JEE MAIN - Physics (2019 - 10th January Evening Slot - No. 19)
The electric field of a plane polarized electromagnetic wave in free space at time t = 0 is given by an expression $$\overrightarrow E \left( {x,y} \right) = 10\widehat j\cos \left[ {\left( {6x + 8z} \right)} \right].$$ The magnetic field $$\overrightarrow B $$(x,z, t) is given by $$-$$ (c is the velocity of light)
$${1 \over c}\left( {6\hat k + 8\widehat i} \right)\cos \left[ {\left( {6x + 8z - 10ct} \right)} \right]$$
$${1 \over c}\left( {6\widehat k - 8\widehat i} \right)\cos \left[ {\left( {6x + 8z - 10ct} \right)} \right]$$
$${1 \over c}\left( {6\hat k + 8\widehat i} \right)\cos \left[ {\left( {6x - 8z + 10ct} \right)} \right]$$
$${1 \over c}\left( {6\hat k - 8\widehat i} \right)\cos \left[ {\left( {6x - 8z + 10ct} \right)} \right]$$
Explanation
$$\overrightarrow E = 10\widehat j\cos \left[ {\left( {6\widehat i + 8\widehat k} \right).\left( {x\widehat i + z\widehat k} \right)} \right]$$
$$ = 10\widehat j\,\cos \left[ {\overrightarrow K .\overrightarrow r } \right]$$
$$ \therefore $$ $$\overrightarrow K = 6\widehat i + 8\widehat k;$$ direction of waves travel.
i.e., direction of 'c'.
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$$ \therefore $$ Direction of $$\widehat B$$ will be along
$$\widehat C \times \widehat E = {{ - 4\widehat i + 3\widehat k} \over 5}$$
Mag. of $$\overrightarrow B $$ will be along
$$\widehat C \times \widehat E = {{ - 4\widehat i + 3\widehat k} \over 5}$$
Mag. of $$\overrightarrow B = {E \over C} = {{10} \over C}$$
$$ \therefore $$ $$\overrightarrow B = {{10} \over C}\left( {{{ - 4\widehat i + 3\widehat k} \over 5}} \right) = {{\left( { - 8\widehat i + 6\widehat k} \right)} \over C}$$
$$ = 10\widehat j\,\cos \left[ {\overrightarrow K .\overrightarrow r } \right]$$
$$ \therefore $$ $$\overrightarrow K = 6\widehat i + 8\widehat k;$$ direction of waves travel.
i.e., direction of 'c'.
$$ \therefore $$ Direction of $$\widehat B$$ will be along
$$\widehat C \times \widehat E = {{ - 4\widehat i + 3\widehat k} \over 5}$$
Mag. of $$\overrightarrow B $$ will be along
$$\widehat C \times \widehat E = {{ - 4\widehat i + 3\widehat k} \over 5}$$
Mag. of $$\overrightarrow B = {E \over C} = {{10} \over C}$$
$$ \therefore $$ $$\overrightarrow B = {{10} \over C}\left( {{{ - 4\widehat i + 3\widehat k} \over 5}} \right) = {{\left( { - 8\widehat i + 6\widehat k} \right)} \over C}$$
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