JEE Main 2020 (Online) 4th September Evening Slot | Electromagnetic Waves Question 97 | Physics

JEE Main 2020 (Online) 4th September Evening Slot

MCQ (Single Correct Answer)

-1

The electric field of a plane electromagnetic wave is given by
$$\overrightarrow E = {E_0}\left( {\widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)$$
Its magnetic field will be given by :

$${{{E_0}} \over c}\left( {\widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)$$

$${{{E_0}} \over c}\left( {\widehat x - \widehat y} \right)\sin \left( {kz - \omega t} \right)$$

$${{{E_0}} \over c}\left( {\widehat x - \widehat y} \right)\cos \left( {kz - \omega t} \right)$$

$${{{E_0}} \over c}\left( { - \widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)$$

JEE Main 2020 (Online) 4th September Morning Slot

MCQ (Single Correct Answer)

-1

Choose the correct option relating wave lengths of different parts of electromagnetic wave spectrum:

$$\lambda $$_{radio waves} > $$\lambda $$_{micro waves} > $$\lambda $$_visible > $$\lambda $$_x-rays

$$\lambda $$_visible > $$\lambda $$_x-rays > $$\lambda $$_{radio waves} > $$\lambda $$_{micro waves}

$$\lambda $$_visible < $$\lambda $$_{micro waves} < $$\lambda $$_{radio waves} < $$\lambda $$_x-rays

$$\lambda $$_x-rays < $$\lambda $$_{micro waves} < $$\lambda $$_{radio waves} < $$\lambda $$_visible

JEE Main 2020 (Online) 3rd September Evening Slot

MCQ (Single Correct Answer)

-1

The electric field of a plane electromagnetic wave propagating along the x direction in vacuum is
$$\overrightarrow E = {E_0}\widehat j\cos \left( {\omega t - kx} \right)$$.
The magnetic field $$\overrightarrow B $$ , at the moment t = 0 is :

$$\overrightarrow B = {{{E_0}} \over {\sqrt {{\mu _0}{ \in _0}} }}\cos \left( {kx} \right)\widehat j$$

$$\overrightarrow B = {{{E_0}} \over {\sqrt {{\mu _0}{ \in _0}} }}\cos \left( {kx} \right)\widehat k$$

$$\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos \left( {kx} \right)\widehat k$$

$$\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos \left( {kx} \right)\widehat j$$

JEE Main 2020 (Online) 3rd September Morning Slot

MCQ (Single Correct Answer)

-1

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 :

$$\overrightarrow E = - {10^{ - 6}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$$ 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

$$\overrightarrow E = 3 \times {10^{ - 8}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$$

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