A Plane Wave, Anatomised

Maxwell's equations show that in an electromagnetic wave, E⃗\vec{E} and B⃗\vec{B} are perpendicular to each other AND to the direction of propagation — EM waves are transverse. (Plausible from the capacitor picture: E points between the plates, the B it generates circles parallel to them — perpendicular.)

The standard plane wave moving along z (NCERT Fig. 8.3):

Ex=E0sin⁡(kz−ωt),By=B0sin⁡(kz−ωt)E_x = E_0\sin(kz - \omega t), \qquad B_y = B_0\sin(kz - \omega t)

  • E⃗\vec{E} along x, B⃗\vec{B} along y, propagation along z — and E^×B^=k^\hat{E}\times\hat{B} = \hat{k}: E cross B points along the propagation direction.
  • k=2πλk = \dfrac{2\pi}{\lambda} — the magnitude of the wave vector k⃗\vec{k}, whose direction is the propagation direction.
  • ω\omega = angular frequency; speed =ω/k= \omega/k.
  • E and B oscillate in phase: they peak together, vanish together.

Plane electromagnetic wave with perpendicular E and B fields along propagation

The Speed and the Ratio

Feeding the plane wave into Maxwell's equations yields two golden relations:

ω=ck,c=1μ0ε0\omega = ck, \qquad c = \frac{1}{\sqrt{\mu_0\varepsilon_0}}

equivalently νλ=c\nu\lambda = c. And the field magnitudes are locked together:

B0=E0c\boxed{B_0 = \frac{E_0}{c}}

In a material medium of permittivity ε\varepsilon and permeability μ\mu, the speed becomes

v=1μεv = \frac{1}{\sqrt{\mu\varepsilon}}

— the velocity of light depends on the medium's electric and magnetic properties (this underlies the refractive index, next chapter: n=c/v=μrεrn = c/v = \sqrt{\mu_r\varepsilon_r}).

Three more NCERT essentials:

  1. EM waves are self-sustaining oscillations of fields — no material medium is required (unlike sound).
  2. The vacuum speed c is the same for all wavelengths (to within experimental precision) — a fundamental constant, so well established that it defines the standard of length.
  3. EM waves carry energy — radio signals, sunlight powering life on earth. (Beyond NCERT, for JEE: they carry momentum too, exerting radiation pressure.)

[NEET Important] B0=E0/cB_0 = E_0/c means the magnetic amplitude is 3×1083\times10^8 times smaller in SI numbers — e.g. 300 V/m of E rides with just 1 μ\muT of B.

Reading and Writing Wave Equations (Exam Drill)

Given any wave like E=E0sin⁡(kz−ωt)E = E_0\sin(kz - \omega t), extract everything:

  1. Wavelength: λ=2π/k\lambda = 2\pi/k. Frequency: ν=ω/2π\nu = \omega/2\pi. Speed check: ω/k\omega/k should be c (vacuum).
  2. Direction of travel: (kz−ωt)(kz - \omega t) moves towards +z; (kz+ωt)(kz + \omega t) towards −z-z.
  3. The partner field: magnitude from B0=E0/cB_0 = E_0/c; direction from E^×B^=\hat{E}\times\hat{B} = propagation direction.

[JEE Tip] The direction-finding rule is a permanent JEE fixture: given E along j^\hat{j} and propagation along i^\hat{i}, B must satisfy j^×B^=i^\hat{j}\times\hat{B} = \hat{i}, so B^=k^\hat{B} = \hat{k}. Work the cross product, never guess.

Key Point: E and B carry the same phase, the same frequency (the source's), and amplitudes locked by c. One field determines the other completely — an EM wave is a single entity, not two waves travelling together.

Solved Examples

Example 1: Finding B from E

A plane EM wave of frequency 25 MHz travels along x. At a point, E⃗=6.3 j^\vec{E} = 6.3\,\hat{j} V/m. Find B⃗\vec{B} there.

Solution:

  1. Magnitude: B=E/c=6.33×108=2.1×10−8B = E/c = \frac{6.3}{3\times10^8} = 2.1 \times 10^{-8} T.
  2. Direction: E⃗×B⃗\vec{E}\times\vec{B} must point along propagation (i^\hat{i}). With E along j^\hat{j}: j^×k^=i^\hat{j}\times\hat{k} = \hat{i} — so B is along z.
  3. B⃗=2.1×10−8 k^\vec{B} = 2.1\times10^{-8}\,\hat{k} T.

Example 2: Reading a full wave

The magnetic field of a wave is By=2×10−7sin⁡(0.5×103x+1.5×1011t)B_y = 2\times10^{-7}\sin(0.5\times10^3 x + 1.5\times10^{11} t) T. Find (a) wavelength and frequency, (b) the electric field expression.

