NEET UG PHYSICS MCQ: ELECTROMAGNETIC WAVES

1 Hard

1 of 1

Application +4 | -1

Assume that all the energy from a

1000 \mathrm{~W}

lamp is radiated uniformly, then the amplitude of electric field of radiation at a distance of

2 \mathrm{~m}

from the lamp is

View answer and explanation

Correct option: A

245.01 \mathrm{~V} / \mathrm{m}

Option A (

245.01 \mathrm{~V} / \mathrm{m}

) is correct because it is calculated using the relationship between power, intensity, and electric field amplitude for electromagnetic waves.

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{E_X}

{E_R}

{E_M}

60\,\,\frac{V}{m}

{E_1} = {E_{01}}\hat x\cos \left[ {2\pi v\left( {\frac{z}{c} - t} \right)} \right]

{E_2} = {E_{02}}\hat x\cos \left[ {k\left( {2z - ct} \right)} \right]

{\varepsilon _{{r_1}}}

{\varepsilon _{{r_2}}}

E = 50\sin \,(500x - 10 \times {10^{10}}t)\,v/m.

c

\mathrm{\varepsilon }

\mathrm{\mu }

\sqrt {\frac{\mathrm{\mu }}{\mathrm{\varepsilon }}}