JEE MAIN - Physics (2020 - 7th January Evening Slot - No. 8)
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 :
parallel to $$\widehat k$$
parallel to $${{\widehat i + \widehat j} \over {\sqrt 2 }}$$
antiparallel to $${{\widehat i + \widehat j} \over {\sqrt 2 }}$$
zero
Explanation
$$\overrightarrow E = {E_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos \left( {kz + \omega t} \right)$$
$$\overrightarrow E $$ at t = 0 at z = $${\pi \over k}$$ is given by
$$\overrightarrow E = $$$${E_0}\left( {{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)\cos \left( {k{\pi \over k} + \omega \left( 0 \right)} \right)$$
= $$ - {E_0}\left( {{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)$$
As $$\overrightarrow E \times \overrightarrow B = \overrightarrow c $$
Force due to magnetic field is in direction $$q\left( {\overrightarrow v \times \overrightarrow B } \right)$$ and given $${\overrightarrow v }$$ parallel to $$\widehat k$$.
$$\overrightarrow F = $$ $$q\left( {\overrightarrow E + \overrightarrow v \times \overrightarrow B } \right)$$
$$ \therefore $$ $$\overrightarrow F $$ is in direction of $${\overrightarrow E }$$
which is antiparallel to $${{\widehat i + \widehat j} \over {\sqrt 2 }}$$
$$\overrightarrow E $$ at t = 0 at z = $${\pi \over k}$$ is given by
$$\overrightarrow E = $$$${E_0}\left( {{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)\cos \left( {k{\pi \over k} + \omega \left( 0 \right)} \right)$$
= $$ - {E_0}\left( {{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)$$
As $$\overrightarrow E \times \overrightarrow B = \overrightarrow c $$
Force due to magnetic field is in direction $$q\left( {\overrightarrow v \times \overrightarrow B } \right)$$ and given $${\overrightarrow v }$$ parallel to $$\widehat k$$.
$$\overrightarrow F = $$ $$q\left( {\overrightarrow E + \overrightarrow v \times \overrightarrow B } \right)$$
$$ \therefore $$ $$\overrightarrow F $$ is in direction of $${\overrightarrow E }$$
which is antiparallel to $${{\widehat i + \widehat j} \over {\sqrt 2 }}$$
Comments (0)
