavelength.
  • Draw the two curves in phase — both crossing zero at the same values of zz and peaking at the same values of zz — and check that E⃗×B⃗\vec E\times\vec B points along +z+z.
  • Label E0E_0, B0B_0, one wavelength λ\lambda, and the axes. For a wave along +x+x with E⃗\vec E along +y+y, the same picture rotates so that B⃗\vec B is along +z+z.

Reading a wave equation quickly

GivenWhat to doResult
Ex=E0sin⁡(kz−ωt)E_x = E_0\sin(kz-\omega t)Opposite signs on kzkz and ωt\omega tWave travels along +z+z
Ex=E0sin⁡(ωt+kz)E_x = E_0\sin(\omega t + kz)Same signs on kzkz and ωt\omega tWave travels along −z-z
Coefficient of the position variableIt is kkλ=2π/k\lambda = 2\pi/k
Coefficient of ttIt is ω\omegaν=ω/2π\nu = \omega/2\pi
E⃗\vec E direction and propagation directionUse E⃗×B⃗\vec E\times\vec B along propagationGives the B⃗\vec B direction
B0B_0 givenMultiply by ccE0=cB0E_0 = cB_0

8.4 Electromagnetic spectrum

  • The classification by frequency (or wavelength) is the electromagnetic spectrum. There are no sharp boundaries; the classification is based roughly on how the waves are produced and detected.
  • In order of increasing wavelength: gamma rays, X-rays, ultraviolet, visible, infrared, microwaves, radio waves — from about 10−1210^{-12} m to about 10610^{6} m.
  • All bands travel at the same speed cc in vacuum; they differ only in wavelength/frequency, and therefore in how they interact with matter.
  • Radio waves — accelerated electrons in aerials; radio, TV, cellular communication.
  • Microwaves — klystrons, magnetrons, Gunn diodes; radar and aircraft navigation, speed guns, microwave ovens (frequency matched to the resonant frequency of water molecules).
  • Infrared — hot bodies and vibrating atoms and molecules; heat waves, physical therapy, remote switches, greenhouse effect, satellite crop monitoring, seeing through fog and haze.
  • Visible light — electrons in atoms dropping to a lower energy level; the only band the human eye detects, 700700 nm to 400400 nm.
  • Ultraviolet — special lamps, very hot bodies, the Sun, welding arcs; water purifiers, LASIK eye surgery, food preservation. Absorbed by ordinary glass and by the ozone layer.
  • X-rays — bombarding a metal target with high-energy electrons; medical diagnosis, cancer treatment, crystal structure.
  • Gamma rays — radioactive decay of nuclei; radiotherapy for cancer, sterilisation.
  • Photon energy E=hνE = h\nu increases from radio to gamma; the more energetic the band, the more energetic the process producing it (circuit currents, molecular vibrations, outer electrons, inner-shell electrons, nuclei).

Comparisons CBSE asks for

PointConduction currentDisplacement current
CauseActual flow of charges (electrons) through a conductorTime-varying electric field / electric flux in a region
Charge transportCharge physically crosses the surfaceNo charge crosses the surface
ExpressionIC=dQdtI_C = \dfrac{dQ}{dt} (flow of charge)ID=ε0dϕEdtI_D = \varepsilon_0\dfrac{d\phi_E}{dt}
Steady conditionsExists even for a steady currentZero when the field is constant in time
Where it dominatesIn the connecting wiresIn the gap between capacitor plates
Magnetic effectProduces a magnetic fieldProduces a magnetic field in exactly the same way
PointGamma raysRadio waves
OriginNuclear — radioactive decay of nuclei, decay of subatomic particlesAccelerated motion of charges in conducting wires and aerials
WavelengthShortest, less than 10−310^{-3} nmLongest, greater than 0.10.1 m
Photon energyHighest, of order MeVLowest
Main useRadiotherapy for cancer, sterilisationRadio, TV and satellite communication
Speed in vacuum3×1083\times10^{8} m s−1^{-1}3×1083\times10^{8} m s−1^{-1} — the same

