Question

The magnetic field vector of an electromagnetic wave is given by \( \vec{B} = B_0 \frac{\hat{i} + \hat{j}}{\sqrt{2}} \cos(kz - \omega t) \); where \(\hat{i}, \hat{j}\) represents unit vector along x and y-axis respectively. At \(t = 0\) s, two electric charges \(q_1\) of \(4\pi\) coulomb and \(q_2\) of \(2\pi\) coulomb located at \((0, 0, \frac{\pi}{k})\) and \((0, 0, \frac{3\pi}{k})\), respectively, have the same velocity of \(0.5\,c\,\hat{i}\), (where c is the velocity of light). The ratio of the force acting on charge \(q_1\) to \(q_2\) is :

• A. \(\sqrt{2} : 1\)
• B. \(1 : \sqrt{2}\)
• C. \(2 : 1\)
• D. \(2\sqrt{2} : 1\)

Hint:

Even if the exact vector components look complex, look for common factors. Here, both positions resulted in \(|\cos(kz)| = 1\), so the force magnitude depended solely on the charge magnitude.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The Lorentz force acting on a charge is given by \(\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})\). In an electromagnetic wave, the electric and magnetic fields are related and perpendicular to each other and the direction of propagation.
Step 2: Key Formula or Approach:
1. Propagation direction: \(\hat{k}\) (from \(kz - \omega t\)).
2. Electric field \(\vec{E}\) is such that \(\hat{E}, \hat{B}, \hat{k}\) form a right-handed system: \(\vec{E} = c(\vec{B} \times \hat{k})\).
3. Total Force: \(\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})\).
Step 3: Detailed Explanation:
At \(t = 0\), \(\vec{B} = B_0 \frac{\hat{i} + \hat{j}}{\sqrt{2}} \cos(kz)\).
The electric field is \(\vec{E} = \frac{c B_0 \cos(kz)}{\sqrt{2}} [(\hat{i} + \hat{j}) \times \hat{k}] = \frac{c B_0 \cos(kz)}{\sqrt{2}} (\hat{i} - \hat{j})\).
Given \(\vec{v} = 0.5c\,\hat{i}\), we find \(\vec{v} \times \vec{B}\): \[ \vec{v} \times \vec{B} = (0.5c\,\hat{i}) \times \frac{B_0 \cos(kz)}{\sqrt{2}} (\hat{i} + \hat{j}) = \frac{0.5c B_0 \cos(kz)}{\sqrt{2}} \hat{k} \] Total force: \(\vec{F} = q \frac{c B_0 \cos(kz)}{\sqrt{2}} [\hat{i} - \hat{j} + 0.5\hat{k}] \).
Magnitude of force \(|\vec{F}| \propto q |\cos(kz)|\). For \(q_1\) at \(z = \frac{\pi}{k}\): \(|\cos(\pi)| = 1\). Force \(F_1 \propto q_1 = 4\pi\).
For \(q_2\) at \(z = \frac{3\pi}{k}\): \(|\cos(3\pi)| = 1\). Force \(F_2 \propto q_2 = 2\pi\).
Ratio \(F_1 / F_2 = 4\pi / 2\pi = 2 : 1\).
Step 4: Final Answer:
The ratio of the forces is \(2 : 1\).