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Questions  >>  CBSE XII  >>  Physics  >>  Electromagnetic waves
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Suppose that the electric field amplitude of an electromagnetic wave is $E_0 = 120 N/C$ and its frequency is $\gamma = 50.0 \;MHz$ . Determine $B_0, w,k$ and $\lambda$

$\begin{array}{1 1} (i) B_0 = 4 \times 10^{-7}T (ii) w= 3.14 \times 10^8\;rads^{-1}(iii) k=1.05\;rad\;m^{-1} (iv) \lambda= 6\;m \\ (i) B_0 = 2 \times 10^{-7}T (ii) w= 1.57 \times 10^8\;rads^{-1}(iii) k=10.5\;rad\;m^{-1} (iv) \lambda= 6\;m \\(i) B_0 = 4 \times 10^{-7}T (ii) w= 3.14 \times 10^8\;rads^{-1}(iii) k=1.05\;rad\;m^{-1} (iv) \lambda= 3\;m\\ (i) B_0 = 4 \times 10^{-7}T (ii) w= 3.14 \times 10^8\;rads^{-1}(iii) k=1.05\;rad\;m^{-1} (iv) \lambda= 3\;m \end{array} $

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A)
Solution :
Given $E_0=120 N/C$
$\alpha = 50\;MHz$
$C= 3 \times 10^8\;ms^{-1}$
(i) $B_0 = \large\frac{E_0}{C}=\frac{120}{3 \times 10^8}$$ = 4 \times 10^7\;T$
(ii) $w= 2 \pi \alpha = 2 \times 3.14 \times (50 \times 10^6)$
$\qquad= 1.05\;rad m^{-1}$
(iii) $k= \large\frac{w}{c}=\frac{3.14 \times 10^8}{3 \times 10^8}$$ =1.05 rad\;m^{-1}$
(iv) $\lambda = \large\frac{c}{\lambda}=\frac{3 \times 10^8}{50 \times 10^6}$$=6\;m$
$ (i) B_0 = 4 \times 10^{-7}T (ii) w= 3.14 \times 10^8\;rads^{-1}(iii) k=1.05\;rad\;m^{-1} (iv) \lambda= 6\;m$
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