Worked examples · Sheet 3

Intensity of electromagnetic waves

Theory used: Intensity of a wave Poynting vector Intensity of an isotropic source Relation between the electric field and the magnetic field

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

A laser pointer emits a beam of power $P = 5.0 mW$ with a circular cross section of diameter $D = 2.0 mm$ . Assuming that the power is distributed uniformly over the cross section and that the beam is a harmonic plane wave in vacuum, compute the intensity of the beam and the amplitudes of the electric field and of the magnetic field.

I = \frac{P}{π ( \frac{D}{2} ) ^{2}}, I = \frac{1}{2} c ε_{0} E_{0}^{2} ⟹ E_{0} = \frac{2 I}{c ε _{0}}, B_{0} = \frac{E _{0}}{c}

I E_{0} B_{0} = \frac{5.0 \cdot 1 0 ^{- 3}}{π \cdot ( 1.0 \cdot 1 0 ^{- 3} ) ^{2}} W/ m^{2} \approx 1.6 \cdot 1 0^{3} W/ m^{2} = \frac{2 \cdot 1.59 \cdot 1 0 ^{3}}{3.00 \cdot 1 0 ^{8} \cdot 8.85 \cdot 1 0 ^{- 12}} V/m \approx 1.1 \cdot 1 0^{3} V/m = \frac{1.09 \cdot 1 0 ^{3}}{3.00 \cdot 1 0 ^{8}} T \approx 3.6 \cdot 1 0^{- 6} T

The intensity is the average power per unit area, and the power is distributed uniformly over a disk of radius $1.0 mm$ , so it suffices to divide the power by the area of the cross section. For a harmonic wave the intensity is related to the amplitude of the electric field by $I = \frac{1}{2} c ε_{0} E_{0}^{2}$ , and solving for $E_{0}$ gives about $1.1 \cdot 1 0^{3} V/m$ ; the amplitude of the magnetic field follows from $E_{0} = c B_{0}$ . The beam of a pointer of a few milliwatts has an intensity comparable to that of sunlight, because the power is concentrated on a very small cross section.

Exercise 2

A lamp, which can be treated as an isotropic point source, emits electromagnetic waves with average power $P = 100 W$ . Assuming that air does not absorb energy, compute at distance $r = 2.0 m$ from the lamp the intensity, the amplitude and the rms value of the electric field, and the distance at which the intensity is halved.

I (r) = \frac{P}{4 π r ^{2}}, E_{0} = \frac{2 I}{c ε _{0}}, E_{rms} = \frac{E _{0}}{2}

\frac{I ( r ^{'} )}{I ( r )} = \frac{r ^{2}}{r ^{'2}} = \frac{1}{2} ⟹ r^{'} = 2 r

I E_{0} E_{rms} r^{'} = \frac{100}{4 π \cdot 2. 0 ^{2}} W/ m^{2} \approx 1.99 W/ m^{2} = \frac{2 \cdot 1.99}{3.00 \cdot 1 0 ^{8} \cdot 8.85 \cdot 1 0 ^{- 12}} V/m \approx 38.7 V/m = \frac{38.7}{2} V/m \approx 27.4 V/m = 2 \cdot 2.0 m \approx 2.8 m

The source is isotropic and the medium does not absorb energy, so the formula for the intensity of an isotropic source applies: the power is distributed uniformly over the sphere of radius $r$ , of area $4 π r^{2}$ . From the intensity the amplitude is obtained as for harmonic waves, and the rms value is the amplitude divided by $2$ , so that $I = c ε_{0} E_{rms}^{2}$ . The intensity is proportional to $\frac{1}{r ^{2}}$ , hence it is halved when $r^{2}$ doubles, that is, at the distance $2 r \approx 2.8 m$ ; at that point the amplitude of the field has decreased only by the factor $2$ .

Exercise 3

The intensity of solar radiation, measured outside the Earth’s atmosphere at the distance $d_{T} = 1.50 \cdot 1 0^{11} m$ from the Sun, is $I_{T} = 1.36 kW/ m^{2}$ . Treating the Sun as an isotropic source and neglecting absorption in interplanetary space, compute the total power emitted by the Sun, the amplitudes of the fields of a harmonic wave with the same intensity, and the intensity of solar radiation at the orbit of Mars, at the distance $d_{M} = 2.28 \cdot 1 0^{11} m$ from the Sun.

P = 4 π d_{T}^{2} I_{T}, E_{0} = \frac{2 I _{T}}{c ε _{0}}, B_{0} = \frac{E _{0}}{c}, I_{M} = I_{T} (\frac{d _{T}}{d _{M}})^{2}

P E_{0} B_{0} I_{M} = 4 π \cdot (1.50 \cdot 1 0^{11})^{2} \cdot 1.36 \cdot 1 0^{3} W \approx 3.85 \cdot 1 0^{26} W = \frac{2 \cdot 1.36 \cdot 1 0 ^{3}}{3.00 \cdot 1 0 ^{8} \cdot 8.85 \cdot 1 0 ^{- 12}} V/m \approx 1.01 \cdot 1 0^{3} V/m = \frac{1.01 \cdot 1 0 ^{3}}{3.00 \cdot 1 0 ^{8}} T \approx 3.37 \cdot 1 0^{- 6} T = 1.36 \cdot 1 0^{3} \cdot (\frac{1.50}{2.28})^{2} W/ m^{2} \approx 589 W/ m^{2}

By the intensity of an isotropic source, $I_{T} = \frac{P}{4 π d _{T}^{2}}$ , so the emitted power is the intensity multiplied by the area of the sphere whose radius is the Earth-Sun distance. Sunlight contains many frequencies, hence the computed amplitudes are those of a harmonic wave that carries the same intensity, and they give the order of magnitude of the fields: about $1 kV/m$ and $3 μ T$ . The ratio of the intensities at two different distances is the inverse of the ratio of the squares of the distances, because the power $P$ is the same; Mars therefore receives less than half of the intensity that the Earth receives.