Question

The magnetic field of a plane electromagnetic wave is
$\vec{ B }=3 \times 10^{-8} \sin [200 \pi( y + ct )] \hat{ i } T$
Where $c=3 \times 10^{8} ms ^{-1}$ is the speed of light. The corresponding electric field is :

• A. $\vec{ E }=-10^{-6} \sin [200 \pi( y + ct )] \hat{ k } V / m$
• B. $\vec{ E }=-9 \sin [200 \pi( y + ct )] \hat{ k } V / m$
• C. $\vec{ E }=9 \sin [200 \pi( y + ct )] \hat{ k } \quad V / m$
• D. $\vec{ E }=3 \times 10^{-8} \sin [200 \pi( y + ct )] \hat{ k } V / m$

Answer 1

Answer: (B)

$\overrightarrow{ B }=3 \times 10^{-8} \sin [200 \pi( y + ct )] \hat{ i } T$
$E _{0}= CB _{0} \Rightarrow E _{0}=3 \times 10^{8} \times 3 \times 10^{-8}=9 V / m$
and direction of wave propagation is given as
$(\overrightarrow{ E } \times \overrightarrow{ B }) \| \overrightarrow{ C }$
$\hat{ B }=\hat{ i },\,\,\,\, \&, \,\,\,\, \hat{ C }=-\hat{ j }$
so $\hat{ E }=-\hat{ k }$
$\therefore \overrightarrow{ E }= E _{0} \sin [200 \pi( y + ct )](-\hat{ k }) V / m$