JEE Physics Electromagnetic Waves Notes Set 03

Displacement Current

A steady current gives the same magnetic field whichever surface is used in Ampere's law. For a charging capacitor this fails: a flat surface cutting the wire has current through it, but a bowl-shaped surface passing between the plates has none.
The gap is filled by the changing electric field between the plates. With plate charge Q, the field gives flux \( \Phi_E=\frac{Q}{\varepsilon_0} \), so \( \varepsilon_0\frac{d\Phi_E}{dt}=\frac{dQ}{dt}=i \). This quantity is called the displacement current \( i_d \).
Total current for any surface: \( i=i_c+i_d=i_c+\varepsilon_0\frac{d\Phi_E}{dt} \). Outside the plates only \( i_c \) exists; between them only \( i_d \) exists and it equals the circuit current. The displacement current has the same magnetic effect as a conduction current.

Maxwell's Equations

1. Gauss's law for electricity: \( \oint \vec E\cdot d\vec A=\frac{Q}{\varepsilon_0} \).
2. Gauss's law for magnetism: \( \oint \vec B\cdot d\vec A=0 \) (there are no isolated magnetic poles).
3. Faraday's law: \( \oint \vec E\cdot d\vec l=-\frac{d\Phi_B}{dt} \) (a changing magnetic field produces an electric field).
4. Ampere-Maxwell law: \( \oint \vec B\cdot d\vec l=\mu_0 i_c+\mu_0\varepsilon_0\frac{d\Phi_E}{dt} \) (a changing electric field produces a magnetic field).
Because each changing field produces the other, a disturbance can sustain itself and travel through empty space as an electromagnetic wave.

Production and Nature of Electromagnetic Waves

Static charges and steady currents do not radiate. Only accelerated charges, for example charges oscillating in an antenna, emit waves, and the wave frequency equals the oscillation frequency. The energy carried away comes from the source.
The fields are transverse: \( \vec E \), \( \vec B \) and the direction of travel are mutually perpendicular, and the wave moves along \( \vec E\times\vec B \). For a wave along z, \( E_x=E_0\sin(kz-\omega t) \) and \( B_y=B_0\sin(kz-\omega t) \); E and B are in phase.
Relations: \( k=\frac{2\pi}{\lambda} \), \( \omega=2\pi\nu \), \( c=\nu\lambda=\frac{\omega}{k}=\frac{1}{\sqrt{\mu_0\varepsilon_0}} \), and \( \frac{E_0}{B_0}=c \). No material medium is needed; the ether idea was dropped after the Michelson-Morley experiment.
In a medium the speed is \( v=\frac{1}{\sqrt{\mu\varepsilon}} \) and the refractive index is \( n=\frac{c}{v}=\sqrt{\frac{\mu\varepsilon}{\mu_0\varepsilon_0}} \). The frequency does not change on entering a medium; the wavelength shrinks by the factor n.

Energy, Intensity and Momentum

Energy densities: \( u_E=\frac12\varepsilon_0E^2 \) and \( u_B=\frac{B^2}{2\mu_0} \). They are equal at every point, so \( u=\varepsilon_0E^2 \). Their time averages are \( \frac14\varepsilon_0E_0^2 \) each, and the energy density oscillates at twice the wave frequency.
Intensity (average energy crossing unit area per second): \( I=\frac12\varepsilon_0cE_0^2=\frac{cB_0^2}{2\mu_0} \) in terms of peak values, or \( I=\varepsilon_0cE_{rms}^2 \); for a point source of power P, \( I=\frac{P}{4\pi r^2} \).
Momentum: for energy U fully absorbed, the momentum delivered is \( p=\frac Uc \); for full reflection it is \( \frac{2U}{c} \). So radiation exerts pressure \( \frac Ic \) on an absorber.

