Question 126: EM waves

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 :

Accepted Answer

$$\overrightarrow E  = {E_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos \left( {kz + \omega t} ight)$$
$$\overrightarrow E $$ at t = 0 at z = $${\pi  \over k}$$ is given by
$$\overrightarrow E  = $$$${E_0}\left( {{{\widehat i + \widehat j} \over {\sqrt 2 }}} ight)\cos \left( {k{\pi  \over k} + \omega \left( 0 ight)} ight)$$
= $$ - {E_0}\left( {{{\widehat i + \widehat j} \over {\sqrt 2 }}} ight)$$
As $$\overrightarrow E  	imes \overrightarrow B  = \overrightarrow c $$
Force due to magnetic field is in direction $$q\left( {\overrightarrow v  	imes \overrightarrow B } ight)$$ and given $${\overrightarrow v }$$ parallel to  $$\widehat k$$.
$$\overrightarrow F  = $$ $$q\left( {\overrightarrow E  + \overrightarrow v  	imes \overrightarrow B } ight)$$
$$ 	herefore $$ $$\overrightarrow F $$ is in direction of $${\overrightarrow E }$$
which is antiparallel to $${{\widehat i + \widehat j} \over {\sqrt 2 }}$$

Answer

parallel to $$\widehat k$$

Answer

parallel to $${{\widehat i + \widehat j} \over {\sqrt 2 }}$$

Answer

antiparallel to $${{\widehat i + \widehat j} \over {\sqrt 2 }}$$

Answer

zero

EM waves

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