The Timpson Lyceum · University Physics II (2026) – Electricity & Magnetism

Chapter 32 — Electromagnetic Waves

Problems with worked solutions · 14 points · 3 extra credit

Chapter 32 — Electromagnetic Waves

Core: 14 points · Extra credit: 3 points. Show your reasoning; partial credit is given.

1. 1. (1 pt •) Starting from the free-space result $c = 1/\sqrt{\mu_0\varepsilon_0}$, evaluate the speed of light using $\mu_0 = 4\pi\times10^{-7}\,\text{T}\cdot\text{m/A}$ and $\varepsilon_0 = 8.85\times10^{-12}\,\text{C}^2/(\text{N}\cdot\text{m}^2)$, and confirm it agrees with $3.00\times10^{8}\,\text{m/s}$.

   Solution

   1.

   $c = \dfrac{1}{\sqrt{\mu_0\varepsilon_0}} = \dfrac{1}{\sqrt{(4\pi\times10^{-7})(8.85\times10^{-12})}}$.

   The product $\mu_0\varepsilon_0 = (1.2566\times10^{-6})(8.85\times10^{-12}) = 1.1121\times10^{-17}\,\text{s}^2/\text{m}^2$, whose square root is $3.335\times10^{-9}\,\text{s/m}$. Thus $c = 1/(3.335\times10^{-9}) \approx 3.00\times10^{8}\,\text{m/s}$, as expected.

2. 2. (2 pts ••) A sinusoidal plane electromagnetic wave travels through vacuum in the $+x$ direction. Its magnetic field has amplitude $B_0 = 1.5\times10^{-8}\,\text{T}$ and points along $+z$ at the instant the electric field is measured at a certain point.

   (a) Find the amplitude $E_0$ of the electric field.

   (b) In which direction does $\vec{E}$ point at that instant, so that $\vec{E}\times\vec{B}$ gives the propagation direction?

   Solution

   2.

   (a) For a plane wave in vacuum $E = cB$, so $E_0 = cB_0 = (3.00\times10^{8})(1.5\times10^{-8}) = 4.5\,\text{V/m}$.

   (b) The wave moves in $+x$ and $\vec{E}\times\vec{B}$ must point along $+x$ (the direction of $\vec S$). With $\vec{B}$ along $+z$, we need $\vec{E}\times\hat{k} \parallel \hat{\imath}$. Since $\hat{\jmath}\times\hat{k} = \hat{\imath}$, $\vec{E}$ points along $+y$.

3. 3. (2 pts ••) At a point in space, a plane wave in vacuum has an electric-field amplitude $E_0 = 60\,\text{V/m}$.

   (a) Find the peak energy density $u_E$ stored in the electric field.

   (b) Find the peak energy density $u_B$ stored in the magnetic field, and comment on how it compares to $u_E$.

   (c) Find the total time-averaged energy density of the wave.

   Solution

   3.

   (a) $u_E = \tfrac12\varepsilon_0 E_0^2 = \tfrac12(8.85\times10^{-12})(60)^2 \approx 1.6\times10^{-8}\,\text{J/m}^3$.

   (b) $B_0 = E_0/c = 60/(3.00\times10^8) = 2.0\times10^{-7}\,\text{T}$, so $u_B = \dfrac{B_0^2}{2\mu_0} = \dfrac{(2.0\times10^{-7})^2}{2(4\pi\times10^{-7})} \approx 1.6\times10^{-8}\,\text{J/m}^3$. The two peak densities are equal — in an EM wave the field energy is shared equally between $\vec E$ and $\vec B$.

   (c) The instantaneous total density is $u = \varepsilon_0 E^2$ with $E = E_0\cos(\dots)$; its time average uses $\langle\cos^2\rangle = \tfrac12$, giving $u_\text{avg} = \tfrac12\varepsilon_0 E_0^2 \approx 1.6\times10^{-8}\,\text{J/m}^3$. (Equivalently, $u_\text{avg}$ equals the sum of the two averaged parts, each $\tfrac14\varepsilon_0 E_0^2$.)

4. 4. (2 pts ••) A plane electromagnetic wave in vacuum has electric-field amplitude $E_0 = 1000\,\text{V/m}$.

   (a) Find the time-averaged intensity (average magnitude of the Poynting vector), $I = \tfrac12\varepsilon_0 c E_0^2$.

   (b) Find the amplitude $B_0$ of the magnetic field.

   Solution

   4.

   (a) $I = \tfrac12\varepsilon_0 c E_0^2 = \tfrac12(8.85\times10^{-12})(3.00\times10^8)(1000)^2 \approx 1.3\times10^{3}\,\text{W/m}^2$.

   (b) $B_0 = E_0/c = 1000/(3.00\times10^8) \approx 3.3\times10^{-6}\,\text{T}$.

5. 5. (2 pts ••) Just above Earth's atmosphere, sunlight arrives with an average intensity of $I = 1.4\times10^{3}\,\text{W/m}^2$. Treat it as a single plane wave.

   (a) Find the amplitude $E_0$ of the electric field.

   (b) Find the amplitude $B_0$ of the magnetic field.

   Solution

   5.

   (a) Solve $I = \tfrac12\varepsilon_0 c E_0^2$ for $E_0$: $E_0 = \sqrt{\dfrac{2I}{\varepsilon_0 c}} = \sqrt{\dfrac{2(1.4\times10^3)}{(8.85\times10^{-12})(3.00\times10^8)}} \approx 1.0\times10^{3}\,\text{V/m}$.

   (b) $B_0 = E_0/c = (1.03\times10^3)/(3.00\times10^8) \approx 3.4\times10^{-6}\,\text{T}$.

6. 6. (2 pts ••) A laser pointer emits a beam of power $P = 5.0\,\text{mW}$ in a circular spot of radius $0.50\,\text{mm}$.

   (a) Find the intensity of the beam.

   (b) Find the radiation pressure the beam exerts on a surface that completely absorbs it.

   (c) Find the total force the beam exerts if the surface instead perfectly reflects it straight back.

   Solution

   6.

   (a) Beam area $A = \pi r^2 = \pi(0.50\times10^{-3})^2 = 7.85\times10^{-7}\,\text{m}^2$. Intensity $I = P/A = (5.0\times10^{-3})/(7.85\times10^{-7}) \approx 6.4\times10^{3}\,\text{W/m}^2$.

   (b) Absorbing surface: $p_\text{rad} = I/c = (6.37\times10^3)/(3.00\times10^8) \approx 2.1\times10^{-5}\,\text{Pa}$.

   (c) A perfect reflector feels twice the pressure, so the force is $F = \dfrac{2I}{c}\,A = \dfrac{2P}{c} = \dfrac{2(5.0\times10^{-3})}{3.00\times10^8} \approx 3.3\times10^{-11}\,\text{N}$.

7. 7. (2 pts ••) A flat solar sail of area $A = 10\,\text{m}^2$ is oriented face-on to sunlight of intensity $I = 1.4\times10^{3}\,\text{W/m}^2$ and perfectly reflects it.

   (a) Find the radiation force on the sail.

   (b) Find the total momentum the sunlight delivers to the sail in one hour.

   Solution

   7.

   (a) Reflecting sail: $F = \dfrac{2I}{c}\,A = \dfrac{2(1.4\times10^3)(10)}{3.00\times10^8} \approx 9.3\times10^{-5}\,\text{N}$.

   (b) Impulse over $t = 3600\,\text{s}$: $\Delta p = F\,t = (9.33\times10^{-5})(3600) \approx 0.34\,\text{kg}\cdot\text{m/s}$.

8. 8. (1 pt •) A microwave oven sets up a standing electromagnetic wave between its walls. Measurements of where food heats fastest show that adjacent "hot spots" (antinodes of intensity) are $6.1\,\text{cm}$ apart. Using the fact that adjacent antinodes are half a wavelength apart, find the wavelength and the operating frequency of the oven.

   Solution

   8.

   Adjacent antinodes are half a wavelength apart, so $\lambda = 2(6.1\,\text{cm}) = 12.2\,\text{cm} = 0.122\,\text{m}$. Frequency $f = c/\lambda = (3.00\times10^8)/(0.122) \approx 2.5\times10^{9}\,\text{Hz} = 2.5\,\text{GHz}$ — the standard microwave-oven band.

