Question 134: EM waves

Question

The magnetic field of an electromagnetic wave
is given by :-
$$\mathop B\limits^ 	o   = 1.6 	imes {10^{ - 6}}\cos \left( {2 	imes {{10}^7}z + 6 	imes {{10}^{15}}t} ight)\left( {2\mathop i\limits^ \wedge   + \mathop j\limits^ \wedge  } ight){{Wb} \over {{m^2}}}$$
The associated electric field will be :-

Accepted Answer

If we use that direction of light propagation will be along $$\overrightarrow E  	imes \overrightarrow B $$. Then (A) option is
correct.
Magnitude of E = CB
E = 3 ÃÂ 10
8
ÃÂ 1.6 ÃÂ 10
Ã¢ÂÂ6
ÃÂ $$\sqrt 5 $$
E = 4.8 ÃÂ $${10^{2\sqrt 5 }}$$
$$\overrightarrow E $$ and  $$\overrightarrow B $$ are perpendicular to each other
$$ \Rightarrow \overrightarrow E .\overrightarrow B  = 0$$
$$ \Rightarrow $$ Either direction of $$\overrightarrow E $$ $$\widehat i - 2\widehat j$$ or $$-
 \widehat i + 2\widehat j$$ from given option
Also wave propagation direction is parallel to $$\overrightarrow E  	imes \overrightarrow B $$ which is $$- \widehat k$$
$$ \Rightarrow $$ $$\overrightarrow E $$ is along ($$-
 \widehat i + 2\widehat j$$)

Answer

$$\mathop E\limits^ 	o   = 4.8 	imes {10^2}\cos \left( {2 	imes {{10}^7}z - 6 	imes {{10}^{15}}t} ight)\left( -2{\mathop i\limits^ \wedge   + \mathop {j}\limits^ \wedge  } ight){V \over m}$$

Answer

$$\mathop E\limits^ 	o   = 4.8 	imes {10^2}\cos \left( {2 	imes {{10}^7}z - 6 	imes {{10}^{15}}t} ight)\left( 2{\mathop i\limits^ \wedge   + \mathop {j}\limits^ \wedge  } ight){V \over m}$$

Answer

$$\mathop E\limits^ 	o   = 4.8 	imes {10^2}\cos \left( {2 	imes {{10}^7}z + 6 	imes {{10}^{15}}t} ight)\left( {\mathop i\limits^ \wedge   - \mathop {2j}\limits^ \wedge  } ight){V \over m}$$

Answer

$$\mathop E\limits^ 	o   = 4.8 	imes {10^2}\cos \left( {2 	imes {{10}^7}z + 6 	imes {{10}^{15}}t} ight)\left( -{\mathop i\limits^ \wedge   + \mathop {2j}\limits^ \wedge  } ight){V \over m}$$

EM waves

Choose the correct answer: