JEE Mains Physics · Formula sheet

Electromagnetic Waves formulas

8 formulas, 3 reference tables and 39 common traps for JEE Mains Physics Electromagnetic Waves, grouped by subtopic.

Displacement Current, Maxwell's Equations and Wave Speed

Displacement current in a capacitor

Displacement current

i_d = \varepsilon_0\frac{d\Phi_E}{dt} = C\frac{dV}{dt}, \qquad \frac{(j_c)_0}{(j_d)_0} = \frac{\sigma}{\varepsilon_0\omega}

Speed of electromagnetic waves in vacuum and in a medium

Wave speed

c = \frac{1}{\sqrt{\mu_0\varepsilon_0}}, \qquad v = \frac{c}{\sqrt{\mu_r\varepsilon_r}} = \frac{\omega}{k}, \qquad n = \sqrt{\mu_r\varepsilon_r}

Maxwell's four equations and their names

Law	Equation	What it says
Gauss's law for electricity	$\oint \vec E\cdot d\vec A = q/\varepsilon_0$	The electric flux out of a closed surface is the enclosed charge divided by ε₀.
Gauss's law for magnetism	$\oint \vec B\cdot d\vec A = 0$	Magnetic field lines close on themselves; there are no magnetic monopoles.
Faraday's law of induction	$\oint \vec E\cdot d\vec l = -\dfrac{d\Phi_B}{dt}$	A changing magnetic flux induces an electric field around a loop.
Ampere-Maxwell law	$\oint \vec B\cdot d\vec l = \mu_0 i_c + \mu_0\varepsilon_0\dfrac{d\Phi_E}{dt}$	A conduction current and a changing electric flux both produce a magnetic field.
Ampere's circuital law	$\oint \vec B\cdot d\vec l = \mu_0 I$	The steady-current special case, with no changing electric flux. Without the displacement term it fails across the gap of a charging capacitor.

A closed-surface integral (dA) means a Gauss law; a loop integral (dl) means Faraday or Ampere-Maxwell.

Common traps

The displacement current equals the conduction current

Between the plates of a capacitor the displacement current is exactly the current flowing in the leads, at every instant and in rms value. An option that makes one of them ten times the other is wrong.

A surface that covers part of the gap takes its share

The field between the plates is uniform, so a surface of area A₀ inside the gap carries only the fraction A₀/A of the displacement current, not all of it.

Use ω, not f, in σ/ε₀ω

The displacement current density peaks at ε₀ωE₀, with ω = 2πf. Writing f in place of ω makes the ratio of conduction to displacement current 2π times too small.

Plain Ampere's law is the steady-current law

The circulation of B equals μ₀I only for steady currents. The law that holds when fields change in time carries the extra term μ₀ε₀ dΦ_E/dt.

Read the integral before the symbols

In a match list, a closed-surface integral of E or B is one of the two Gauss laws, and a loop integral is Faraday's law or the Ampere-Maxwell law. Sorting by dA or dl first leaves only two choices to make.

A steady source gives a steady field

A permanent magnet, a direct current and an electric field that grows at a constant rate all produce a magnetic field that does not change with time. A changing magnetic field needs a changing current or an accelerating charge.

Take the speed from the phase, not c

When ω/k in the phase is not 3 × 10⁸ m/s, the wave is in a medium. Use v = ω/k for the refractive index and for E₀ = vB₀; using c there gives a wrong amplitude.

A magnetic medium needs μᵣ too

The refractive index is √(μᵣεᵣ). If the permeability is given as a multiple of μ₀, it goes under the root with the dielectric constant. Dropping it leaves √εᵣ only.

Square the index, do not double it

For a non-magnetic medium the dielectric constant is n². A wave that travels at half of c has n = 2 and a dielectric constant of 4, not 2.

The frequency stays the same in a medium

Entering a medium changes the speed and the wavelength by the same factor n. The frequency is fixed by the source and does not change.

E and B Fields of a Plane Wave

Learn this subtopic in the notes

Writing the magnetic field of a wave from its electric field

One field from the other

\vec B = \frac{\vec k \times \vec E}{\omega}, \qquad B_0 = \frac{E_0}{c}, \qquad \hat n = \hat E \times \hat B

Electric and magnetic fields at one point of a wave

Fields at a point

|\vec B| = \frac{|\vec E|}{c}, \qquad \hat B = \hat n \times \hat E, \qquad \frac{E_0}{H_0} = \sqrt{\frac{\mu_0}{\varepsilon_0}} \approx 377\ \Omega

Electric and magnetic forces on a charge in a wave

Forces on a moving charge

F_e = qE_0, \qquad F_m = qvB_0 = \frac{qvE_0}{c}, \qquad \frac{F_e}{F_m} = \frac{c}{v}

Common traps

The order of the cross product matters

The magnetic field is along k × E, not E × k. Reversing the order gives the right axis with the wrong sign, and the options usually offer both.

B keeps the phase of E

The two fields oscillate together, so B has the same argument as E. An option that changes kx − ωt to kx + ωt describes a wave going the other way.

Divide by c, do not multiply

In SI units B₀ = E₀/c is tiny, of order 10⁻⁷ T for fields of tens of volts per metre. An option that gives B the same number as E has skipped the division.

Travel along −x flips a sign

When the phase is kx + ωt, put −î in the cross product. Using +î out of habit reverses the direction of the field you are finding.

The frequency is a distractor

At a point and an instant, B = E/c. The frequency given in the question changes nothing in this calculation.

B over E is 1/c, not c

The magnetic amplitude is the electric amplitude divided by c. A statement that B₀/E₀ equals the speed of light has the ratio upside down.

