The oscillating electric field of an electromagnetic wave is given by $E_y = 30\sin\left(2\times10^{11}\,t + 300\pi\,x\right)\ \text{V m}^{-1}$. (a) Obtain the wavelength of the electromagnetic wave. (b) Write down the expression for the oscillating magnetic field.

(a) $\lambda = \dfrac{2}{300} \approx 6.7\times10^{-3}$ m. (b) $B_z = 10^{-7}\sin\left(2\times10^{11}\,t + 300\pi\,x\right)$ T.

CBSE Class 12 · 2008 · EM WavesShort Answer

The oscillating electric field of an electromagnetic wave is given by

E_{y} = 30 sin (2 \times 1 0^{11} t + 300 π x) V m^{- 1}

.
(a) Obtain the wavelength of the electromagnetic wave.
(b) Write down the expression for the oscillating magnetic field.

Compare with

E_{y} = E_{0} sin (ω t + k x)

E_{0} = 30

V m

^{- 1}

ω = 2 \times 1 0^{11}

rad s

^{- 1}

and

k = 300 π

^{- 1}

.
(a) Since

k = 2 π / λ

λ = \frac{2 π}{k} = \frac{2 π}{300 π} = \frac{2}{300} \approx 6.7 \times 1 0^{- 3} m

(b) The magnetic field amplitude is

B_{0} = \frac{E _{0}}{c} = \frac{30}{3 \times 1 0 ^{8}} = 1 \times 1 0^{- 7} T

The magnetic field oscillates along z, in phase with

E_{y}

B_{z} = 1 0^{- 7} sin (2 \times 1 0^{11} t + 300 π x) T

λ \approx 6.7 \times 1 0^{- 3} m, B_{0} = 1 0^{- 7} T

CBSE Class 12 Board · 2008 · Subjective