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

The given problem involves determining the magnetic field \(\vec{B}\) of a plane electromagnetic wave moving through free space. Given that the electric field \(\vec{E}\) is expressed as \(\vec{E} = 54\sin(kz-\omega t)\,\hat{j}\ \text{V/m}\), we need to find the expression for the magnetic field \(\vec{B}\).

First, we recognize a few key points about plane electromagnetic waves:

• The direction of wave propagation is perpendicular to both the electric and magnetic fields.
• The relation between the magnitudes of the electric field \(E\_0\) and the magnetic field \(B\_0\) is given by \(c = \frac{E\_0}{B\_0}\), where \(c = 3 \times 10^8\ \text{m/s}\) is the speed of light in vacuum.

Step 1: Calculate the magnitude of the magnetic field:

From the relation:

\(E\_0 = 54\ \text{V/m}\)

We find \(B\_0\) using:

\(B\_0 = \frac{E\_0}{c} = \frac{54}{3 \times 10^8} = 1.8 \times 10^{-7}\ \text{T}\)

Step 2: Determine the direction of the magnetic field:

Given that the electric field \(\vec{E}\) is along \(\hat{j}\) and the wave propagates along the \(\hat{k}\) direction, using the right-hand rule, the magnetic field \(\vec{B}\) should be along the \(\hat{i}\) direction.

Thus, the expression for the magnetic field is:

\(\vec{B} = 1.8 \times 10^{-7} \sin(kz-\omega t)\,\hat{i}\ \text{T}\)

Therefore, the correct answer is:

\(+1.8\times10^{-7}\sin(kz-\omega t)\,\hat{i}\ \text{T}\)