JEE MAIN - Physics (2019 - 12th April Evening Slot - No. 9)
A plane electromagnetic wave having a frequency v = 23.9 GHz propagates along the positive z-direction in
free space. The peak value of the Electric Field is 60 V/m. Which among the following is the acceptable
magnetic field component in the electromagnetic wave ?
$$\overrightarrow B $$ = 2 × 10–7
sin(1.5 × 102
x + 0.5 × 1011t) $$\widehat j$$
$$\overrightarrow B $$ = 60
sin(0.5 × 103x + 0.5 × 1011t) $$\widehat k$$
$$\overrightarrow B $$ = 2 × 10–7
sin(0.5 × 103
z + 1.5 × 1011t) $$\widehat i$$
$$\overrightarrow B $$ = 2 × 10–7
sin(0.5 × 103
z - 1.5 × 1011t) $$\widehat i$$
Explanation
Since the wave is propagating in positive z-direction
So acceptable magnetic field component will be
$$\overrightarrow B = {B_0}\sin \left( {kz - \omega t} \right)$$$$\widehat i$$
$$ \because $$ C = $${{{E_o}} \over {{B_o}}}$$
$$ \Rightarrow $$ $${B_o} = {{{E_o}} \over C}$$ = $${{60} \over {3 \times {{10}^8}}}$$ = 2 $$ \times $$ 10-7
and $$\omega = 2\pi f$$ = 2$$\pi $$$$ \times $$23.9$$ \times $$109 = 1.5 × 1011 Hz
and k = $${\omega \over c}$$ = $${{1.5 \times {{10}^{11}}} \over {3 \times {{10}^8}}}$$ = 0.5 × 103
Note : When wave is propagating in positive z-direction then sign of kz and $${\omega t}$$ should be opposite.
From option you can see only option D can be correct.
So acceptable magnetic field component will be
$$\overrightarrow B = {B_0}\sin \left( {kz - \omega t} \right)$$$$\widehat i$$
$$ \because $$ C = $${{{E_o}} \over {{B_o}}}$$
$$ \Rightarrow $$ $${B_o} = {{{E_o}} \over C}$$ = $${{60} \over {3 \times {{10}^8}}}$$ = 2 $$ \times $$ 10-7
and $$\omega = 2\pi f$$ = 2$$\pi $$$$ \times $$23.9$$ \times $$109 = 1.5 × 1011 Hz
and k = $${\omega \over c}$$ = $${{1.5 \times {{10}^{11}}} \over {3 \times {{10}^8}}}$$ = 0.5 × 103
Note : When wave is propagating in positive z-direction then sign of kz and $${\omega t}$$ should be opposite.
From option you can see only option D can be correct.
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