9. Extra Credit (3 pts •••)

   Consider a linearly polarized plane wave traveling in the $+x$ direction in vacuum, with

   $$\vec{E}(x,t) = E_0\cos(kx-\omega t)\,\hat{\jmath}, \qquad \vec{B}(x,t) = B_0\cos(kx-\omega t)\,\hat{k}.$$

   (a) Write the instantaneous Poynting vector $\vec{S} = \dfrac{1}{\mu_0}\,\vec{E}\times\vec{B}$ and show it points along $+x$.

   (b) By integrating over one full period $T = 2\pi/\omega$, evaluate the time average $\langle S\rangle = \dfrac{1}{T}\displaystyle\int_0^T S\,dt$, and use $B_0 = E_0/c$ together with $c = 1/\sqrt{\mu_0\varepsilon_0}$ to reduce your result to $I = \tfrac12\varepsilon_0 c E_0^2$.

   (c) A perfectly absorbing surface of area $A$ is placed face-on to this wave. Argue from the fact that electromagnetic energy $U$ carries momentum $p = U/c$ that the time-averaged radiation pressure on the surface is $\langle p_\text{rad}\rangle = I/c$, and explain why a perfect reflector feels twice this pressure.

   Solution

   Extra Credit

   (a) With $\vec E = E_0\cos(kx-\omega t)\,\hat{\jmath}$ and $\vec B = B_0\cos(kx-\omega t)\,\hat{k}$,

   $$\vec{S} = \frac{1}{\mu_0}\,\vec{E}\times\vec{B} = \frac{E_0 B_0}{\mu_0}\cos^2(kx-\omega t)\,(\hat{\jmath}\times\hat{k}) = \frac{E_0 B_0}{\mu_0}\cos^2(kx-\omega t)\,\hat{\imath}.$$

   Because $\cos^2 \ge 0$, $\vec S$ always points along $+x$ — energy flows in the propagation direction.

   (b) The magnitude is $S(t) = \dfrac{E_0 B_0}{\mu_0}\cos^2(kx-\omega t)$. At fixed $x$, average over one period:

   $$\langle S\rangle = \frac{1}{T}\int_0^T \frac{E_0 B_0}{\mu_0}\cos^2(kx-\omega t)\,dt = \frac{E_0 B_0}{\mu_0}\cdot\frac{1}{T}\int_0^T \cos^2(kx-\omega t)\,dt.$$

   Using $\cos^2\theta = \tfrac12[1+\cos 2\theta]$, the $\cos 2\theta$ term integrates to zero over a full period, leaving $\dfrac{1}{T}\int_0^T\cos^2\,dt = \tfrac12$. Hence

   $$I = \langle S\rangle = \frac{E_0 B_0}{2\mu_0}.$$

   Substitute $B_0 = E_0/c$:

   $$I = \frac{E_0^2}{2\mu_0 c}.$$

   Finally use $c = 1/\sqrt{\mu_0\varepsilon_0}$, i.e. $\dfrac{1}{\mu_0 c} = \dfrac{c}{\mu_0 c^2} = \varepsilon_0 c$ (since $\mu_0 c^2 = 1/\varepsilon_0$). Therefore

   $$I = \tfrac12\varepsilon_0 c\,E_0^2,$$

   the stated result.

   (c) In time $\Delta t$ the wave delivers energy $U = I A\,\Delta t$ to the absorbing surface of area $A$. Electromagnetic energy carries momentum $p = U/c$, so the momentum absorbed is $\Delta p = U/c = I A\,\Delta t / c$. The (time-averaged) force is $F = \Delta p/\Delta t = I A/c$, and the pressure is

   $$\langle p_\text{rad}\rangle = \frac{F}{A} = \frac{I}{c}.$$

   For a perfect reflector, the incoming photons' momentum is reversed rather than merely stopped: the surface must supply an impulse to halt the incident momentum $+IA\Delta t/c$ and then send it back with $-IA\Delta t/c$, a total change of $2IA\Delta t/c$. The momentum transferred is therefore doubled, giving $\langle p_\text{rad}\rangle = 2I/c$.