E₀/B₀ is c, but E₀/H₀ is 377 Ω

The factor √(μ₀/ε₀) links E with the magnetic intensity H = B/μ₀, not with B. Writing E₀ = √(μ₀/ε₀)B₀ mixes the two.

Neither field lies along the travel direction

A wave along y can have E and B only along x and z, one each. Any pair that puts a field along y, or both fields on the same axis, is not a plane electromagnetic wave.

Use B₀ = E₀/c in the magnetic force

The magnetic force is qvB₀, and B₀ is the electric amplitude divided by c. Putting E₀ in place of B₀ makes the magnetic force c times too large.

The electric force is the larger one

The ratio of electric to magnetic force is c/v. Inverting it gives a ratio less than 1, which would mean the magnetic force wins; it never does for a charge slower than light.

Energy, Intensity and Radiation Pressure

Learn this subtopic in the notes

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

Common traps

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.

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.

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.

Electromagnetic Spectrum: Order, Sources and Uses

Learn this subtopic in the notes

Order and wavelength ranges of the electromagnetic spectrum

Band	Wavelength range	Frequency range	Photon energy
γ-rays	shorter than 10⁻³ nm	above 3 × 10²⁰ Hz	above 1.24 MeV
X-rays	1 nm to 10⁻³ nm	3 × 10¹⁷ to 3 × 10²⁰ Hz	1.24 keV to 1.24 MeV Shorter than ultraviolet, longer than γ-rays.
Ultraviolet	400 nm to 1 nm	7.5 × 10¹⁴ to 3 × 10¹⁷ Hz	3.1 eV to 1.24 keV
Visible	700 nm to 400 nm	4.3 × 10¹⁴ to 7.5 × 10¹⁴ Hz	1.8 eV to 3.1 eV
Infrared	1 mm to 700 nm	3 × 10¹¹ to 4.3 × 10¹⁴ Hz	1.24 meV to 1.8 eV
Microwaves	0.1 m to 1 mm	3 × 10⁹ to 3 × 10¹¹ Hz	12.4 µeV to 1.24 meV
Radio waves	longer than 0.1 m	below 3 × 10⁹ Hz	below 12.4 µeV

Wavelength rises down the table; frequency and photon energy fall.

Sources, detectors and uses of each band

Band	Produced by	Detected by	Main uses
Radio waves	Rapid acceleration and deceleration of electrons in aerials	Receiving aerials	Radio and television broadcasting, mobile communication
Microwaves	Klystron valve, magnetron valve, Gunn diode	Point-contact diodes	Radar, aircraft navigation, microwave ovens Klystron and magnetron both mean microwaves.
Infrared	Vibrations of atoms and molecules; hot bodies	Thermopiles, bolometers, infrared photographic film	Physiotherapy heat lamps, greenhouse effect, remote controls, seeing through fog
Visible light	Electrons in atoms dropping between outer energy levels	The eye, photocells, photographic film	Vision, photography, optical communication
Ultraviolet	Electron transitions in atoms; the sun and arc lamps	Photocells, photographic film	Sterilising surgical instruments, purifying water, Lasik eye surgery
X-rays	Inner-shell electron transitions; fast electrons striking a metal target	Photographic film, Geiger tubes, ionisation chambers	Medical diagnosis, cancer treatment, study of crystal structure A metal target hit by fast electrons gives X-rays, not γ-rays.
γ-rays	Radioactive decay of nuclei	Photographic film, Geiger tubes, ionisation chambers	Destroying cancer cells, sterilising medical equipment

The source sets the band: the deeper inside the atom, the shorter the wavelength.

Common traps

γ-rays are shorter than X-rays

In an order question γ-rays come first, with the shortest wavelength. Swapping γ-rays and X-rays is the commonest wrong option.

More energetic means shorter wavelength

A higher photon energy means a higher frequency and a shorter wavelength. Infrared is more energetic than microwaves, so light used in optical fibres has a shorter wavelength than radar microwaves.

Convert each end of a band separately

Wavelength goes as 1/f, so a range of frequencies does not turn into a range of wavelengths by converting the difference. Find λ = c/f at each end, then subtract.

Inner shells give X-rays, the nucleus gives γ-rays

Both bands are very short, but X-rays come from electrons falling into inner shells of an atom, and γ-rays come from the decay of a nucleus. A match list pairs each with the other's source.

Radar uses microwaves

Radar and aircraft navigation work with microwaves, not radio waves. The short wavelength gives a sharp beam and a clear echo.

The greenhouse effect is infrared

The earth radiates infrared, and the atmosphere traps part of it. The greenhouse effect belongs with infrared, not ultraviolet.

Ultraviolet sterilises, X-rays image

Ultraviolet kills germs on instruments and in water but does not pass through the body. X-rays pass through soft tissue, so they are the band for medical images.

Electromagnetic Waves formulas

Displacement Current, Maxwell's Equations and Wave Speed

Displacement current in a capacitor

Speed of electromagnetic waves in vacuum and in a medium

Maxwell's four equations and their names

Common traps

E and B Fields of a Plane Wave

Writing the magnetic field of a wave from its electric field

Electric and magnetic fields at one point of a wave

Electric and magnetic forces on a charge in a wave

Common traps

Energy, Intensity and Radiation Pressure

Energy density of an electromagnetic wave

Intensity of an electromagnetic wave

Momentum and radiation pressure of light

Common traps

Electromagnetic Spectrum: Order, Sources and Uses

Order and wavelength ranges of the electromagnetic spectrum

Sources, detectors and uses of each band

Common traps

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