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Questions  >>  CBSE XII  >>  Physics  >>  Electromagnetic waves
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Q)

The magnetic field of a beam emerging from a filter facing a flood light is given by $B = 12 \times 10^{-8} \sin (1.20 \times 10^7\;z -3.60 \times 10^{15}t)T$ What is the average intensity of the beam ?

$\begin{array}{1 1} 1.71 \times 10^{-2} w/m^2 \\ 17.1\;w/m^2 \\ 1.71\;w/m^2 \\ 0.171\;w/m^2 \end{array} $

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A)
Solution :
given $B= 12 \times 10^{-8} \sin (1.20 \times 10^7 z - 3.6 \times 10^{15})T$
$B= B_0 \sin (kz-wt)$
Comparing both the equation we get ,
$B_0 = 12 \times 10^{-8}T$
$I_{av}= \large\frac{1}{2} \frac{B_0^2}{\mu_0}$$c$
$\qquad = \large\frac{1}{2} \times \frac{12 \times 10^{-8}}{4 \pi \times 10^{-7}}$
$\qquad= 1.71\;w/m^2$
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