JEE MAIN - Physics (2020 - 3rd September Morning Slot - No. 21)
The magnetic field of a plane electromagnetic wave is
$$\overrightarrow B = 3 \times {10^{ - 8}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat i$$ T
where c = 3 $$ \times $$ 108 ms–1 is the speed of light. The corresponding electric field is :
$$\overrightarrow B = 3 \times {10^{ - 8}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat i$$ T
where c = 3 $$ \times $$ 108 ms–1 is the speed of light. The corresponding electric field is :
$$\overrightarrow E = - {10^{ - 6}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$$ V/m
$$\overrightarrow E = - 9\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$$ V/m
$$\overrightarrow E = 9\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$$ V/m
$$\overrightarrow E = 3 \times {10^{ - 8}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$$
Explanation
Given, $$\overrightarrow B = 3 \times {10^{ - 8}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat i$$ T
$$ \therefore $$ E0 = CB0
= 3 × 108 × 3 × 10–8 = 9 V/m
We know, $$\left( {\overrightarrow E \times \overrightarrow B } \right)||\overrightarrow C $$
And here $$\widehat B = \widehat i\& \widehat C = - \widehat j$$
$$ \therefore $$ $$\widehat E = - \widehat k$$
So, $$\overrightarrow E = - 9\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$$ V/m
$$ \therefore $$ E0 = CB0
= 3 × 108 × 3 × 10–8 = 9 V/m
We know, $$\left( {\overrightarrow E \times \overrightarrow B } \right)||\overrightarrow C $$
And here $$\widehat B = \widehat i\& \widehat C = - \widehat j$$
$$ \therefore $$ $$\widehat E = - \widehat k$$
So, $$\overrightarrow E = - 9\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$$ V/m
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