JEE MAIN - Physics (2020 - 9th January Evening Slot - No. 22)
A plane electromagnetic wave is propagating
along the direction
$${{\widehat i + \widehat j} \over {\sqrt 2 }}$$
, with its polarization
along the direction $$\widehat k$$ . The correct form of the
magnetic field of the wave would be (here B0
is an appropriate constant) :
$${B_0}{{\widehat i - \widehat j} \over {\sqrt 2 }}\cos \left( {\omega t - k{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)$$
$${B_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos \left( {\omega t - k{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)$$
$${B_0}{{\widehat j - \widehat i} \over {\sqrt 2 }}\cos \left( {\omega t + k{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)$$
$${B_0}\widehat k\cos \left( {\omega t - k{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)$$
Explanation
Direction of propagation = $${{\widehat i + \widehat j} \over {\sqrt 2 }}$$
Electric field is in direction = $$\widehat k$$
As $$\overrightarrow E \times \overrightarrow B $$ = $${{\widehat i + \widehat j} \over {\sqrt 2 }}$$
Propagation direction of $$\overrightarrow B = {{\widehat i - \widehat j} \over {\sqrt 2 }}$$
Electric field is in direction = $$\widehat k$$
As $$\overrightarrow E \times \overrightarrow B $$ = $${{\widehat i + \widehat j} \over {\sqrt 2 }}$$
Propagation direction of $$\overrightarrow B = {{\widehat i - \widehat j} \over {\sqrt 2 }}$$
Comments (0)
