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An electromagnetic wave travelling in the x-direction has frequency of $ 2\times {{10}^{14}}Hz $ and electric field amplitude of $ 27V{{m}^{-1}} $ , from the options below, which one describes the magnetic field for this
(A) $ \overrightarrow{B}\left( x,t \right)=\left( 3\times {{10}^{-8}}\text{ T} \right)\hat{j}\text{ }\sin \left[ 2\pi \left( 1\cdot 5\times {{10}^{-8}}x-2\times {{10}^{14}}t \right) \right] $
(B) $ \overrightarrow{B}\left( x,t \right)=\left( 9\times {{10}^{-8}}T \right)\hat{j}\text{ }\sin \left[ \left( 1\cdot 5\times {{10}^{-6}}x-2\times {{10}^{14}}t \right) \right] $
(C) $ \overrightarrow{B}\left( x,t \right)=\left( 9\times {{10}^{-8}}T \right)\hat{k}\text{ }\sin \left[ 2\pi \left( 0\cdot 66\times {{10}^{6}}x-2\times {{10}^{14}}t \right) \right] $
(D) $ \overrightarrow{B}\left( x,t \right)=\left( 9\times {{10}^{-8}}T \right)\hat{i}\text{ }\sin \left[ 2\pi \left( 1\cdot 5\times {{10}^{-8}}x-2x\times {{10}^{14}}t \right) \right] $

seo-qna
Last updated date: 13th Jun 2024
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Answer
VerifiedVerified
373.8k+ views
Hint: Wave function for a given plane electromagnetic wave is written as
 $ \text{B=Bo}\sin \left( kx-wt \right) $
In case of plane electromagnetic waves, direction of propagation is along the cross product $ \overrightarrow{E}\times \overrightarrow{B} $ . Hence, if direction of propagation is along x axis then $ \overrightarrow{E} $ is along y-axis and $ \overrightarrow{B} $ is along Z axis. From this, the direction of the magnetic field is found.

Complete step by step solution
We know that
 $ \text{Bo}=\dfrac{\text{Eo}}{\text{c}} $
Now,
 $ \text{Eo=27} $ …. Given
 $ \text{c}=3\times {{10}^{8}} $
Hence
 $ \text{Bo}=\dfrac{27}{3\times {{10}^{8}}}=9\times {{10}^{-8}} $
As the direction of propagation is along the x-axis, therefore $ \text{Bo} $ is along the z axis.
So $ \overrightarrow{\text{Bo}}=\left( 9\times {{10}^{-8}} \right)\hat{k} $
Now $ k=\dfrac{2\pi }{\lambda } $ and $ w=\dfrac{2\pi }{T} $
So putting these values in equation
 $ \text{B=Bo }\sin \left( kx-wt \right) $
Now
 $ \begin{align}
  & \lambda =\dfrac{c}{v} \\
 & =\dfrac{3\times {{10}^{8}}}{2\times {{10}^{14}}}=1\cdot 5\times {{10}^{-6}}m \\
\end{align} $
And
 $ T=\dfrac{1}{v}=\dfrac{1}{2\times {{10}^{14}}}=0\cdot 5\times {{10}^{-14}}\text{ sec} $
Putting these values in equation (1), we get
 $ \begin{align}
  & \text{B}=\left( 9\times {{10}^{-8}} \right)\hat{k}\text{ sin}\left[ 2\pi \left( \dfrac{x}{1\cdot 5\times {{10}^{-6}}}-\dfrac{t}{0\cdot 5\times {{10}^{14}}} \right) \right] \\
 & \text{B=}\left( 9\times {{10}^{-8}} \right)\hat{k}\text{ }\sin \left[ 2\pi \left( 0\cdot 666\times {{10}^{6}}x-2\times {{10}^{14}}t \right) \right] \\
\end{align} $ .

Note
 $ E_o $ and $ B_o $ are the maximum magnitude of electric and magnetic fields in case of an electromagnetic wave.
Number of cycles of electric field or magnetic field completed in one second is called frequency of electromagnetic wave.
For mathematical representation, we define angular frequency by multiplying frequency with $ 2\pi $
 $ w=2\pi v $
Time period of electromagnetic oscillation can be written as:
 $ T=\dfrac{1}{v} $
Distance travelled by wave in one time period is called wavelength and is represented by $ \lambda $
 $ \begin{align}
  & \lambda =CT \\
 & \lambda =\dfrac{C}{V} \\
\end{align} $ .