JEE Mains Physics · Electromagnetic Waves

Energy, Intensity and Radiation Pressure

A wave stores equal energy in its electric and magnetic fields, carries it at c as an intensity ½cε₀E₀², and pushes on a surface with a pressure I/c, or 2I/c when it is reflected.

All Electromagnetic Waves notes JEE Mains Physics strategy

Why this matters

Thirty PYQs, six of them asking for a number, and four from 2026. Ten are about energy density: the two fields' shares, the average over a cycle, the energy held in a volume. Thirteen find an intensity, or a peak field from an intensity, often for a lamp that radiates only part of its power. Seven are about the momentum of light and the pressure it puts on an absorbing or a reflecting surface.

Concept 1 of 3: Energy density of an electromagnetic wave

Energy is stored in both fields of the wave, and because B = E/c the two shares come out exactly equal at every instant. The energy density goes as E², so it is never negative and pulses at twice the frequency of the field. Averaged over a cycle, sin² is one half.

Definition

Electric share $u_E = \tfrac{1}{2}\varepsilon_0E^{2}$ ; magnetic share $u_B = \dfrac{B^{2}}{2\mu_0}$ . With $B = E/c$ and $c^{2} = 1/(\mu_0\varepsilon_0)$ they are equal at every instant.
Total at an instant: $u = u_E + u_B = \varepsilon_0E^{2}$ .
Average over a cycle: $\langle u\rangle = \tfrac{1}{2}\varepsilon_0E_0^{2} = \dfrac{B_0^{2}}{2\mu_0}$ . Each field's average share is $\tfrac{1}{4}\varepsilon_0E_0^{2}$ , half of the total.
Energy held in a volume V: $U = \langle u\rangle V$ . Holding the same energy in a smaller volume needs a larger field, since $E_0^{2}$ scales as $1/V$ .
The energy density oscillates at $2\omega$ , because $\sin^{2}\omega t = \tfrac{1}{2}(1 - \cos 2\omega t)$ .
The frequency of the wave does not enter $\langle u\rangle$ .

Energy density

u_E = \tfrac{1}{2}\varepsilon_0E^{2} = u_B = \frac{B^{2}}{2\mu_0}, \qquad \langle u\rangle = \tfrac{1}{2}\varepsilon_0E_0^{2} = \frac{B_0^{2}}{2\mu_0}

Worked example

A wave in vacuum has an electric amplitude of 120 V/m. Find the average energy density, the average electric share, and the energy held in a volume of

2 \times 10^{-3}

m³. (

\varepsilon_0 = 8.85 \times 10^{-12}

F/m)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 31 January 2024 · Q10Moderate

Example 1 · Electromagnetic Waves · Energy, Intensity and Radiation Pressure

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of

5 \times10^{10}\text{ }Hz

and an amplitude of

50Vm^{- 1}

. The total average energy density of the electromagnetic field of the wave is : [Use

\varepsilon_{0}= 8.85 \times10^{- 12}C^{2}/Nm^{2}

]

½ε₀E₀² is the total average, not the electric share

Averaged over a cycle, the whole wave holds ½ε₀E₀². The electric field's share is half of that, ¼ε₀E₀², and the magnetic field holds the other quarter.

Do not add the magnetic term again

½ε₀E₀² already includes both fields. Adding B₀²/2μ₀ on top of it doubles the answer, and that doubled value is usually one of the options.

The magnetic share is B²/2μ₀, not μ₀B²/2

The permeability goes in the denominator of the magnetic energy density, just as the permittivity goes in the numerator of the electric one. Check with B = E/c: B²/2μ₀ becomes ½ε₀E².

The frequency does not change the average energy

The average energy density depends only on the amplitude. A frequency given in the question is there to tempt a calculation that is not needed.

Drill 9 more on energy density of an electromagnetic wave

Concept 2 of 3: Intensity of an electromagnetic wave

Intensity is the energy crossing one square metre each second. The energy density moves along at c, so the intensity is simply the average energy density times c. For a lamp, the power it actually radiates spreads over a sphere, so the intensity falls as 1/r².

Definition

$I = \langle u\rangle c = \tfrac{1}{2}c\varepsilon_0E_0^{2} = \dfrac{cB_0^{2}}{2\mu_0}$ . In terms of 377 Ω: $I = \dfrac{E_0^{2}}{2 \times 377}$ .
The other way round: $E_0 = \sqrt{\dfrac{2I}{c\varepsilon_0}}$ and $B_0 = E_0/c$ .
A point source of power P that radiates a fraction $\eta$ of it as waves: $I = \dfrac{\eta P}{4\pi r^{2}}$ . Every point at the same distance gets the same intensity, whatever its direction.
Unit W/m², dimensions $MT^{-3}$ .
With rms fields: $I = c\varepsilon_0E_{rms}^{2}$ , since $E_{rms} = E_0/\sqrt{2}$ .

