Question

An electromagnetic wave has electric and magnetic fields given by \(\vec{E}(t) = \vec{E}_m \sin(kx - \omega t)\), \(\vec{B}(t) = \vec{B}_m \sin(kx - \omega t)\). If the direction of \(\vec{E}_m\) and \(\vec{B}_m\) are \((i + 2j)\) and \((-i + \frac{1}{2} j)\) respectively, then the direction of propagation of the wave is:

• A. \(\mathbf{k}\)
• B. \(-\mathbf{k}\)
• C. \(i + j\)
• D. \(i - j\)

Hint:

The propagation direction of an EM wave is always along \(\vec{E} \times \vec{B}\).

Answer 1

The Correct Option is A

Step 1: Use the fact that $\vec{E}$, $\vec{B}$ and the propagation direction $\vec{k}$ are mutually perpendicular in an electromagnetic wave.
A true propagation direction must give zero when dotted with both $\vec{E}_m$ and $\vec{B}_m$. Testing each option against this rule rules out most of them right away.

Step 2: Test $i+j$.
\[ \vec{E}_m \cdot (i+j) = (1)(1) + (2)(1) = 3 \neq 0 \] Not perpendicular to $\vec{E}_m$, so ruled out.

Step 3: Test $i-j$.
\[ \vec{E}_m \cdot (i-j) = (1)(1) + (2)(-1) = -1 \neq 0 \] Also ruled out.

Step 4: Test $\pm k$.
\[ \vec{E}_m \cdot k = 0, \qquad \vec{B}_m \cdot k = 0 \] Both vanish since neither vector has a $k$ component, so the direction must be $+k$ or $-k$; only the sign is left to fix.

Step 5: Fix the sign with the right hand rule, curling fingers from $\vec{E}$ towards $\vec{B}$.
Turning from $\vec{E}_m = i+2j$ towards $\vec{B}_m = -i+\frac{1}{2}j$ the short way points the thumb out of the page, along $+k$, matching $\vec{k} \propto \vec{E}\times\vec{B}$.

Final Answer:
\[ \boxed{k} \]