Question

The electric field of an electromagnetic wave has the form \( \vec{E} = E_0 \cos(\omega t - kz)\hat{i} \). At \( z = 0 \), a test particle of charge \( q \) and velocity \( \vec{v} = 0.5c \hat{k} \) (where \( c \) is the speed of light) is placed. The total instantaneous force on the particle is:

• A. \( \frac{qE_0}{2} \hat{i} \)
• B. \( \frac{qE_0}{\sqrt{2}} (\hat{i} + \hat{j}) \)
• C. \( \frac{qE_0}{2} (\hat{i} - \hat{k}) \)
• D. Zero

Hint:

Always consider both electric and magnetic field effects when calculating Lorentz force in electromagnetic waves.

Answer 1

The Correct Option is A

Step 1: Determine the magnetic field.
For an electromagnetic wave propagating along \( z \)-axis, \[ \vec{B} = \frac{E_0}{c} \cos(\omega t - kz) \hat{j}. \] Step 2: Calculate the Lorentz force.
\[ \vec{F} = q(\vec{E} + \vec{v} \times \vec{B}). \] At \( z = 0 \): \[ \vec{E} = E_0 \cos(\omega t) \hat{i}, \quad \vec{B} = \frac{E_0}{c} \cos(\omega t) \hat{j}. \] Since \( \vec{v} = 0.5c \hat{k} \): \[ \vec{v} \times \vec{B} = 0.5c \hat{k} \times \frac{E_0}{c}\hat{j} = -0.5E_0 \hat{i}. \] Step 3: Total force.
\[ \vec{F} = qE_0 \cos(\omega t)\hat{i} + q(-0.5E_0 \cos(\omega t)\hat{i}) = \frac{qE_0}{2}\cos(\omega t)\hat{i}. \] Step 4: Final Answer.
The instantaneous force is \( \frac{qE_0}{2}\hat{i}. \)