Intensity

I = \tfrac{1}{2}c\varepsilon_0E_0^{2} = \frac{cB_0^{2}}{2\mu_0}, \qquad I = \frac{\eta P}{4\pi r^{2}}

Worked example

A 125 W lamp radiates 4% of its power as light, equally in all directions. Find the intensity and the peak electric and magnetic fields 2 m away. (

\varepsilon_0 = 8.85 \times 10^{-12}

F/m,

c = 3 \times 10^{8}

m/s)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 27 Jan 2024 · Q16Moderate

Example 2 · Electromagnetic Waves · Energy, Intensity and Radiation Pressure

A plane electromagnetic wave propagating in

x

-direction is described by

E_{y}=\left( 200Vm^{- 1} \right)\sin\left\lbrack 1.5 \times10^{7}t - 0.05x \right\rbrack

; The intensity of the wave is: (Use

\in_{0}= 8.85 \times10^{- 12}C^{2}{\text{ }N}^{- 1}{\text{ }m}^{- 2}

)

Only the radiated power counts

A bulb rated at some wattage with a stated efficiency radiates only that fraction as light. Divide the radiated power, not the rated power, by 4πr².

Intensity depends on distance, not direction

A point source spreads its power evenly over a sphere. Moving a detector around the sphere at the same distance leaves the intensity unchanged.

The ½ goes with the peak field

I = ½cε₀E₀² uses the amplitude E₀. With the rms field the ½ is already inside, and I = cε₀E_rms². Using the ½ with an rms value halves the answer.

The field goes as the square root of the intensity

Doubling the intensity raises the peak field by √2, not by 2. Doubling the field makes the intensity four times as large.

Drill 12 more on intensity of an electromagnetic wave

Concept 3 of 3: Momentum and radiation pressure of light

Light carries momentum as well as energy: energy U comes with momentum U/c. A surface that absorbs the light takes that momentum; a mirror sends it back and so takes twice as much. Momentum delivered per second is a force, and force per area is the radiation pressure.

Definition

Momentum carried by energy U: $p = U/c$ . Light has momentum even though a photon has zero rest mass.
At normal incidence: a perfect absorber feels a pressure $I/c$ ; a perfect reflector feels $2I/c$ .
Force on the surface = pressure × area: $F = IA/c$ for an absorber. The force is steady while the light falls, so the exposure time does not enter it.
Momentum (or energy) delivered in a time t is the force (or power) times t.
For a point source, first find I at the surface from $I = P/(4\pi r^{2})$ .
On a curved surface around the source, the sideways pushes cancel; what remains is the pressure acting over the area the surface presents to the light, its flat projection.

Radiation pressure

p = \frac{U}{c}, \qquad P_{\text{absorbed}} = \frac{I}{c}, \qquad P_{\text{reflected}} = \frac{2I}{c}, \qquad F = P_{\text{rad}}A

Worked example

Light of intensity 1.2 kW/m² falls normally on a sheet of area 0.5 m² for 10 minutes. Find the force on the sheet and the momentum it receives if it (a) absorbs all of the light and (b) reflects all of it. (

c = 3 \times 10^{8}

m/s)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 3 Apr 2025 · Q4Moderate

Example 3 · Electromagnetic Waves · Energy, Intensity and Radiation Pressure

The radiation pressure exerted by a 450 W light source on a perfectly reflecting surface placed at 2 m away from it, is :

A reflector feels twice the push

Reflected light reverses its momentum, so a mirror receives 2U/c and feels a pressure 2I/c. Using I/c for a reflecting surface halves the answer.

The exposure time does not change the force

While light falls on a surface, the force on it is steady: pressure times area. A time given in the question matters only for the total momentum or energy delivered.

Zero rest mass does not mean zero momentum

A photon has no rest mass, but light still carries momentum U/c. That momentum is what produces radiation pressure.

Use the area the light sees

For a curved surface wrapped around a source, the sideways pushes cancel. The net force is the pressure times the flat area the surface presents to the light, not its full curved area.

Drill 6 more on momentum and radiation pressure of light

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Energy density of an electromagnetic wave
Energy density
$u_E = \tfrac{1}{2}\varepsilon_0E^{2} = u_B = \frac{B^{2}}{2\mu_0}, \qquad \langle u\rangle = \tfrac{1}{2}\varepsilon_0E_0^{2} = \frac{B_0^{2}}{2\mu_0}$
Intensity of an electromagnetic wave
Intensity
$I = \tfrac{1}{2}c\varepsilon_0E_0^{2} = \frac{cB_0^{2}}{2\mu_0}, \qquad I = \frac{\eta P}{4\pi r^{2}}$
Momentum and radiation pressure of light
Radiation pressure
$p = \frac{U}{c}, \qquad P_{\text{absorbed}} = \frac{I}{c}, \qquad P_{\text{reflected}} = \frac{2I}{c}, \qquad F = P_{\text{rad}}A$

Watch out for (12)

½ε₀E₀² is the total average, not the electric share
→ Energy density of an electromagnetic wave
Do not add the magnetic term again
→ Energy density of an electromagnetic wave
The magnetic share is B²/2μ₀, not μ₀B²/2
→ Energy density of an electromagnetic wave
The frequency does not change the average energy
→ Energy density of an electromagnetic wave
Only the radiated power counts
→ Intensity of an electromagnetic wave
Intensity depends on distance, not direction
→ Intensity of an electromagnetic wave
The ½ goes with the peak field
→ Intensity of an electromagnetic wave
The field goes as the square root of the intensity
→ Intensity of an electromagnetic wave
A reflector feels twice the push
→ Momentum and radiation pressure of light
The exposure time does not change the force
→ Momentum and radiation pressure of light
Zero rest mass does not mean zero momentum
→ Momentum and radiation pressure of light
Use the area the light sees
→ Momentum and radiation pressure of light

Test yourself on Electromagnetic Waves

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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