JEE MAIN - Physics (2019 - 8th April Evening Slot - No. 6)
The magnetic field of an electromagnetic wave
is given by :-
$$\mathop B\limits^ \to = 1.6 \times {10^{ - 6}}\cos \left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( {2\mathop i\limits^ \wedge + \mathop j\limits^ \wedge } \right){{Wb} \over {{m^2}}}$$
The associated electric field will be :-
$$\mathop B\limits^ \to = 1.6 \times {10^{ - 6}}\cos \left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( {2\mathop i\limits^ \wedge + \mathop j\limits^ \wedge } \right){{Wb} \over {{m^2}}}$$
The associated electric field will be :-
$$\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z - 6 \times {{10}^{15}}t} \right)\left( -2{\mathop i\limits^ \wedge + \mathop {j}\limits^ \wedge } \right){V \over m}$$
$$\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z - 6 \times {{10}^{15}}t} \right)\left( 2{\mathop i\limits^ \wedge + \mathop {j}\limits^ \wedge } \right){V \over m}$$
$$\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( {\mathop i\limits^ \wedge - \mathop {2j}\limits^ \wedge } \right){V \over m}$$
$$\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( -{\mathop i\limits^ \wedge + \mathop {2j}\limits^ \wedge } \right){V \over m}$$
Explanation
If we use that direction of light propagation will be along $$\overrightarrow E \times \overrightarrow B $$. Then (A) option is
correct.
Magnitude of E = CB
E = 3 × 108 × 1.6 × 10–6 × $$\sqrt 5 $$
E = 4.8 × $${10^{2\sqrt 5 }}$$
$$\overrightarrow E $$ 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 $$\overrightarrow E \times \overrightarrow B $$ which is $$- \widehat k$$
$$ \Rightarrow $$ $$\overrightarrow E $$ is along ($$- \widehat i + 2\widehat j$$)
Magnitude of E = CB
E = 3 × 108 × 1.6 × 10–6 × $$\sqrt 5 $$
E = 4.8 × $${10^{2\sqrt 5 }}$$
$$\overrightarrow E $$ 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 $$\overrightarrow E \times \overrightarrow B $$ which is $$- \widehat k$$
$$ \Rightarrow $$ $$\overrightarrow E $$ is along ($$- \widehat i + 2\widehat j$$)
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