Answer to Question #88565 in Electricity and Magnetism for Priyanshu

Question

A uniform plane wave of 10 kHz travelling in free space strikes a large block of a material having ε=9ε_0,μ=4μ_0 and σ=0 normal to the surface.If the incident magnetic field vector is given by B→=10^(-3) sin(ωt-βy)k^ tesla
Write the complete expression for the incident,reflected and transmitted field vectors.
Please explain in detail.

Answer 1

The incident wave E_i has the magnitude

$$E_0=\frac {B_0} {\sqrt{\mu_0 ɛ_0} }= 10^{-3} \times3 \times10^8= 3 \times 10^5 (1)$$

$$β=\frac {ω} {c}=\frac {2\pi\times10 \times10^3} {c}=\frac {\pi} {150} (2)$$

The incident wave fields:

$$E_i=3 \times 10^5 \times {\cos (2\pi \times 10^4- \frac {\pi} {150} y)} \widehat{-x}$$

We assume that the incident electric field is reflected with a reflection coefficient Γ and transmitted with a transmitted with a transmission coefficient τ. That implies that if the electric field intensity of the incident, reflected and transmitted waves at the boundary (z = 0) are E_i0, E_r0 and E_t0 respectively, then E_r0= rE_i0 and E_t0= t E_i0

$$r= \frac {k_2-k_1} { k_2+k_1} (3)$$ $$k_1=\sqrt{\frac {\mu_0} { ɛ_0}}=120\pi$$

$$k_2=\sqrt{\frac {\mu_2} { ɛ_2}}=180\pi$$

We got: r=0.2

$$t=\frac {2k_2} {k_2+k_1}$$

We got: t =1.2

Reflected vector

$$E_r=0.6 \times 10^5 \times \cos{(2\pi \times 10^4 t - \frac {\pi} {150} y)} \widehat{x}$$

Transmitted vector

$$E_t =3.6 \times 10^5 \times \cos{(2\pi \times 10^4 - \frac {\pi} {150} y)} \widehat{-x}$$

Answer:

The incident vector

$$E_i=3 \times 10^5 \times {\cos (2\pi \times 10^4- \frac {\pi} {150} y)} \widehat{-x}$$

Reflected vector

$$E_r=0.6 \times 10^5 \times \cos{(2\pi \times 10^4 t - \frac {\pi} {150} y)} \widehat{x}$$

Transmitted vector

$$E_t =3.6 \times 10^5 \times \cos{(2\pi \times 10^4 - \frac {\pi} {150} y)} \widehat{-x}$$