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ધારોકે એક વિદ્યુતચુંબકીય તરંગના વિદ્યુતક્ષેત્રની કંપવિસ્તાર $E_{0}=120\; N / C$ અને તેની આવૃત્તિ $v=50.0\; MHz$ છે.
$(a)$ $B_{0}, \omega, k,$ અને $\lambda .$ શોધો.
$(b)$ $E$ અને $B$ માટેના સૂત્રો શોધો.
Solution
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:
$\vec{B}=B_{0} \sin (k x-\omega t) k$
$\vec{B}=\left(4 \times 10^{-7}\right) \sin \left[1.05 x-3.14 \times 10^{8} t\right] k$