Question

The magnetic field of a plane electromagnetic wave is given by $\vec{ B }=3 \times 10^{-8} \cos \left(1.6 \times 10^3 x +48 \times 10^{10} t \right) \hat{ j }$, then the associated electric field will be :

• A. $3 \times 10^{-8} \cos \left(1.6 \times 10^3 x +48 \times 10^{10} t \right) \hat{ i } V / m$
• B. $3 \times 10^{-8} \sin \left(1.6 \times 10^3 x +48 \times 10^{10} t \right) \hat{ i } V / m$
• C. $9 \sin \left(1.6 \times 10^3 x -48 \times 10^{10} t \right) \hat{ k } V / m$
• D. $9 \cos \left(1.6 \times 10^3 x +48 \times 10^{10} t \right) \hat{ k } V / m$

Answer 1

Answer: (D)

$B=3 \times 10^{-8} \cos \left(1.6 \times 10^3 x +48 \times 10^{10} t \right)$
$C =\frac{\omega}{ k }=\frac{48 \times 10^{10}}{1.6 \times 10^3}=3 \times 10^8 m / s$
$C= E _0 / B _0$
$E =3 \times 10^{-8} \times 3 \times 10^8=9 N / C$
$\therefore E=9 \cos \left(1.6 \times 10^3 x +48 \times 10^{10} t \right)$