Solution:

  1. (a) k=0.5×103k = 0.5\times10^3 m−1^{-1}: λ=2π/k=1.26\lambda = 2\pi/k = 1.26 cm. ω=1.5×1011\omega = 1.5\times10^{11} rad/s: ν=ω/2π=23.9\nu = \omega/2\pi = 23.9 GHz (a microwave).
  2. (b) E0=B0c=2×10−7×3×108=60E_0 = B_0 c = 2\times10^{-7} \times 3\times10^8 = 60 V/m.
  3. The (kx+ωt)(kx + \omega t) form travels along −x-x; with B along y, E must satisfy E^×j^=−i^\hat{E}\times\hat{j} = -\hat{i}, giving E along z: Ez=60sin⁡(0.5×103x+1.5×1011t)E_z = 60\sin(0.5\times10^3x + 1.5\times10^{11}t) V/m.

Example 3: Writing the partner field [JEE Numerical]

A wave travels along +z with Ex=120sin⁡(kz−ωt)E_x = 120\sin(kz-\omega t) V/m at 50 MHz. Write ByB_y completely.

Solution:

  1. B0=E0/c=1203×108=4×10−7B_0 = E_0/c = \frac{120}{3\times10^8} = 4\times10^{-7} T.
  2. ω=2πν=3.14×108\omega = 2\pi\nu = 3.14\times10^8 rad/s; k=ω/c=1.05k = \omega/c = 1.05 rad/m.
  3. Same phase, perpendicular direction: By=4×10−7sin⁡(1.05 z−3.14×108 t)B_y = 4\times10^{-7}\sin(1.05\,z - 3.14\times10^8\,t) T. (i^×j^=k^\hat{i}\times\hat{j} = \hat{k} — checks.)

Example 4: Speed in a medium [JEE Numerical]

Light enters a medium with εr=4\varepsilon_r = 4 and μr=1\mu_r = 1. Find its speed and the refractive index.

Solution:

  1. v=1με=cμrεr=3×1084v = \frac{1}{\sqrt{\mu\varepsilon}} = \frac{c}{\sqrt{\mu_r\varepsilon_r}} = \frac{3\times10^8}{\sqrt{4}}.
  2. v=1.5×108v = 1.5\times10^8 m/s; n=c/v=2n = c/v = 2.
  3. The medium's electric and magnetic properties set light's speed — Maxwell's Eq. (8.11) in action.

Example 5: From wavelength to everything [NEET Numerical]

A 6 m wavelength wave travels in vacuum. Find ν\nu, kk and ω\omega.

Solution:

  1. ν=c/λ=3×108/6=5×107\nu = c/\lambda = 3\times10^8/6 = 5\times10^7 Hz =50= 50 MHz.
  2. k=2π/λ=1.05k = 2\pi/\lambda = 1.05 rad/m.
  3. ω=2πν=3.14×108\omega = 2\pi\nu = 3.14\times10^8 rad/s. (Check: ω/k=3×108\omega/k = 3\times10^8 m/s ✓.)

Example 6: The amplitude lock [NEET Numerical]

Sunlight at a point has electric amplitude 300 V/m. What is the magnetic amplitude?

Solution:

  1. B0=E0/c=3003×108B_0 = E_0/c = \frac{300}{3\times10^8}.
  2. B0=10−6B_0 = 10^{-6} T = 1 μ\muT — about fifty times weaker than the earth's magnetic field, though the E field is sizeable.
  3. The ratio is always exactly c.

Example 7: Which way does it go? [JEE Numerical]

A wave has EE along −j^-\hat{j} and BB along +i^+\hat{i} at some instant. Find the propagation direction.

Solution:

  1. Propagation is along E^×B^\hat{E}\times\hat{B}.
  2. (−j^)×(i^)=−(j^×i^)=−(−k^)=+k^(-\hat{j})\times(\hat{i}) = -(\hat{j}\times\hat{i}) = -(-\hat{k}) = +\hat{k}.
  3. The wave travels along +z. (Cross products, not memory!)

Example 8: Are E and B in phase?

In a plane EM wave, when E is maximum at a point, what is B doing there — and why does it matter?

Solution:

  1. E and B share the factor sin⁡(kz−ωt)\sin(kz-\omega t): B is also maximum — they oscillate in phase.
  2. They vanish together too: the wave's energy arrives in synchronised pulses of both fields.
  3. (A common misconception puts them 90 degrees apart like an LC exchange — wrong for a travelling wave. In phase!)

Example 9: No medium needed

Sound cannot cross a vacuum but light can. Explain via the nature of EM waves.

Solution:

  1. Sound is a mechanical wave — oscillations OF a medium; no molecules, no wave.
  2. An EM wave is a self-sustaining oscillation of fields: changing E regenerates B (Ampere-Maxwell) and changing B regenerates E (Faraday).
  3. The fields need no carrier — so sunlight crosses 1.5×10111.5\times10^{11} m of vacuum to reach us.

Example 10: c as the metre's anchor

Why is the speed of light used to define the standard of length?

Solution:

  1. Experiments across wavelengths show c in vacuum is the same for all EM waves to within a few m/s out of 3×1083\times10^8 m/s.
  2. Its constancy is so strongly supported, and its value so precisely known, that the metre is defined through c (distance light travels in a fixed fraction of a second).
  3. A fundamental constant became a ruler — physics at its most practical.