The clectric field for a sinusoidal electromagnetic wave is with E.-139 V/m and

Question

The clectric field for a sinusoidal electromagnetic wave is with E.-139 V/m and k-1.49E+8 m elect the direction of propagation for this wave. Submt Answer Tries 0/100 Find the wave number of the wave. Submit Answer Tries 0/100 Find the wavelength of the wave. Submit Answer Tries 0/100 Find the p of the wave. Submit Answer Tries o/100 Find the frequency of the wave. Submit Answer Tries 0/100 Find the angular frequency of the wave. Submit Answer Tries 0/100 d of the wave Submil Answer Tries o/100 Find th Find the amplitude of the magnetic field. Submt Answer Tries o/100 For sinusoidal electromagnetic waves, the magnetic field and the electric field are in phase, so they both have the same trigonometeric function assocoiated with them (no phase shift). For the wave described above, which unit vector will be associated with the magnetie field?

Explanation / Answer

a) direction propogation of wave: towards -y axis

b) wave number, k = 1.49*10^8 m^-1

c) we know, k = 2*pi/lamda

==> lamda = 2*pi/k

= 2*pi/(1.49*10^8)

= 4.22*10^-8 m or 42.2 nm

d) use, c = lamda*f

1/f = lamda/c

T = lamda/c

= 4.22*10^-8/(3*10^8)

= 1.407*10^-16 s

e) f = 1/T

= 1/(1.407*10^-16)

= 7.11*10^15 Hz

f) angular frequency, w = 2*pi*f

= 2*pi*7.11*10^15

= 4.47*10^16 rad/s

g) speed of the wave, c = 3*10^8 m/s

h) Amplitude of magnetic field, Bmax = Emax/c

= 139/(3*10^8)

= 4.63*10^-7 T

i) B = Bmax*sin(k*y + w*t) k

= 4.63*10^-7*sin(1.49*10^8*y + 4.47*10^16*t) k

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.