Question

A plane electromagnetic wave is moving in free space with velocity \[ c=3\times10^{8}\ \text{m/s} \] and its electric field is given as \[ \vec E = 54\sin(kz-\omega t)\,\hat{j}\ \text{V/m}, \] where \(\hat{j}\) is the unit vector along the \(y\)-axis. The magnetic field \(\vec B\) of the wave is:

• A. \(-1.8\times10^{-7}\sin(kz-\omega t)\,\hat{i}\ \text{T}\)
• B. \(1.4\times10^{-7}\sin(kz-\omega t)\,\hat{k}\ \text{T}\)
• C. \(1.4\times10^{-7}\sin(kz-\omega t)\,\hat{i}\ \text{T}\)
• D. \(+1.8\times10^{-7}\sin(kz-\omega t)\,\hat{i}\ \text{T}\)

Hint:

In an EM wave, always remember: \(\vec E \perp \vec B \perp\) direction of propagation and \(E=cB\).

Answer 1

The Correct Option is D

Concept: For a plane electromagnetic wave propagating in free space:
The electric field \(\vec E\), magnetic field \(\vec B\), and direction of propagation are mutually perpendicular.
The magnitudes of \(\vec E\) and \(\vec B\) are related by: \[ E = cB \]
The direction of propagation is given by \(\vec E \times \vec B\).

Step 1: Determine direction of propagation The phase of the wave is \((kz-\omega t)\), which indicates propagation along the \(+z\)-direction. Given: \[ \vec E \parallel \hat{j}\ (\text{along }y\text{-axis}) \] For propagation along \(+z\), \[ \vec E \times \vec B \parallel \hat{k} \] Hence, \(\vec B\) must be along the \(+x\)-direction (\(\hat{i}\)).
Step 2: Calculate magnitude of magnetic field \[ B_0=\frac{E_0}{c} =\frac{54}{3\times10^{8}} =1.8\times10^{-7}\ \text{T} \]
Step 3: Write magnetic field expression \[ \vec B = 1.8\times10^{-7}\sin(kz-\omega t)\,\hat{i}\ \text{T} \] Final Answer: \[ \boxed{+1.8\times10^{-7}\sin(kz-\omega t)\,\hat{i}\ \text{T}} \]