Q.An electromagnetic wave travelling along $z$-axis is given as: $\mathbf{E} = E_0\cos(kz - \omega t)$. Choose the correct options from the following:

Concept: Electromagnetic Wave Relation — transverse nature of E and B. For a plane wave $\mathbf E=E_0\cos(kz-\omega t)$ travelling along $z$ in vacuum, check each statement:

(a) $\mathbf B=\dfrac1c(\hat k\times\mathbf E)$ is correct on its own, but the stated equality $=\dfrac1\omega(\hat k\times\mathbf E)$ is wrong — since $c=\omega/k$, we have $\dfrac1c=\dfrac{k}{\omega}$, not $\dfrac1\omega$ (a missing factor of the wavenumber $k$). False as printed.

(b) $\mathbf E=c(\mathbf B\times\hat k)$ — starting from $\mathbf B=\dfrac1c(\hat k\times\mathbf E)$ and using the vector identity for mutually perpendicular unit vectors, this rearranges correctly to give back $\mathbf E$. True. …

The governing relation

In vacuum, a plane electromagnetic wave travelling along $\hat k$ satisfies

and $\mathbf E$, $\mathbf B$, $\hat k$ form a mutually perpendicular, right-handed triad.

Checking statement (a)

The first equality is the standard, correct relation. But the second equality requires $\dfrac1c=\dfrac1\omega$, i.e. $c=\omega$ — which is false in general; the correct relation is $c=\omega/k$, so $\dfrac1c=\dfrac{k}{\omega}$, not $\dfrac1\omega$. The statement is missing a factor of the wavenumber $k$. (a) is false as printed.

Checking statement (b)

Starting from $\mathbf B=\dfrac1c(\hat k\times\mathbf E)$, take $\mathbf B\times\hat k$ on both sides:

For $\mathbf E$ perpendicular to $\hat k$ (a transverse wave) and $\hat k$ a unit vector, the vector triple product identity gives $(\hat k\times\mathbf E)\times\hat k=\mathbf E$ (since $\hat k\times\mathbf E$ is perpendicular to both $\hat k$ and $\mathbf E$, and crossing it again with $\hat k$ rotates it back onto $\mathbf E$, scaled by $|\hat k|^2=1$). So

This matches statement (b) exactly. A quick concrete check confirms it: if $\mathbf E=E_0\hat\imath\cos(kz-\omega t)$, then $\mathbf B=\dfrac{E_0}{c}\hat\jmath\cos(kz-\omega t)$ (using $\hat k\times\hat\imath=\hat\jmath$), and $\mathbf B\times\hat k=\dfrac{E_0}{c}(\hat\jmath\times\hat k)\cos(\cdot)=\dfrac{E_0}{c}\hat\imath\cos(\cdot)$, so $c(\mathbf B\times\hat k)=E_0\hat\imath\cos(\cdot)=\mathbf E$. (b) is true.

Checking statement (c)

An electromagnetic wave is transverse: both $\mathbf E$ and $\mathbf B$ oscillate in the plane perpendicular to the propagation direction $\hat k$, with no component along $\hat k$ itself. This is exactly what the dot products express: …

Method: Verifying Statements About the E–B–k Relations of a Plane Wave

Use this method for MCQ-style questions that list several claimed relations among $\mathbf{E}$, $\mathbf{B}$, and the propagation direction $\hat{k}$, and ask which are correct.

Steps

Step 1: Recall the two governing facts that every valid statement must be consistent with

1. $\mathbf{B}=\dfrac{1}{c}(\hat{k}\times\mathbf{E})$, with $c=\omega/k$ — check any stated coefficient (like $1/\omega$ or $1/k$) dimensionally against this.
2. The wave is transverse: $\hat{k}\cdot\mathbf{E}=0$ and $\hat{k}\cdot\mathbf{B}=0$ always, which is equivalent to saying $\hat{k}\times\mathbf{E}$ and $\hat{k}\times\mathbf{B}$ are generally non-zero (a zero cross product would mean the field points along $\hat{k}$, i.e. a longitudinal wave — never true for light). Step 2: Rearrange the governing relation to check any inverted or rewritten statement To test a claim like $\mathbf{E}=c(\mathbf{B}\times\hat{k})$, start from $\mathbf{B}=\dfrac1c(\hat{k}\times\mathbf{E})$ and apply the vector triple-product identity $(\hat{k}\times\mathbf{E})\times\hat{k}=\mathbf{E}$ (valid because $\mathbf{E}\perp\hat{k}$ and $|\hat{k}|=1$) — cross both sides with $\hat{k}$ and simplify, rather than guessing whether the rearrangement is valid. $\mathbf{E}=c\,(\mathbf{B}\times\hat{k})$ …