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Electromagnetic Waves Formulas For JEE 2026
Another short but significant chapter in terms of JEE Mains is Electromagnetic Waves, which generally provides 1–2 questions per year. It covers such important issues as displacement current, Maxwell equations, electromagnetic waves, electromagnetic spectrum, energy and intensity, and pressure of radiation. It is easy to be scored on most questions whose questions are concept-oriented with few direct formula applications.
This chapter is not difficult and with correct revision and practice; you are able to cover it within a short time and earn marks with little effort. The use of a JEE Mains Physics Formula PDF can assist you in revising the key formulas in a single place and enhance the speed and accuracy during the exam.
Displacement Current: Maxwell's Correction
In the 1860s, James Clerk Maxwell noticed a gap in the existing laws of electromagnetism. Ampere's law (which relates magnetic fields to electric currents) worked perfectly for steady currents, but it gave contradictory results when the current was changing — for example, in a circuit with a charging capacitor. Maxwell resolved this by introducing a new concept: displacement current.
Consider a parallel plate capacitor being charged. Current flows in the wire, creating a magnetic field around it. But between the plates there is no actual flow of charge — only a changing electric field. Maxwell realized that this changing electric field produces a magnetic field, just as a real current does. He called this effect the displacement current.
Displacement Current ($$I_d$$): An equivalent "current" produced by a time-varying electric field, even in the absence of actual charge flow. It has the same magnetic effect as a conduction current.
Maxwell's Displacement Current
$$$I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$$$where:
- $$I_d$$ = displacement current (A)
- $$\varepsilon_0 = 8.854 \times 10^{-12}$$ F/m (permittivity of free space)
- $$\frac{d\Phi_E}{dt}$$ = rate of change of electric flux (V·m/s)
Modified Ampere–Maxwell Law:
$$$\oint \vec{B} \cdot d\vec{l} = \mu_0 (I_c + I_d) = \mu_0 I_c + \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt}$$$where $$I_c$$ is the conduction current (actual current in wires).
Worked Example
A parallel plate capacitor with plate area $$A = 0.5$$ m$$^2$$ has an electric field that increases at a rate of $$\frac{dE}{dt} = 10^{12}$$ V/m·s. Find the displacement current.
Electric flux: $$\Phi_E = EA$$
$$I_d = \varepsilon_0 \frac{d\Phi_E}{dt} = \varepsilon_0 A \frac{dE}{dt} = 8.854 \times 10^{-12} \times 0.5 \times 10^{12}$$
$$I_d = 4.43$$ A
Note: Displacement current is not a "real" current — no charges actually flow. It is the effect of a changing electric field. However, it produces a magnetic field exactly like a real current does, and it ensures that Ampere's law is consistent everywhere.
Electromagnetic Wave Properties and Formulas
Maxwell's equations predict that when electric and magnetic fields change with time, they can sustain each other and travel through space as a wave — even in vacuum, with no medium required. These are electromagnetic (EM) waves. Light, radio waves, X-rays, and microwaves are all examples of EM waves.
Properties of EM Waves
Key Properties of Electromagnetic Waves
- EM waves are transverse — the electric field ($$\vec{E}$$), magnetic field ($$\vec{B}$$), and direction of propagation are all mutually perpendicular
- They do not require a medium — they can travel through vacuum
- In vacuum, all EM waves travel at the speed of light: $$c = 3 \times 10^8$$ m/s
- They carry energy and momentum
- They can be polarized (because they are transverse)
- They obey the wave equation: $$c = f\lambda$$ (where $$f$$ = frequency, $$\lambda$$ = wavelength)
Speed of EM Waves
$$$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \approx 3 \times 10^8 \text{ m/s}$$$where:
- $$\mu_0 = 4\pi \times 10^{-7}$$ T·m/A (permeability of free space)
- $$\varepsilon_0 = 8.854 \times 10^{-12}$$ F/m (permittivity of free space)
In a medium with permittivity $$\varepsilon$$ and permeability $$\mu$$:
$$$v = \frac{1}{\sqrt{\mu\varepsilon}} = \frac{c}{n}$$$where $$n = \sqrt{\mu_r \varepsilon_r}$$ is the refractive index of the medium.
Tip: The relation $$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$$ is one of the most beautiful results in physics — it connects electricity ($$\varepsilon_0$$), magnetism ($$\mu_0$$), and optics ($$c$$) in a single equation. This was Maxwell's prediction that light is an electromagnetic wave.
Relation Between E and B
In an EM wave, the electric and magnetic fields oscillate together, always maintaining a fixed ratio.
E–B Relationship in EM Waves
For an EM wave travelling along the $$x$$-axis:
$$$E = E_0 \sin(kx - \omega t), \quad B = B_0 \sin(kx - \omega t)$$$The amplitudes are related by:
$$$\frac{E_0}{B_0} = c \quad \text{or equivalently} \quad E = cB$$$where:
- $$E_0$$ = amplitude of the electric field (V/m)
- $$B_0$$ = amplitude of the magnetic field (T)
- $$k = \frac{2\pi}{\lambda}$$ = wave number (rad/m)
- $$\omega = 2\pi f$$ = angular frequency (rad/s)
$$\vec{E}$$ and $$\vec{B}$$ are always in phase (they reach their maxima and minima simultaneously).
