Standard calculations and where books differ

One method for the usual questions

Almost every calculation of this chapter follows the same short path. Learn it as a ladder.

1. Find the intensity $I$. It is power over area: $I = P/A$. For a point source, $A = 4\pi r^2$. For a beam, $A = \pi a^2$ (radius $a$).
2. Find the fields. $E_0 = \sqrt{2I/\varepsilon_0 c}$ and $B_0 = E_0/c$.
3. Find the push. $F = P/c$ for a black target and $2P/c$ for a mirror, straight on.
4. Find the energy. In a length $L$ of a beam, $U = PL/c$.

A pulsed laser

A pulse of a laser may carry only a fraction of a joule, but it lasts for a few nanoseconds. Its power is the energy divided by the duration, and this can be very large.

Make the beam smaller and watch the peak electric field. Does it double when the radius is halved? It does, since $E_0 \propto 1/a$ for a fixed power.

A worked example. A laser gives out a pulse of $80$ mJ in $12$ ns. The beam is circular, with a radius of $2.5$ mm. Find the power, the intensity, $E_0$, $B_0$, the momentum and the force on an absorbing target.

1. The power. $P = \dfrac{U}{\tau} = \dfrac{0.080\ \text{J}}{12 \times 10^{-9}\ \text{s}} = 6.67\ \text{MW}$.
2. The intensity. $I = \dfrac{P}{\pi a^2} = \dfrac{6.667 \times 10^{6}\ \text{W}}{\pi \times (0.0025\ \text{m})^2} = 3.40 \times 10^{11}\ \text{W/m}^2$.
3. The fields. $E_0 = \sqrt{\dfrac{2I}{\varepsilon_0 c}} = 16.0 \times 10^{6}\ \text{V/m}$ and $B_0 = \dfrac{E_0}{c} = 0.053\ \text{T}$.
4. The momentum. $p = \dfrac{U}{c} = \dfrac{0.080\ \text{J}}{3.00 \times 10^{8}\ \text{m/s}} = 2.67 \times 10^{-10}\ \text{kg m/s}$.
5. The force. $F = \dfrac{P}{c} = 22.22\ \text{mN}$, for as long as the pulse lasts.

The magnetic field of the pulse, a few hundredths of a tesla, is like that of a strong magnet. The electric field is millions of volts for each metre.

The displacement current from $dE/dt$

From the first course, the displacement current is $I_d = \varepsilon_0 A\,dE/dt$. For discs of radius $R$, the magnetic field at the edge is $B = \mu_0 I_d/2\pi R$.

A worked example. The plates are discs of radius $5.0$ cm, and the field between them grows at $1.5 \times 10^{12}$ V/(m s).

1. The displacement current. $I_d = \varepsilon_0 \pi R^2\dfrac{dE}{dt} = 8.85 \times 10^{-12} \times \pi \times (0.050\ \text{m})^2 \times 1.5 \times 10^{12} = 0.1043\ \text{A}$.
2. The field at the edge. $B = \dfrac{\mu_0 I_d}{2\pi R} = \dfrac{2 \times 10^{-7}\ \text{T m/A} \times 0.1043\ \text{A}}{0.050\ \text{m}} = 0.417\ \mu\text{T}$.

Energy in a length of beam

A beam of power $P$ moves at $c$. In a length $L$ it holds the energy $U = PL/c$, because that length passes a point in the time $L/c$.

A worked example. A beam of $1.0$ mW holds $U = \dfrac{0.0010\ \text{W} \times 1.0\ \text{m}}{3.00 \times 10^{8}\ \text{m/s}} = 3.33\ \text{pJ}$ in each metre of its length.

Where books differ

• Intensity. NCERT and Class 12 books write $I$. University books, including OpenStax, write $S$ or $\langle S\rangle$ for the magnitude of the average Poynting vector.
• Amplitude. The peak field is written $E_0$, $E_{max}$ or $E_{peak}$. A-Level and IB use the amplitude for waves in general and write $I \propto A^2$ for the intensity.
• A-Level and IB. These courses treat the intensity of a wave as power per unit area, $I = P/A$, and use the inverse square law $I \propto 1/r^2$. They do not use $\varepsilon_0 c E_0^2$ for light, and they do not use the Poynting vector.
• Radiation pressure is not asked in A-Level or IB. In the Indian syllabus and in university physics the pressure is $I/c$ (absorbing) and $2I/c$ (reflecting).
• rms values. Use $E_0$ with the factor $\tfrac{1}{2}$, or $E_{rms}$ with no factor. Never use both.
• Units. $E$ in V/m (or N/C), $B$ in T, $I$ in W/m², pressure in Pa.

Practise with new numbers

Try it in Python

The program takes three laser pulses. For each it finds the power, the intensity, the peak fields and the force on an absorbing target.

Common mistakes

• Dividing the energy by the area and calling it the intensity. The intensity is power over area. Divide the energy by the duration first.
• Using the diameter as the radius. The area is $\pi a^2$ with the radius $a$. Check which of the two a question gives.
• Forgetting that the force is $P/c$ only while the pulse lasts. Over a long time the momentum is $U/c$.
• Using $E_0$ in place of $E_{rms}$ in $I = \varepsilon_0 cE_{rms}^2$. Check which of the two the question gives.

Remember these

• Intensity is power over area: $P/4\pi r^2$ for a point source, $P/\pi a^2$ for a beam.
• $E_0 = \sqrt{2I/\varepsilon_0 c}$ and $B_0 = E_0/c$.
• Pulse: $P = U/\tau$, $p = U/c$, $F = P/c$ (absorbing).
• $I_d = \varepsilon_0 A\,dE/dt$. In a length $L$ of a beam, $U = PL/c$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A laser beam of power $2.0$ mW has a radius of $1.0$ mm. What is its intensity?

2. A pulse carries $60$ mJ and lasts $6$ ns. What is its power, and what force does it exert if a target absorbs it?

3. Which formula has the amplitude of the field in it: $I = \tfrac{1}{2}\varepsilon_0 c E^2$ with $E$ the peak value or with $E$ the rms value?

4. What does an A-Level or IB course call the quantity that this course calls the intensity of a wave?

References

• OpenStax, University Physics Volume 2, Chapter 16 (Electromagnetic Waves): energy and momentum, with the Poynting vector.
• NCERT Physics Class 12, Chapter 8 (Electromagnetic Waves): the energy and the momentum of a wave.