Question:
The electric field of a plane electromagnetic wave is given by
\( \overrightarrow E = {E_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos \left( {kz + \omega t} \right) \)
At t = 0, a positively charged particle is at the point (x, y, z) = \( \left( {0,0,{\pi \over k}} \right) \).
If its instantaneous velocity at (t = 0) is \( {v_0}\widehat k \) , the force acting on it due to the wave is :
The electric field of a plane electromagnetic wave is given by
\( \overrightarrow E = {E_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos \left( {kz + \omega t} \right) \)
At t = 0, a positively charged particle is at the point (x, y, z) = \( \left( {0,0,{\pi \over k}} \right) \).
If its instantaneous velocity at (t = 0) is \( {v_0}\widehat k \) , the force acting on it due to the wave is :
\( \overrightarrow E = {E_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos \left( {kz + \omega t} \right) \)
At t = 0, a positively charged particle is at the point (x, y, z) = \( \left( {0,0,{\pi \over k}} \right) \).
If its instantaneous velocity at (t = 0) is \( {v_0}\widehat k \) , the force acting on it due to the wave is :
Show Hint
Find E at the given point and time (cos at phase \(\pi\)), find the direction of propagation (here \(-\hat k\)), then use \(\vec F=q(\vec E+\vec v\times\vec B)\) with \(\vec B=\dfrac{\hat n\times\vec E}{c}\).
Updated On: Oct 31, 2026
- parallel to \( \widehat k \)
- parallel to \( {{\widehat i + \widehat j} \over {\sqrt 2 }} \)
- antiparallel to \( {{\widehat i + \widehat j} \over {\sqrt 2 }} \)
- zero
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: The phase is \(kz+\omega t\), so the wave travels along \(-\hat z\). At \(t=0\), \(z=\pi/k\): \(\cos(kz+\omega t)=\cos\pi=-1\).
Step 2: \(\vec E=-E_0\dfrac{\hat i+\hat j}{\sqrt2}\), so the electric force \(q\vec E\) is antiparallel to \(\dfrac{\hat i+\hat j}{\sqrt2}\).
Step 3: Magnetic field: \(\vec B=\dfrac{1}{c}(\hat n\times\vec E)\) with \(\hat n=-\hat k\). Magnetic force \(q\vec v\times\vec B=\dfrac{qv_0}{c}\,\hat k\times(-\hat k\times\vec E)=\dfrac{qv_0}{c}\vec E\) (since \(\hat k\cdot\vec E=0\)), which is also along \(\vec E\).
Step 4: Total force \(=q\vec E\left(1+\dfrac{v_0}{c}\right)\), antiparallel to \(\dfrac{\hat i+\hat j}{\sqrt2}\).
Answer: (C) antiparallel to \(\dfrac{\hat i+\hat j}{\sqrt2}\)
Step 2: \(\vec E=-E_0\dfrac{\hat i+\hat j}{\sqrt2}\), so the electric force \(q\vec E\) is antiparallel to \(\dfrac{\hat i+\hat j}{\sqrt2}\).
Step 3: Magnetic field: \(\vec B=\dfrac{1}{c}(\hat n\times\vec E)\) with \(\hat n=-\hat k\). Magnetic force \(q\vec v\times\vec B=\dfrac{qv_0}{c}\,\hat k\times(-\hat k\times\vec E)=\dfrac{qv_0}{c}\vec E\) (since \(\hat k\cdot\vec E=0\)), which is also along \(\vec E\).
Step 4: Total force \(=q\vec E\left(1+\dfrac{v_0}{c}\right)\), antiparallel to \(\dfrac{\hat i+\hat j}{\sqrt2}\).
Answer: (C) antiparallel to \(\dfrac{\hat i+\hat j}{\sqrt2}\)
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