# tsl319 - ∂ ∂t(F and ∂ ∂x(A ∂ 2 E y ∂t∂x =-∂...

This preview shows pages 1–4. Sign up to view the full content.

Electromagnetic Plane Wave (1) Maxwell’s equations for electric and magnetic fields in free space (no sources): Gauss’ laws: I ~ E · d ~ A = 0 , I ~ B · d ~ A = 0 . Faraday’s and Ampère’s laws: I ~ E · d ~ = - d Φ B dt , I ~ B · d ~ = μ 0 ² 0 d Φ E dt . Consider fields of particular directions and dependence on space: ~ E = E y ( x, t ) ˆ j, ~ B = B z ( x, t ) ˆ k. Gauss’ laws are then automatically satisfied. Use the cubic Gaussian surface to show that the net electric flux Φ E is zero, the net magnetic flux Φ B is zero. E y x B tsl319 – p.1/4

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Electromagnetic Plane Wave (2) Faraday’s law, I ~ E · d ~ = - d Φ B dt , applied to loop in ( x, y ) -plane becomes [ E y ( x + dx, t ) - E y ( x, t )] dy = - ∂t B z ( x, t ) dxdy ∂x E y ( x, t ) = - ∂t B z ( x, t ) (F) Ampère’s law, I ~ B · d ~ = μ 0 ² 0 d Φ E dt , applied to loop in ( x, z ) -plane becomes [ - B z ( x + dx, t ) + B z ( x, t )] dz = μ 0 ² 0 ∂t E y ( x, t ) dxdz ⇒ - ∂x B z ( x, t ) = μ 0 ² 0 ∂t E y ( x, t ) (A) z B E y x dx dz dy dx tsl320 – p.2/4
Electromagnetic Plane Wave (3) Take partial derivatives ∂x (F) and ∂t (A): 2 E y ∂x 2 = - 2 B z ∂t∂x , - 2 B z ∂t∂x = μ 0 ² 0 2 E y ∂t 2 . 2 E y ∂t 2 = c 2 2 E y ∂x 2 (E) (wave equation for electric field). Take partial derivatives

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ∂ ∂t (F) and ∂ ∂x (A): ∂ 2 E y ∂t∂x =-∂ 2 B z ∂t 2 ,-∂ 2 B z ∂x 2 = μ ² ∂ 2 E y ∂t∂x . ⇒ ∂ 2 B z ∂t 2 = c 2 ∂ 2 B z ∂x 2 (B) (wave equation for magnetic field). c = 1 √ ² μ (speed of light). Sinusoidal solution: • E y ( x, t ) = E max sin( kx-ωt ) • B z ( x, t ) = B max sin( kx-ωt ) tsl321 – p.3/4 Electromagnetic Plane Wave (4) For given wave number k the angular frequency ω is determined, for example by substitution of E max sin( kx-ωt ) into (E). For given amplitude E max the amplitude B max is determined, for example, by substituting E max sin( kx-ωt ) and B max sin( kx-ωt ) into (A) or (F). ⇒ ω k = E max B max = c. The direction of wave propagation is determind by the Poynting vector: ~ S = 1 μ ~ E × ~ B. tsl322 – p.4/4...
View Full Document

{[ snackBarMessage ]}

### Page1 / 4

tsl319 - ∂ ∂t(F and ∂ ∂x(A ∂ 2 E y ∂t∂x =-∂...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online