Question 86: 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: