कल्पना कीजिए कि एक वैध्युतचुंबकीय तरंग के विध्युत क्षेत्र का आयाम $E_{o}=120 N / C$ है तथा इसकी आवृत्ति $v=50.0 \,MHz$ है। $(a)$ $B_{0}, \omega, k$ तथा $\lambda$ ज्ञात कीजिए, $(b)$ $E$ तथा $B$ के लिए व्यंजक प्राप्त कीजिए।

Electric field amplitude, $E _{0}=120 N / C$

Frequency of source, $v=50.0 MHz =50 \times 10^{6} Hz$

Speed of light, $c=3 \times 10^{8} m / s$

$(a)$ Magnitude of magnetic field strength is given as:

$B_{0}=\frac{E_{0}}{c}$

$=\frac{120}{3 \times 10^{8}}$

$=4 \times 10^{-7} T=400 nT$

Angular frequency of source is given as:

$\omega=2 \pi \nu=2 \pi \times 50 \times 10^{6}$

$=3.14 \times 10^{8} rad / s$

Propagation constant is given as:

$k=\frac{\omega}{c}$

$=\frac{3.14 \times 10^{8}}{3 \times 10^{8}}=1.05 rad / m$

Wavelength of wave is given as:

$\lambda=\frac{c}{v}$

$=\frac{3 \times 10^{8}}{50 \times 10^{6}}=6.0 m$

$(b)$ Suppose the wave is propagating in the positive $x$ direction. Then, the electric field vector will be in the positive y direction and the magnetic field vector will be in the positive z direction. This is because all three vectors are mutually perpendicular. Equation of electric field vector is given as:

$\vec{E}=E_{0} \sin (k x-\omega t) j$

$=120 \sin \left[1.05 x-3.14 \times 10^{8} t\right] j$

And, magnetic field vector is given as:

$\vec{B}=B_{0} \sin (k x-\omega t) k$

$\vec{B}=\left(4 \times 10^{-7}\right) \sin \left[1.05 x-3.14 \times 10^{8} t\right] k$