Question

The electric and magnetic field vectors of a plane electromagnetic wave propagating along the positive x-axis are represented by \( \vec{E} \) and \( \vec{B} \) respectively. Which of the following vector expressions correctly defines the direction of propagation of this wave?

• A. \( \vec{B} \times \vec{E} \)
• B. \( \vec{E} \times \vec{B} \)
• C. \( \vec{E} \cdot \vec{B} \)
• D. \( \vec{E} + \vec{B} \)

Hint:

Always remember alphabetical order to avoid getting mixed up: E comes before B, so the cross product is always E \(\times\) B. Swapping them will result in a negative sign error!

Answer 1

The Correct Option is B

Concept: Electromagnetic waves are transverse in nature, meaning that the oscillating electric field vector \( \vec{E} \), the oscillating magnetic field vector \( \vec{B} \), and the wave's direction of propagation are all mutually perpendicular to one another. The directional vector flow of energy in an electromagnetic field is mathematically described by the Poynting vector, which depends on the vector cross product of the field components.

Step 1: Apply the right-hand vector cross product rule to transverse fields. By Maxwell's equations, the direction of propagation of an electromagnetic wave is always parallel to the cross product of the electric field vector and the magnetic field vector. \[ \hat{n} = \frac{\vec{E} \times \vec{B}}{|\vec{E} \times \vec{B}|} \] This vector product follows the standard right-hand rule. If the wave travels along the positive x-axis (\(\hat{i}\)), and the electric field oscillates along the y-axis (\(\hat{j}\)), the magnetic field must oscillate along the z-axis (\(\hat{k}\)) because: \[ \hat{j} \times \hat{k} = \hat{i} \]

Step 2: Eliminate incorrect vector operations.
• \( \vec{B} \times \vec{E} \) points in the opposite direction (\(-\hat{i}\)) due to the anti-commutative property of cross products.
• \( \vec{E} \cdot \vec{B} \) is a scalar dot product, which equals zero since the fields are perpendicular (\(\theta = 90^\circ\)), providing no directional vector information. Therefore, \( \vec{E} \times \vec{B} \) uniquely maps out the correct forward propagation path.