Question 114: EM waves

Question

For a plane electromagnetic wave, the magnetic field at a point x and time t is
$$\overrightarrow B \left( {x,t} ight)$$ = $$\left[ {1.2 	imes {{10}^{ - 7}}\sin \left( {0.5 	imes {{10}^3}x + 1.5 	imes {{10}^{11}}t} ight)\widehat k} ight]$$ T
The instantaneous electric field $$\overrightarrow E $$
 corresponding to $$\overrightarrow B $$
 is :
(speed of light c = 3 × 108
 ms–1)

Accepted Answer

Given,
$$\overrightarrow B \left( {x,t} ight)$$ = $$\left[ {1.2 	imes {{10}^{ - 7}}\sin \left( {0.5 	imes {{10}^3}x + 1.5 	imes {{10}^{11}}t} ight)\widehat k} ight]$$ T
Wave is travelling along (Ã¢ÂÂx) axis and $$\overrightarrow B $$ is along
+z axis.
We know, Magnitude of electric field
E = BC
= 1.2 $$ 	imes $$ 10
-7
sin ( 0.5 10 + 1.5$$ 	imes $$10
11
t ) $$ 	imes $$ 3 $$ 	imes $$ 10
8
= 36 sin (0.5$$ 	imes $$10
3
x+1.5$$ 	imes $$10
11
t) V/m
Also, $$\overrightarrow s  = {{\overrightarrow E  	imes \overrightarrow B } \over {{\mu _0}}}$$
$$ \Rightarrow $$ $$ - \widehat i = {{\overrightarrow E  	imes \widehat k} \over {{\mu _0}}}$$
$$ 	herefore $$ Direction of $${\overrightarrow E  =  - \widehat j}$$
$$ 	herefore $$ Instantaneous electric field,
$$\overrightarrow E \left( {x,t} ight) = \left[ { - 36\sin \left( {0.5 	imes {{10}^3}x + 1.5 	imes {{10}^{11}}t} ight)\widehat j} ight]{V \over m}$$

Answer

$$\overrightarrow E \left( {x,t} ight) = \left[ {36\sin \left( {1 	imes {{10}^3}x + 1.5 	imes {{10}^{11}}t} ight)\widehat i} ight]$$ $${V \over m}$$

Answer

$$\overrightarrow E \left( {x,t} ight) = \left[ {36\sin \left( {0.5 	imes {{10}^3}x + 1.5 	imes {{10}^{11}}t} ight)\widehat k} ight]{V \over m}$$

Answer

$$\overrightarrow E \left( {x,t} ight) = \left[ {36\sin \left( {1 	imes {{10}^3}x + 0.5 	imes {{10}^{11}}t} ight)\widehat j} ight]{V \over m}$$

Answer

$$\overrightarrow E \left( {x,t} ight) = \left[ { - 36\sin \left( {0.5 	imes {{10}^3}x + 1.5 	imes {{10}^{11}}t} ight)\widehat j} ight]{V \over m}$$

EM waves

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