Question 88: EM waves

Question

The magnetic field of a plane electromagnetic wave is given by :
$
\overrightarrow{\mathrm{B}}=2 	imes 10^{-8} \sin \left(0.5 	imes 10^{3} x+1.5 	imes 10^{11} \mathrm{t}ight) \,\hat{j} \mathrm{~T}$.
The amplitude of the electric field would be :

Accepted Answer

Speed of light $c = {\omega  \over k} = {{1.5 	imes {{10}^{11}}} \over {0.5 	imes {{10}^3}}} = 3 	imes {10^8}$ m/sec
So, ${E_0} = {B_0}c$
$ = 2 	imes {10^{ - 8}} 	imes 3 	imes {10^8}$
$ = 6$ V/m
Direction will be along z-axis.

Answer

$6\, \mathrm{Vm}^{-1}$ along $x$-axis

Answer

$3\, \mathrm{Vm}^{-1}$ along $z$-axis

Answer

$6\, \mathrm{Vm}^{-1}$ along $z$-axis

Answer

$2 	imes 10^{-8} \,\mathrm{Vm}^{-1}$ along $z$-axis

EM waves

The magnetic field of a plane electromagnetic wave is given by : \overrightarrow{\mathrm{B}}=2 \times 10^{-8} \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} \mathrm{t}\right) \,\hat{j} \mathrm{~T}. The amplitude of the electric field would be :

Choose the correct answer: