NEET UG PHYSICS MCQ: ELECTROMAGNETIC WAVES

1 Easy

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Application +4 | -1

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of

5 \times 10^{10} \mathrm{~Hz}

and an amplitude of

50 \mathrm{Vm}^{-1}

. The total average energy density of the electromagnetic field of the wave is : [Use

\varepsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2

]

View answer and explanation

Correct option: C

1.106 \times 10^{-8} \mathrm{Jm}^{-3}

Option C (

1.106 \times 10^{-8} \mathrm{Jm}^{-3}

) is correct because it is the total average energy density calculated using the formula

u = \varepsilon_0 E_0^2 / 2

More Questions from 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 }}}