- Home
- Standard 12
- Physics
The magnetic field of a plane electromagnetic wave is given by $\overrightarrow{ B }=3 \times 10^{-8} \cos \left(1.6 \times 10^3 x +48 \times 10^{10} t \right) \hat{ j }$, then the associated electric field will be :
$3 \times 10^{-8} \cos \left(1.6 \times 10^3 x +48 \times 10^{10} t \right) \hat{ i }\,V / m$
$3 \times 10^{-8} \sin \left(1.6 \times 10^3 x +48 \times 10^{10} t \right) \hat{ i }\,V / m$
$9 \sin \left(1.6 \times 10^3 x -48 \times 10^{10} t \right) \hat{ k}\,V / m$
$9 \cos \left(1.6 \times 10^3 x +48 \times 10^{10} t \right) \hat{ k }\, V / m$
Solution
$B=3 \times 10^{-8} \cos \left(1.6 \times 10^3 x +48 \times 10^{10} t \right)$ $C=\frac{\omega}{ k }=\frac{48 \times 10^{10}}{1.6 \times 10^3}=3 \times 10^8\,m / s$ $C=E_0 / B_0$
$E=3 \times 10^{-8} \times 3 \times 10^8=9\,N / C$
$\therefore E=9 \cos \left(1.6 \times 10^3 x +48 \times 10^{10} t \right)$
