Chapter 8: Electromagnetic Waves (EM Waves)

(a) $A=\pi r^2=\pi (0.12{)}^2=4.52\times 10^{-2}$ m², $d=0.05$ m.

$$C=\frac{{\epsilon }_{0}A}{d}=\frac{8.854\times 10^{-12}\times 4.52\times 10^{-2}}{0.05}\approx 8.0\times 10^{-12}\text{F}=8.0\text{pF}$$

$Q=CV$, so $I=C\frac{dV}{dt}$ and

$$\frac{dV}{dt}=\frac{I}{C}=\frac{0.15}{8.0\times 10^{-12}}\approx 1.9\times 10^{10}\text{V/s}$$

(b) The displacement current between the plates equals the conduction current in the wires: $I_d=0.15$ A.

(c) Yes, provided we count the displacement current. At each plate, the conduction current in the wire equals the displacement current in the gap, so the current "in" equals the current "out" and the junction rule holds.

(a) C ≈ 8.0 pF, dV/dt ≈ $1.9\times 10^{10}$ V/s. (b) 0.15 A. (c) Yes, if displacement current is included.

(a) $X_C=\frac{1}{\omega C}$, so

$$I_{\text{rms}}=V_{\text{rms}}\omega C=230\times 300\times 100\times 10^{-12}=6.9\times 10^{-6}\text{A}=6.9\mu \text{A}$$

(b) Yes. The displacement current in the gap equals the conduction current in the wires at every instant.

(c) Between the plates the displacement current is spread evenly over the plate area. A circle of radius $r=3.0$ cm encloses the fraction $\frac{r^2}{R^2}$ of it. By the Ampere–Maxwell law, $B\cdot 2\pi r={\mu }_0{I}_d\frac{r^2}{R^2}$, so

$$B=\frac{{\mu }_{0}rI}{2\pi R^2}$$

With the peak current $I_0=\sqrt{2}\times 6.9\times 10^{-6}=9.76\times 10^{-6}$ A:

$$B_0=\frac{2\times 10^{-7}\times 0.03\times 9.76\times 10^{-6}}{(0.06{)}^2}\approx 1.6\times 10^{-11}\text{T}$$

(a) 6.9 μA. (b) Yes. (c) $B_0\approx 1.6\times 10^{-11}$ T.

All are electromagnetic waves, so in vacuum they all travel with the same speed, $c=3\times 10^8$ m/s. (Their frequencies and wavelengths are very different.)

Their speed in vacuum, $3\times 10^8$ m/s.

$\vec{E}$ and $\vec{B}$ both lie in the x-y plane, perpendicular to each other and to the direction of travel (z); $\vec{E}\times \vec{B}$ points along +z.

$$\lambda =\frac{c}{\nu }=\frac{3\times 10^8}{30\times 10^6}=10\text{m}$$

E and B are in the x-y plane, perpendicular to each other and to z; λ = 10 m.

$\lambda =\frac{c}{\nu }$:

• at 7.5 MHz: $\lambda =\frac{3\times 10^8}{7.5\times 10^6}=40$ m
• at 12 MHz: $\lambda =\frac{3\times 10^8}{12\times 10^6}=25$ m

From 40 m to 25 m.

An oscillating charge radiates electromagnetic waves of the same frequency as its oscillation: $10^9$ Hz.

$10^9$ Hz.

$$E_0=c{B}_0=3\times 10^8\times 510\times 10^{-9}=153\text{N/C}$$

$E_0=153$ N/C.

(a)

• $B_0=\frac{E_0}{c}=\frac{120}{3\times 10^8}=4.0\times 10^{-7}$ T $=400$ nT
• $\omega =2\pi \nu =2\pi \times 50\times 10^6\approx 3.14\times 10^8$ rad/s
• $k=\frac{\omega }{c}=\frac{3.14\times 10^8}{3\times 10^8}\approx 1.05$ rad/m
• $\lambda =\frac{c}{\nu }=\frac{3\times 10^8}{50\times 10^6}=6.0$ m

(b) Taking the wave to travel along x, with $\vec{E}$ along y and $\vec{B}$ along z:

$$\vec{E}=120\sin (1.05x-3.14\times 10^{8}t)\hat{j}\text{N/C}$$

$$\vec{B}=4.0\times 10^{-7}\sin (1.05x-3.14\times 10^{8}t)\hat{k}\text{T}$$

(a) $B_0$ = 400 nT, ω ≈ $3.14\times 10^8$ rad/s, k ≈ 1.05 rad/m, λ = 6.0 m. (b) As above, E along y and B along z, in phase.

$E=h\nu =\frac{hc}{\lambda }$, with $hc\approx 1.24\times 10^{-6}$ eV m. Using a typical wavelength for each part:

Link to the sources: the photon energy matches the energy changes inside the source.

• Radio waves come from currents oscillating in circuits and antennas (tiny energies).
• Microwaves and infrared come from rotations and vibrations of molecules.
• Visible and UV light come from transitions of outer electrons in atoms (a few eV).
• X-rays come from inner-shell electrons of heavy atoms (keV).
• Gamma rays come from changes inside the nucleus (MeV).

Photon energy rises from about $10^{-6}$ eV (radio) to about $10^6$ eV (gamma); it matches the energy level spacing of the source: circuits, molecules, outer electrons, inner electrons, nuclei.

(a) $\lambda =\frac{c}{\nu }=\frac{3\times 10^8}{2.0\times 10^{10}}=1.5\times 10^{-2}$ m.

(b) $B_0=\frac{E_0}{c}=\frac{48}{3\times 10^8}=1.6\times 10^{-7}$ T.

(c) Average energy densities: $u_E=\frac{1}{4}{\epsilon }_0{E}_0^2$ and $u_B=\frac{1}{4}\frac{B_0^2}{{\mu }_0}$.

Put $B_0=\frac{E_0}{c}$ and $c^2=\frac{1}{{\mu }_0{\epsilon }_0}$:

$$u_B=\frac{1}{4}\frac{E_0^2}{{\mu }_0{c}^2}=\frac{1}{4}\frac{E_0^2{\mu }_0{\epsilon }_0}{{\mu }_0}=\frac{1}{4}{\epsilon }_0{E}_0^2=u_E$$

(a) 1.5 cm. (b) $1.6\times 10^{-7}$ T. (c) Using $B_0=E_0/c$ and $c^2=1/{\mu }_0{\epsilon }_0$, both equal $\frac{1}{4}{\epsilon }_0{E}_0^2$.