Worked Example
An EM wave has an electric field amplitude of $$E_0 = 600$$ V/m. Find the magnetic field amplitude.
$$B_0 = \frac{E_0}{c} = \frac{600}{3 \times 10^8} = 2 \times 10^{-6}$$ T $$= 2\;\mu$$T
Note how small $$B_0$$ is compared to $$E_0$$ — this is because $$c$$ is very large.
Energy, Intensity, and Radiation Pressure of EM Waves
EM waves carry energy as they travel through space. This energy is shared equally between the electric and magnetic fields. The rate at which energy flows per unit area is described by the Poynting vector.
Poynting Vector ($$\vec{S}$$): The vector that describes the rate of energy flow (power per unit area) in an electromagnetic wave. It points in the direction of wave propagation.
Energy Density and Poynting Vector
Energy density (energy per unit volume) of an EM wave:
$$$u = \frac{1}{2}\varepsilon_0 E^2 + \frac{B^2}{2\mu_0}$$$At any instant, the electric and magnetic contributions are equal:
$$$u_E = \frac{1}{2}\varepsilon_0 E^2 = u_B = \frac{B^2}{2\mu_0}$$$Total: $$u = \varepsilon_0 E^2 = \frac{B^2}{\mu_0}$$
Average energy density:
$$$\langle u \rangle = \frac{1}{2}\varepsilon_0 E_0^2 = \frac{B_0^2}{2\mu_0}$$$Poynting vector (instantaneous power per unit area):
$$$\vec{S} = \frac{1}{\mu_0}(\vec{E} \times \vec{B})$$$Magnitude: $$S = \frac{EB}{\mu_0} = \frac{E^2}{\mu_0 c} = \frac{cB^2}{\mu_0}$$
Intensity ($$I$$): The average power delivered per unit area by an EM wave, measured in W/m$$^2$$.
Intensity of an EM Wave
$$$I = \langle S \rangle = \frac{1}{2}\frac{E_0^2}{\mu_0 c} = \frac{1}{2}\varepsilon_0 c E_0^2 = \frac{E_0 B_0}{2\mu_0} = \frac{c B_0^2}{2\mu_0}$$$Also: $$I = \langle u \rangle \cdot c$$ (average energy density $$\times$$ speed)
Worked Example
A laser beam has an intensity of $$10^6$$ W/m$$^2$$. Find the amplitudes of the electric and magnetic fields.
$$I = \frac{1}{2}\varepsilon_0 c E_0^2$$
$$E_0 = \sqrt{\frac{2I}{\varepsilon_0 c}} = \sqrt{\frac{2 \times 10^6}{8.854 \times 10^{-12} \times 3 \times 10^8}}$$
$$E_0 = \sqrt{\frac{2 \times 10^6}{2.656 \times 10^{-3}}} = \sqrt{7.53 \times 10^8} \approx 2.74 \times 10^4$$ V/m
$$B_0 = \frac{E_0}{c} = \frac{2.74 \times 10^4}{3 \times 10^8} \approx 9.13 \times 10^{-5}$$ T
Radiation Pressure
EM waves carry momentum and exert pressure on surfaces they hit:
- For a perfectly absorbing surface: $$P = \frac{I}{c}$$
- For a perfectly reflecting surface: $$P = \frac{2I}{c}$$
where $$P$$ is the radiation pressure (in Pa) and $$I$$ is the intensity (W/m$$^2$$).
Tip: Radiation pressure is extremely small for ordinary light, but it becomes significant for high-intensity laser beams. JEE problems typically give intensity and ask you to find the force on a surface of area $$A$$: $$F = PA$$.
Electromagnetic Spectrum: All Bands and Uses
Electromagnetic waves span an enormous range of frequencies and wavelengths — from radio waves with wavelengths of kilometres to gamma rays with wavelengths smaller than atoms. This entire range is called the electromagnetic spectrum. All these waves travel at the speed of light in vacuum; they differ only in frequency (and hence wavelength and energy).