Electromagnetic Spectrum

The spectrum has no sharp boundaries. In order of decreasing wavelength:
1. Radio waves (above about 0.1 m): produced by accelerated charges in circuits; AM band 530 kHz to 1710 kHz, FM 88 to 108 MHz; used in broadcasting and mobile communication.
2. Microwaves (about 1 mm to 0.1 m, GHz frequencies): made by klystrons, magnetrons and Gunn diodes; used in radar and microwave ovens (matched to the natural frequency of water molecules).
3. Infrared (about 700 nm to 1 mm): from vibrating atoms and molecules; called heat waves; used in physiotherapy, remote controls and night vision.
4. Visible light (about 400 nm to 700 nm): violet to red, with red having the longest wavelength and lowest frequency.
5. Ultraviolet (about 1 nm to 400 nm): from inner-shell electron transitions and the Sun; absorbed by the ozone layer; used in sterilising and LASIK surgery.
6. X-rays (about 10 pm to 1 nm): from sudden stopping of fast electrons and inner transitions; used in medical imaging and crystal studies.
7. Gamma rays (below about 10 pm): emitted by radioactive nuclei; the most penetrating radiation, used in cancer treatment.
All of them travel at c in vacuum and cannot be deflected by electric or magnetic fields, because they carry no charge. Photon energy \( E=h\nu \) rises from radio to gamma.

Worked Examples

Worked Example 1. A plane electromagnetic wave of frequency 25 MHz travels along x in free space. At a point, \( \vec E=6.3\hat j \) V/m. Find \( \vec B \).
Answer:
\( B=\frac Ec=\frac{6.3}{3\times10^8}=2.1\times10^{-8} \) T.
The wave moves along +x and E is along +y, so B must satisfy \( \hat j\times\hat k=\hat i \): B is along +z. Hence \( \vec B=2.1\times10^{-8}\hat k \) T.

Worked Example 2. The magnetic field in a plane electromagnetic wave is \( B_y=2\times10^{-7}\sin(0.5\times10^3x+1.5\times10^{11}t) \) T. Find the wavelength, frequency and the electric field.
Answer:
\( \lambda=\frac{2\pi}{0.5\times10^3}=1.26 \) cm; \( \nu=\frac{1.5\times10^{11}}{2\pi}=23.9 \) GHz.
\( E_0=cB_0=3\times10^8\times2\times10^{-7}=60 \) V/m, along z: \( E_z=60\sin(0.5\times10^3x+1.5\times10^{11}t) \) V/m.

Worked Example 3. Light of energy flux 18 W/cm\( ^2 \) falls on a non-reflecting surface of area 20 cm\( ^2 \) at normal incidence. Find the average force during 30 minutes.
Answer:
Energy \( U=18\times20\times1800=6.48\times10^5 \) J.
Momentum \( p=\frac Uc=2.16\times10^{-3} \) kg m/s, and force \( F=\frac pt=\frac{2.16\times10^{-3}}{1800}=1.2\times10^{-6} \) N.

Worked Example 4. A 100 W bulb is 2.5 percent efficient in producing light. Find the peak electric and magnetic fields at 3 m, treating it as a point source.
Answer:
\( I=\frac{100\times0.025}{4\pi(3)^2}=0.022 \) W/m\( ^2 \).
\( E_{rms}=\sqrt{\frac{I}{\varepsilon_0c}}=2.9 \) V/m, so \( E_0=\sqrt2E_{rms}=4.1 \) V/m and \( B_0=\frac{E_0}{c}=1.4\times10^{-8} \) T.

Worked Example 5. A capacitor of circular plates (radius 1 m, C = 1 nF) is charged at t = 0 through R = 1 M\( \Omega \) from a 2 V battery. Find B at a point halfway between centre and edge at t = 1 ms.
Answer:
\( \tau=RC=10^{-3} \) s. Charge \( q=CV(1-e^{-t/\tau}) \), so \( i=\frac VRe^{-t/\tau}=2\times10^{-6}e^{-1}=7.36\times10^{-7} \) A.
The loop of half the radius encloses a quarter of the plate area, so \( i_d=\frac i4=1.84\times10^{-7} \) A.
\( B(2\pi r)=\mu_0i_d \) with r = 0.5 m gives \( B=\frac{4\pi\times10^{-7}\times1.84\times10^{-7}}{\pi}=7.4\times10^{-14} \) T.