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कल्पना कीजिए कि निर्वात में एक वैध्यूतचुंबकीय तरंग का विध्यूत क्षेत्र $E =\left\{(3.1 N / C ) \cos \left[(1.8 rad / m ) y +\left(5.4 \times 10^{6} rad / s \right) t\right]\right\} \hat{ i }$ है।
$(a)$ तरंग संचरण की दिशा क्या है?
$(b)$ तरंगदैर्घ्य $\lambda$ कितनी है?
$(c)$ आवृति $v$ कितनी है?
$(d)$ तरंग के चुंबकीय क्षेत्र सदिश का आयाम कितना है?
$(e)$ तरंग के चुंबकीय क्षेत्र के लिए व्यंजक लिखिए।
Solution
$(a)$ From the given electric field vector, it can be inferred that the electric field is directed along the negative $x$ direction. Hence, the direction of motion is along the negative y direction i.e., – $j$
$(b)$ It is given that, $\vec{E}=3.1 N / C \cos \left[(1.8 rad / m ) y+\left(5.4 \times 10^{8} rad / s \right) t\right] \hat{i}\dots(i)$
The general equation for the electric field vector in the positive x direction can be written
as:
$\vec{E}=E_{0} \sin (k x-\omega t) \hat{i}\ldots(ii)$
On comparing equations $(i)$ and $(ii)$, we get Electric field amplitude,
$E _{0}=3.1 N / C$
Angular frequency, $\omega=5.4 \times 10^{8} rad / s$
Wave number, $k =1.8 rad / m$
Wavelength, $\lambda=\frac{2 \pi}{1.8}=3.490 m$
$(c)$ Frequency of wave is given as
$v=\frac{\omega}{2 \pi}$
$=\frac{5.4 \times 10^{8}}{2 \pi}=8.6 \times 10^{7} Hz$
$(d)$ Magnetic field strength is given as
$B_{0}=\frac{E_{0}}{c}$
Where,
$c=$ Speed of light $=3 \times 10^{8} m / s$
$\therefore B_{0}=\frac{3.1}{3 \times 10^{8}}=1.03 \times 10^{-7} T$
$(e)$ On observing the given vector field, it can be observed that the magnetic field vector is directed along the negative $z$ direction. Hence, the general equation for the magnetic field vector is written as:
$\vec{B}-B_{0} \cos (k y+\omega t) k$
$=\left\{\left(1.03 \times 10^{-7} T\right) \cos \left[(1.8 r a d / m) y+\left(5.4 \times 10^{6} r a d / s\right) t\right]\right\} k$
