NEET UG PHYSICS MCQ: ELECTROMAGNETIC WAVES

1 Easy

1 of 1

Application +4 | -1

In a place e.m. wave, the electric field oscillates sinusoidally at a frequency of

2.5 \times 10^{10} {~Hz}

and amplitude

480 {~V} / {m}

. The amplitude of oscillating magnetic field will be :-

View answer and explanation

Correct option: C

1.6 \times 10^{-6} {~Wb} / {m}^2

Option C (

1.6 \times 10^{-6} {~Wb} / {m}^2

) is correct because the amplitude of the oscillating magnetic field is calculated using the formula

B_0 = \frac{E_0}{c}

, where c is the speed of light.

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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 }}}