ધારોકે શૂન્યાવકાશમાં રહેલ વિદ્યુતચુંબકીય તરંગનું વિદ્યુતક્ષેત્ર,

$E =\left\{(3.1 \;N / C ) \text { cos }\left[(1.8 \;rad / m ) y+\left(5.4 \times 10^{6} \;rad / s \right) t\right]\right\} \hat{ i }$ છે.

$(a)$ પ્રસરણ દિશા કઈ છે ?

$(b)$ તરંગલંબાઈ કેટલી છે ?

$(c)$ આવૃત્તિ $v$ કેટલી છે ?

$(d)$ તરંગના ચુંબકીય ક્ષેત્રની કંપવિસ્તાર કેટલો છે?

$(e)$ તરંગના ચુંબકીયક્ષેત્ર માટેનું સમીકરણ લખો.

$(a)$ 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:

$\vec{B}-B_{0} \cos (k y+\omega t) k$

$=\left\{\left(1.03 \times 10^{-7} T\right) \cos \left[(1.8 r a d / m) y+\left(5.4 \times 10^{6} r a d / s\right) t\right]\right\} k$