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  • A plane electromagnetic wave travels in free space along - direction. If the value of <img alt="\math

A plane electromagnetic wave travels in free space along \mathrm{X} - direction. If the value of \mathrm{\overrightarrow{B}} (in tesla) at a particular point in space and time is \mathrm{1.2 \times 10^{-8} \hat{\mathrm{k}}}, the value of \mathrm{\overrightarrow{\mathrm{E}}\left(\mathrm{inVm}^{-1}\right)} at that point is:
 

Option: 1

1.2 \hat{\mathrm{j}}

 


Option: 2

3.6 \hat{\mathrm{k}}
 


Option: 3

1.2 \hat{\mathrm{k}}
 


Option: 4

3.6 \hat{\mathrm{j}}


Answers (1)

best_answer

Here, \mathrm{\overrightarrow{\mathrm{B}}=1.2 \times 10^{-8} \hat{\mathrm{k}} \mathrm{T}}
The magnitude of \mathrm{\overrightarrow{\mathrm{E}}} is

\mathrm{ \mathrm{E}=\mathrm{Bc}=\left(1.2 \times 10^{-8} \mathrm{~T}\right)\left(3 \times 10^8 \mathrm{~ms}^{-1}\right)=3.6 \mathrm{Vm}^{-1} }

\mathrm{ \overrightarrow{\mathrm{B}} \text{is along} \mathrm{Z} } - direction and the wave propagates along  \mathrm{X} - direction. Therefore \mathrm{\overrightarrow{\mathrm{E}}} should be in a direction perpendicular to both \mathrm{X \: and \: Z} axes. Using vector algebra \mathrm{\vec{E} \times \vec{B}}should be along \mathrm{X} - direction.

Since \mathrm{(+\hat{\mathbf{j}}) \times(+\hat{\mathbf{k}})=\hat{\mathbf{i}}, \overrightarrow{\mathrm{E}} \text{ is along the} \mathrm{Y}-} direction.

Thus, \mathrm{\overrightarrow{\mathrm{E}}=3.6 \hat{\mathrm{j}} \mathrm{Vm}^{-1}}

Hence option 4 is correct.

Posted by

seema garhwal

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