Results worth memorising

  • ID=ICI_D = I_C for a capacitor being charged; if the charging current is 0.250.25 A, the displacement current across the plates is 0.250.25 A.
  • c=1/μ0ε0=E0/B0=3×108c = 1/\sqrt{\mu_0\varepsilon_0} = E_0/B_0 = 3\times10^{8} m s−1^{-1}.
  • v=c/μrεrv = c/\sqrt{\mu_r\varepsilon_r} in a medium; n=μrεrn = \sqrt{\mu_r\varepsilon_r}.
  • uˉE=uˉB=14ε0E02\bar u_E = \bar u_B = \dfrac14\varepsilon_0E_0^2, total uˉ=12ε0E02\bar u = \dfrac12\varepsilon_0E_0^2.
  • p=U/cp = U/c and radiation pressure I/cI/c (absorbed) or 2I/c2I/c (reflected).
  • Between the plates of a circular parallel-plate capacitor, B=μ0rID2πR2B = \dfrac{\mu_0 r I_D}{2\pi R^2} for r<Rr < R.
  • A charge oscillating at 10910^{9} Hz radiates em waves of frequency 10910^{9} Hz — the frequencies are identical.
  • 2121 cm hydrogen line: microwave region.
  • 5×10195\times10^{19} Hz: gamma-ray region. 101610^{16} to 102010^{20} Hz: X-rays.
  • If the Earth had no atmosphere its average surface temperature would be lower, because greenhouse gases trap re-radiated infrared.

Traps that cost marks

  • Displacement current is not a flow of charge. Never say "electrons jump across the gap"; say the changing electric flux acts as a source of magnetic field.
  • ID=ε0 dϕE/dtI_D = \varepsilon_0\, d\phi_E/dt — do not drop the ε0\varepsilon_0, and do not write dϕB/dtd\phi_B/dt, which belongs to Faraday's law.
  • In the Ampere-Maxwell law the μ0\mu_0 multiplies both terms: μ0(IC+ID)\mu_0(I_C + I_D), so the second term carries μ0ε0\mu_0\varepsilon_0.
  • E⃗\vec E and B⃗\vec B in an em wave are in the same phase, not out of phase by 90∘90^\circ. That confusion is a standard MCQ distractor.
  • E⃗\vec E and B⃗\vec B are perpendicular to each other, never parallel — the other standard distractor.
  • Em waves are transverse, not longitudinal, and need no medium.
  • E0=cB0E_0 = cB_0, so the magnetic amplitude is the smaller number. Multiplying instead of dividing (or the reverse) is the commonest numerical slip.
  • sin⁡(kz−ωt)\sin(kz-\omega t) travels along +z+z; sin⁡(ωt+kz)\sin(\omega t + kz) travels along −z-z. Check the relative signs before naming a direction.
  • When asked for the direction of B⃗\vec B, use E⃗×B⃗\vec E\times\vec B along the propagation direction — and remember to include the sign, e.g. along −y-y, not just "along yy".
  • All em waves have the same speed in vacuum: the ratio of the speeds of 40004000 Å and 80008000 Å light in vacuum is 1:11:1, not 1:21:2.
  • Frequency does not change when a wave enters a medium; the wavelength and the speed do.
  • Ultraviolet, not X-rays, is what a welder's goggles block, what purifies water and what is used in LASIK eye surgery.
  • Infrared is the heat wave, and it lies between visible (shorter λ\lambda side) and microwaves (longer λ\lambda side).
  • Microwaves for radar; do not answer "radio waves". The reason is short wavelength giving a narrow, well-directed beam and good resolution.
  • Gamma rays and X-rays overlap in wavelength; separate them by origin (nucleus versus inner-shell electrons), which is what the marking scheme wants.
  • Write units: displacement current in ampere, energy density in J m−3^{-3}, ε0\varepsilon_0 in C2^2 N−1^{-1} m−2^{-2}, μ0\mu_0 in T m A−1^{-1}.
  • When a band is asked for "with its range", give the range — a band name alone loses the second mark.
Solved Numerical ExamplesWorked numericals to attempt first, each with a hint before the full solution.Coming soon

These are being added chapter by chapter. They are solved numericals from the prescribed book and the reference guides — not previous board questions, which live under Practice — so that you can work through the method before meeting it in a real paper.

Important DerivationsEvery derive / obtain / prove question in this chapter.4
  1. 1
    Derivation 1
    A parallel plate capacitor of plate area AA each and separation dd, is being charged by an ac source. Show that the displacement current inside the capacitor is the same as the current charging the capacitor.

    Asked in: CBSE Class 12 Physics — SA · AI 2019, U · Section B, 2 marks

    Show the derivation

    Let QQ be the instantaneous charge on the plates. The uniform electric field between the plates is E=σε0=QAε0E=\dfrac{\sigma}{\varepsilon_0}=\dfrac{Q}{A\varepsilon_0} Electric flux through the area AA between the plates: ϕE=EA=Qε0\phi_E=EA=\dfrac{Q}{\varepsilon_0} Displacement current: ID=ε0dϕEdt=ε0ddt(Qε0)=dQdtI_D=\varepsilon_0\dfrac{d\phi_E}{dt}=\varepsilon_0\dfrac{d}{dt}\left(\dfrac{Q}{\varepsilon_0}\right)=\dfrac{dQ}{dt} But dQdt=IC\dfrac{dQ}{dt}=I_C, the conduction (charging) current in the wire feeding the plate. Hence ID=ICI_D=I_C — the displacement current inside the capacitor exactly equals the current charging the capacitor.

