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

Question

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

Aicharya · Accepted Answer

Given, $\mathrm{B}=3 \times 10^{-8} \cos \left(1.6 \times 10^3 x\right.$ $\left.+48 \times 10^{10} t\right)j$
On comparing equations (1) & (2), we get;
$\begin{aligned}
\mathrm{B}_0 & =3 \times 10^{-8} \\
c & =\frac{\omega}{k}=\frac{48 \times 10^{10}}{1.6 \times 10^3} \\
& =3 \times 10^8
\end{aligned}$
And also, $c=\frac{\mathrm{E}_0}{\mathrm{~B}_0}$
Hence, $E_0=B_0 \times c$
$\begin{aligned}
& =3 \times 10^{-8} \times 3 \times 10^8 \\
& =9
\end{aligned}$
We know direction of propagation=E $\times$ B
So as wave propagates in $+x$ direction and magnetic field propagates in $+Y$ direction so electric field should propagate in $+Z$ direction.
So, the required equation is:
$\mathrm{E}=9 \cos \left(1.6 \times 10^3 x+48 \times 10^{10} t\right)t$

The magnetic field of a plane electromagnetic wave is given by $\overrightarrow{\mathrm{B}}=3 \times 10^{-8}…

Solution