Question 193: EM waves

Question

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

Accepted Answer

For electromagnetic wave,
$$|\overrightarrow B | = {{|\overrightarrow E |} \over c}$$
Here $$\overrightarrow B $$ is magnetic field associated with EM wave
$$\overrightarrow E $$ is electric field associated with EM wave
c is the speed of EM wave
$$ \Rightarrow |\overrightarrow E | = c|\overrightarrow B |$$
$$ = 3 	imes {10^8} 	imes 3 	imes {10^{ - 8}}\cos (1.6 	imes {10^3}x + 48 	imes {10^{10}}t)$$ V/m
Direction can be determined from
Poynting vector $$ = {{\overrightarrow E  	imes \overrightarrow B } \over {{\mu _0}}}$$
$$\overrightarrow E  = 9\cos (1.6 	imes {10^3}x + 48 	imes {10^{10}}t)\widehat k$$ V/m

Answer

$$9\cos (1.6 	imes {10^3}x + 48 	imes {10^{10}}t)\widehat k$$ V/m

Answer

$$3 	imes {10^{ - 8}}\cos (1.6 	imes {10^3}x + 48 	imes {10^{10}}t)\widehat i$$ V/m

Answer

$$3 	imes {10^{ - 8}}\sin (1.6 	imes {10^3}x + 48 	imes {10^{10}}t)\widehat i$$ V/m

Answer

$$9\sin (1.6 	imes {10^3}x - 48 	imes {10^{10}}t)\widehat k$$ V/m

EM waves

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

Choose the correct answer: