Electromagnetic Waves — Easy Practice (Set 1)

Quick Recap — Nature of EM Waves

• Displacement current (Maxwell): $I_d = \varepsilon_0\dfrac{d\Phi_E}{dt}$ — a changing electric field acts like a current.
• EM waves are transverse: $\vec{E}\perp\vec{B}$, and both are perpendicular to the direction of propagation (which is along $\vec{E}\times\vec{B}$).
• Speed in vacuum: $c = \dfrac{1}{\sqrt{\mu_0\varepsilon_0}} \approx 3\times10^8$ m/s; in a medium $v = \dfrac{c}{n}$.
• Field relation: $E_0 = cB_0$ (so $E/B = c$); $E$ and $B$ oscillate in phase.
• Energy and momentum: EM waves carry both, and the energy is shared equally between the electric and magnetic fields.

Beyond-NCERT JEE Formulae

A consolidated formula sheet for the calculation-heavy part of this chapter -- field links, energy, intensity, radiation pressure and the spectrum -- with when-to-use notes and JEE traps. The single-line recaps in the practice sets stated these; here we collect them, add the beyond-NCERT forms, and flag where marks are lost.

1. Wave speed and the E-B lock

• Vacuum: $c=\dfrac{1}{\sqrt{\mu_0\varepsilon_0}}=3\times10^8$ m/s. In a medium: $v=\dfrac{1}{\sqrt{\mu\varepsilon}}=\dfrac{c}{n}$ with $n=\sqrt{\varepsilon_r\mu_r}$.
• The fields are locked together: $E_0=cB_0$, and at every instant $\dfrac{E}{B}=c$. Inside a medium this becomes $\dfrac{E}{B}=v$, not $c$.
• When to use: any problem that gives one of $E_0,B_0$ and asks for the other, or gives field values and wants the medium's speed or index.
• [JEE Tip] Because $c$ is huge, $B_0$ is tiny -- a few $10^{-7}$ T for visible light. A $B$-amplitude near $10^{-7}$ T is a hint the wave is optical.

2. Energy density (equal electric and magnetic shares)

• Instantaneous: $u=\tfrac12\varepsilon_0E^2+\dfrac{B^2}{2\mu_0}$. Since $E=cB$, the two terms are equal at every instant.
• Time-averaged: $u_{avg}=\tfrac12\varepsilon_0E_0^2=\dfrac{B_0^2}{2\mu_0}=\varepsilon_0E_{rms}^2=\dfrac{B_{rms}^2}{\mu_0}$.
• When to use: whenever the stored energy per unit volume is asked, or as the stepping stone to intensity.
• [JEE Tip] Half of $u_{avg}$ is electric and half magnetic. Given only $B_0$, do not stop at the magnetic average $\dfrac{B_0^2}{4\mu_0}$ -- double it for the total.

3. Intensity and the Poynting vector

• Poynting vector: $\vec S=\dfrac{1}{\mu_0}\vec E\times\vec B$ -- instantaneous energy flow per unit area, pointing along the propagation direction.
• Intensity (time-averaged magnitude of $\vec S$): $I=u_{avg}\,c=\tfrac12\varepsilon_0cE_0^2=\varepsilon_0cE_{rms}^2=\dfrac{cB_0^2}{2\mu_0}$.
• Impedance of free space (beyond-NCERT): $Z_0=\sqrt{\dfrac{\mu_0}{\varepsilon_0}}=\mu_0c\approx377$ ohm, giving the compact form $I=\dfrac{E_{rms}^2}{Z_0}$.
• Isotropic point source: $I=\dfrac{P}{4\pi r^2}$ -- power spreads over a sphere, not a disc.
• When to use: power-per-area problems, source-at-a-distance problems, and any bridge from field amplitude to power.
• [JEE Tip] $E_0$ is the PEAK field; a time-average always carries the factor $\tfrac12$ (equivalently, uses $E_{rms}^2$). Never feed $E_0$ into $\varepsilon_0cE_{rms}^2$ -- that double-counts the half.

4. Radiation pressure and momentum

• A wave of energy $U$ carries momentum $p=\dfrac{U}{c}$ (photon picture: $U=hf$ and $p=\dfrac{h}{\lambda}$).
• Normal incidence on a surface of area $A$ under intensity $I$:
• Perfect absorber: pressure $\dfrac{I}{c}$, force $\dfrac{IA}{c}$, momentum $\dfrac{U}{c}$.
• Perfect reflector: pressure $\dfrac{2I}{c}$, force $\dfrac{2IA}{c}$, momentum $\dfrac{2U}{c}$.
• Partial reflector of reflectivity $R$: pressure $(1+R)\dfrac{I}{c}$.
• When to use: solar-sail, comet-tail, laser-push, or mirror-versus-black-plate questions.
• [JEE Tip] Reflection reverses momentum, so it delivers twice the push of absorption. Read the surface: black or absorbing takes factor $1$, mirror or reflecting takes factor $2$.

5. EM spectrum (radio to gamma)

Ordered by decreasing wavelength, i.e. increasing frequency:

$$\text{radio} > \text{microwave} > \text{IR} > \text{visible} > \text{UV} > \text{X-ray} > \text{gamma}$$

Rough ranges (convert with $c=f\lambda$):

• Radio: $\lambda>0.1$ m, $f<3\times10^9$ Hz.
• Microwave: $\lambda\sim1$ mm to $0.1$ m.
• Infrared: $\lambda\sim700$ nm to $1$ mm.
• Visible: $\lambda\sim400$ to $700$ nm, $f\sim4\times10^{14}$ to $7.5\times10^{14}$ Hz.
• Ultraviolet: $\lambda\sim10$ to $400$ nm.
• X-ray: $\lambda\sim0.01$ to $10$ nm.
• Gamma: $\lambda<0.01$ nm.
• [JEE Tip] Convert with $c=f\lambda$, then place by wavelength. Increasing-frequency mnemonic: Raging Martians Invaded Venus Using X-ray Guns.

Solved Examples -- Beyond-NCERT Formulae

Example 1 -- Magnetic amplitude and rms fields from $E_0$. A plane EM wave in vacuum has peak electric field $E_0=60$ V/m. Find $B_0$, $E_{rms}$ and $B_{rms}$.

• $B_0=\dfrac{E_0}{c}=\dfrac{60}{3\times10^8}=2.0\times10^{-7}$ T.
• $E_{rms}=\dfrac{E_0}{\sqrt2}=\dfrac{60}{1.414}=42.4$ V/m.
• $B_{rms}=\dfrac{B_0}{\sqrt2}=1.41\times10^{-7}$ T.

The magnetic amplitude is minuscule, which is exactly why the electric field does essentially all the work on charges.

Example 2 -- Average energy density. Find $u_{avg}$ for a wave of peak field $E_0=48$ V/m. Take $\varepsilon_0=8.85\times10^{-12}$ F/m.

• $u_{avg}=\tfrac12\varepsilon_0E_0^2=\tfrac12(8.85\times10^{-12})(48)^2=1.02\times10^{-8}$ J/m$^3$.
• Electric share $\tfrac14\varepsilon_0E_0^2=5.10\times10^{-9}$ J/m$^3$; the magnetic share is equal.
• Cross-check via $B_0=\dfrac{E_0}{c}=1.6\times10^{-7}$ T: $\dfrac{B_0^2}{2\mu_0}=\dfrac{(1.6\times10^{-7})^2}{2(4\pi\times10^{-7})}=1.02\times10^{-8}$ J/m$^3$. Same total, as it must be.

Example 3 -- Intensity two ways (impedance of free space). A wave has $E_{rms}=50$ V/m. Find its intensity.

• Standard: $I=\varepsilon_0cE_{rms}^2=(8.85\times10^{-12})(3\times10^8)(50)^2=6.6$ W/m$^2$.
• Impedance form: with $Z_0=\mu_0c\approx377$ ohm, $I=\dfrac{E_{rms}^2}{Z_0}=\dfrac{2500}{377}=6.6$ W/m$^2$.

Both routes agree; the impedance form is a fast shortcut when $E_{rms}$ is known.

Example 4 -- Field amplitude from a point source. A small source radiates $P=100$ W isotropically. Find the intensity and the peak electric field at $r=2.0$ m.

• $I=\dfrac{P}{4\pi r^2}=\dfrac{100}{4\pi(2.0)^2}=1.99$ W/m$^2$.
• $E_0=\sqrt{\dfrac{2I}{\varepsilon_0c}}=\sqrt{\dfrac{2(1.99)}{2.655\times10^{-3}}}=38.7$ V/m.
• Then $B_0=\dfrac{E_0}{c}=1.29\times10^{-7}$ T.

Note the sphere area $4\pi r^2$; using $\pi r^2$, the area of a disc, would inflate $I$ fourfold.

Example 5 -- Radiation force: absorber versus mirror. Sunlight of intensity $I=1360$ W/m$^2$ falls normally on a panel of area $A=10$ m$^2$. Find the force if the panel (a) absorbs and (b) reflects all the light, and the pressure in case (a).

• Absorber: $F=\dfrac{IA}{c}=\dfrac{(1360)(10)}{3\times10^8}=4.5\times10^{-5}$ N; pressure $\dfrac{I}{c}=4.5\times10^{-6}$ Pa.
• Reflector: $F=\dfrac{2IA}{c}=9.1\times10^{-5}$ N, twice the absorber value.

Example 6 -- Momentum delivered. An EM wave delivers $U=500$ J of energy to a surface at normal incidence. Find the momentum transferred if the surface (a) absorbs and (b) reflects the radiation.

• Absorber: $p=\dfrac{U}{c}=\dfrac{500}{3\times10^8}=1.67\times10^{-6}$ kg m/s.
• Reflector: $p=\dfrac{2U}{c}=3.33\times10^{-6}$ kg m/s.

Example 7 -- Partial reflector (beyond-NCERT). A beam of intensity $I=25$ W/m$^2$ strikes a surface of reflectivity $R=0.60$ at normal incidence. Find the radiation pressure.

• $P_{rad}=(1+R)\dfrac{I}{c}=(1.60)\dfrac{25}{3\times10^8}=1.33\times10^{-7}$ Pa.
• It lies between the absorber value $\dfrac{I}{c}=8.3\times10^{-8}$ Pa and the mirror value $\dfrac{2I}{c}=1.67\times10^{-7}$ Pa, as it must.

Example 8 -- Spectrum: wavelength and band. (a) A wave has frequency $f=1.0\times10^{17}$ Hz. Find its wavelength and name the band.

• $\lambda=\dfrac{c}{f}=\dfrac{3\times10^8}{1.0\times10^{17}}=3.0\times10^{-9}$ m, i.e. $3$ nm, which is the X-ray region.

(b) A radar transmitter emits $\lambda=0.10$ m. Then $f=\dfrac{c}{\lambda}=3\times10^9$ Hz, the microwave band.

Ordering by increasing frequency: radio, microwave, infrared, visible, ultraviolet, X-ray, gamma.