  2. 2
    Derivation 2
    Write Maxwell's generalization of Ampere's circuital law. Show that in the process of charging a capacitor, the current produced within the plates of the capacitor is i=ε0dϕEdti=\varepsilon_0\dfrac{d\phi_E}{dt}, where ϕE\phi_E is the electric flux produced during charging of the capacitor plates.

    Asked in: CBSE Class 12 Physics — SA · Delhi 2016, Ev · Section C, 3 marks

    Show the derivation

    Maxwell's generalization of Ampere's circuital law (Ampere-Maxwell law): ∮B⃗⋅dl⃗=μ0(ic+id)=μ0(ic+ε0dϕEdt)\oint\vec B\cdot d\vec l=\mu_0(i_c+i_d)=\mu_0\left(i_c+\varepsilon_0\dfrac{d\phi_E}{dt}\right) where ici_c is the conduction current and id=ε0dϕEdti_d=\varepsilon_0\dfrac{d\phi_E}{dt} is the displacement current. Derivation for a charging capacitor: let the plates (area AA each) carry instantaneous charge QQ. The field between the plates is E=σε0=QAε0E=\dfrac{\sigma}{\varepsilon_0}=\dfrac{Q}{A\varepsilon_0}, so the electric flux between the plates is ϕE=EA=Qε0\phi_E=EA=\dfrac{Q}{\varepsilon_0}. Then i=ε0dϕEdt=ε0ddt(Qε0)=dQdti=\varepsilon_0\dfrac{d\phi_E}{dt}=\varepsilon_0\dfrac{d}{dt}\left(\dfrac{Q}{\varepsilon_0}\right)=\dfrac{dQ}{dt}, which is exactly the conduction current charging the capacitor plates. This confirms that the current "produced" within the plates (the displacement current) equals i=ε0dϕEdti=\varepsilon_0\dfrac{d\phi_E}{dt}, and is numerically equal to the charging current in the connecting wire.

  3. 3
    Derivation 3
    Illustrate by giving suitable examples, how you can show that electromagnetic waves carry both energy and momentum.

    Asked in: CBSE Class 12 Physics — Very Short Answer · AI 2019 · Section A, 1 mark

    Show the derivation

    Energy: when sunlight (an em wave) falls on a surface, it heats the surface up, and it can generate a photocurrent in a photocell — this shows the wave delivers energy to matter. Momentum: an em wave incident on a surface exerts a (radiation) pressure on it — for example, this pressure is what pushes the tail of a comet away from the Sun, and it is also the principle used in devices such as light sails, showing that em waves also carry and transfer momentum.

  4. 4
    Derivation 4
    Prove that the average energy density of the oscillating electric field is equal to that of the oscillating magnetic field.

    Asked in: CBSE Class 12 Physics — SA · Delhi 2019, U · Section C, 3 marks

    Show the derivation

    The average energy density of the electric field is uE=12ε0Erms2=14ε0E02u_E=\dfrac12\varepsilon_0E_{rms}^2=\dfrac14\varepsilon_0E_0^2 (using Erms=E0/2E_{rms}=E_0/\sqrt2). The average energy density of the magnetic field is uB=12μ0Brms2=14μ0B02u_B=\dfrac1{2\mu_0}B_{rms}^2=\dfrac1{4\mu_0}B_0^2 (using Brms=B0/2B_{rms}=B_0/\sqrt2). For an electromagnetic wave in vacuum, the field amplitudes are related by E0=cB0E_0=cB_0, and the speed of the wave is c=1μ0ε0⇒c2=1μ0ε0c=\dfrac1{\sqrt{\mu_0\varepsilon_0}} \Rightarrow c^2=\dfrac1{\mu_0\varepsilon_0}. Substituting E0=cB0E_0=cB_0 into uEu_E:   uE=14ε0(cB0)2=14ε0c2B02=14ε0⋅1μ0ε0⋅B02=14μ0B02\;u_E=\dfrac14\varepsilon_0(cB_0)^2=\dfrac14\varepsilon_0c^2B_0^2=\dfrac14\varepsilon_0\cdot\dfrac1{\mu_0\varepsilon_0}\cdot B_0^2=\dfrac1{4\mu_0}B_0^2 This is exactly uBu_B. Hence uE=uBu_E=u_B — the average energy densities of the electric and magnetic fields in an em wave are equal, and each equals half of the total average energy density.

Finished revising this chapter? Practise Electromagnetic Waves →