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  •  The magnetic field of a plane electromagnetic wave is given by :  <img alt="\overrightarrow{\mathrm{B}}=2 \times 10^{-8} \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} \math

 The magnetic field of a plane electromagnetic wave is given by : 

\overrightarrow{\mathrm{B}}=2 \times 10^{-8} \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} \mathrm{t}\right) \hat{j} \mathrm{~T}

 The amplitude of the electric field would be : 

Option: 1

\mathrm{6 \; \mathrm{Vm}^{-1} \text { along } x \text {-axis }}


Option: 2

\mathrm{3 \: \mathrm{Vm}^{-1} \text { along } z \text {-axis }}


Option: 3

\mathrm{6 \: \mathrm{Vm}^{-1} \text { along } z \text {-axis }}


Option: 4

\mathrm{2 \times 10^{-8} \: \mathrm{Vm}^{-1} \text { along } z \text {-axis }}


Answers (1)

best_answer

\mathrm{\bar{B}=\left(2 \times 10^{-8}\right) \sin \left(0.5 \times 10^3 x+1.5 \times 10^{11} t\right) \hat{\jmath}}

We know that,

\mathrm{\frac{E_0}{B_0}=C}

\mathrm{Eo\: \& \: B_0\: are \: the\: amplitude \: of \: electric\: field \: and \: magntic\: field}

\mathrm{E_0=C \times B_0 =3 \times 10^8 \times 2 \times 10^{-8} }

\mathrm{E =6 \mathrm{~V} / \mathrm{m}}

From the eqn of \mathrm{}\bar{B},  it can be said that the propaget of wave is taking place along negative x-axis

Also, \mathrm{\bar{E} \times \bar{B}} points along the direction of propagation

                

Hence (3) is correct option.

Posted by

Shailly goel

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