| Type | Wavelength Range | Frequency Range | Source |
|---|---|---|---|
| Radio waves | $$> 0.1$$ m | $$< 3 \times 10^9$$ Hz | Oscillating circuits, antennas |
| Microwaves | 0.1 m – 1 mm | $$3 \times 10^9$$ – $$3 \times 10^{11}$$ Hz | Klystron, magnetron |
| Infrared (IR) | 1 mm – 700 nm | $$3 \times 10^{11}$$ – $$4.3 \times 10^{14}$$ Hz | Hot bodies, sun, fire |
| Visible light | 700 – 400 nm | $$4.3 \times 10^{14}$$ – $$7.5 \times 10^{14}$$ Hz | Sun, lamps, lasers |
| Ultraviolet (UV) | 400 – 1 nm | $$7.5 \times 10^{14}$$ – $$3 \times 10^{17}$$ Hz | Sun, mercury lamps, arcs |
| X-rays | 1 nm – 0.01 nm | $$3 \times 10^{17}$$ – $$3 \times 10^{19}$$ Hz | X-ray tubes, synchrotrons |
| Gamma rays | $$< 0.01$$ nm | $$> 3 \times 10^{19}$$ Hz | Radioactive nuclei, nuclear reactions |
Visible Light Spectrum
Visible light is the narrow band of EM waves that our eyes can detect. It ranges from about 400 nm (violet) to 700 nm (red). The order from longest to shortest wavelength is:
Colours of Visible Light (VIBGYOR)
| Colour | Wavelength (nm) | Frequency ($$\times 10^{14}$$ Hz) |
|---|---|---|
| Red | 620 – 700 | 4.3 – 4.8 |
| Orange | 590 – 620 | 4.8 – 5.1 |
| Yellow | 570 – 590 | 5.1 – 5.3 |
| Green | 495 – 570 | 5.3 – 6.1 |
| Blue | 450 – 495 | 6.1 – 6.7 |
| Violet | 380 – 450 | 6.7 – 7.9 |
Tip: Remember the order using VIBGYOR (Violet, Indigo, Blue, Green, Yellow, Orange, Red) from shortest to longest wavelength. For JEE, remember that red has the longest wavelength and violet has the shortest in visible light.
Uses of Different EM Waves
Applications of EM Waves
- Radio waves: Broadcasting (AM, FM radio, TV), communication
- Microwaves: Microwave ovens (heats water molecules), radar, satellite communication
- Infrared: Night vision, remote controls, thermal imaging, greenhouse heating
- Visible light: Vision, photography, optical fibres, lasers
- Ultraviolet: Sterilization, vitamin D production in skin, LASIK eye surgery, detecting forged documents
- X-rays: Medical imaging (bones, teeth), security scanning, studying crystal structures
- Gamma rays: Cancer treatment (radiotherapy), sterilizing medical equipment, nuclear physics
Worked Example
An EM wave has a frequency of $$5 \times 10^{14}$$ Hz. (a) Find its wavelength. (b) Identify the type of wave.
(a) $$\lambda = \frac{c}{f} = \frac{3 \times 10^8}{5 \times 10^{14}} = 6 \times 10^{-7}$$ m $$= 600$$ nm
(b) 600 nm falls in the visible light range (orange-red region).
Worked Example
The electric field of an EM wave in vacuum is $$E = 50\sin(2\pi \times 10^{10}t - kx)$$ V/m. Find: (a) frequency, (b) wavelength, (c) wave number $$k$$, (d) magnetic field amplitude.
(a) $$\omega = 2\pi \times 10^{10}$$ rad/s, so $$f = 10^{10}$$ Hz $$= 10$$ GHz
(b) $$\lambda = \frac{c}{f} = \frac{3 \times 10^8}{10^{10}} = 0.03$$ m $$= 3$$ cm (this is a microwave)
(c) $$k = \frac{2\pi}{\lambda} = \frac{2\pi}{0.03} \approx 209.4$$ rad/m
(d) $$B_0 = \frac{E_0}{c} = \frac{50}{3 \times 10^8} = 1.67 \times 10^{-7}$$ T
Summary of Key Formulas
Quick Reference
| Quantity | Formula |
|---|---|
| Displacement current | $$I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$$ |
| Speed of EM wave | $$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = 3 \times 10^8$$ m/s |
| Wave equation | $$c = f\lambda$$ |
| E–B relation | $$E_0 = cB_0$$ |
| Average energy density | $$\langle u \rangle = \frac{1}{2}\varepsilon_0 E_0^2 = \frac{B_0^2}{2\mu_0}$$ |
| Intensity | $$I = \frac{1}{2}\varepsilon_0 c E_0^2$$ |
| Poynting vector | $$S = \frac{EB}{\mu_0}$$ |
| Radiation pressure (absorbing) | $$P = \frac{I}{c}$$ |
| Radiation pressure (reflecting) | $$P = \frac{2I}{c}$$ |
Electromagnetic Waves Formulas For JEE 2026
Electromagnetic Waves formulae JEE 2026 are quite significant to revise quickly and achieve better performance. As explained, this chapter deals with displacement current, correction of Maxwell, properties of electromagnetic waves, intensity, radiation pressure, and electromagnetic spectrum. Because most of the questions are concept based and mostly direct, knowing the formulas and simple properties will assist you in finding the answers within a short time and in the correct manner. This chapter should be one of the simplest to score in JEE Physics with frequent revision being able to turn it into one of the simplest ones.
During the last preparation phase, aim at memorising the main formulas and wave relations, and an arrangement of the electromagnetic spectrum. Other ideas such as the relation between the electric and magnetic fields, speed of the EM waves, pressure of radiation should also be noted. Electromagnetic Waves formulas JEE 2026 with smart revision and regular practice can also save your time during the exam and help to increase your overall marks. Having a clear basics and a good revision, you can do very well in this chapter.
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