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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Electromagnetic Waves
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A plane electromagnetic wave of frequency $25\: MHz$ travels in free space along the $x$ - direction. At a particular point in space and time, $E= 6.3 \: jV/m$. What is B at this point?

$\begin {array} {1 1} (a)\;2.1 \times 10^{-6}i\: T & \quad (b)\;0.21 \times 10^{-8}i\: T \\ (c)\;2.1 \times 10^{-8}k\: T & \quad (d)\;0.21 \times 10^{-8}j\: T \end {array}$

 

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1 Answer

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the magnitude of $B$ is
$B=\large\frac{E}{c}$$ = \large\frac{6.3}{3 \times 10^8}$$ = 2.1 \times 10^{-8} T$
To find the direction, we note that $E$ is along $y$ - direction and the wave propagates along $x$ - axis.
Therefore, $B$ should be in a direction perpendicular to both $x$ - and $y$ - axes.
Using vector algebra,$ E\times B$ should be along $x$ - direction.
Since, $(+j) \times (+k) = i$ , so, $B$ is along +k or the $z$ - direction.
Ans : (c)

 

answered Feb 27, 2014 by thanvigandhi_1
edited Oct 8, 2014 by thagee.vedartham
 

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