{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Phys 2214 hw6 Solutions

Phys 2214 hw6 Solutions - Phys 2214 Homework#6 Solutions 1...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Phys 2214 Homework #6 Solutions October 26, 2011 1. EM Waves in Vacuum Given Gauss’ law, ∇· ~ E = ρ/ 0 , assume ρ = 0, since we’re in a vacuum. Also, we take the plane wave to be traveling in the + x direction, so ~ E = f ( x - ct x + g ( x - ct y + h ( x - ct z . Note that ~ E can have an x -component and still be a plane wave. But, applying Gauss’ law in vacuum, we get ∇ · ~ E = 0 = ∂f ∂x + ∂g ∂x + ∂h ∂x = ∂f ∂x + 0 + 0 . So, ∂f/∂x = 0 and therefore f = constant. But f = E x = constant is not a wave. So, E x = 0. So, there is no component of ~ E in the direction of propagation and the wave must be transverse. 2. Faraday’s law and EM waves Given Faraday’s law, ∇ × ~ E = - ~ B/∂t , with ~ B = B y ˆ y ; then - ∂B y ∂t ˆ y = ∇ × ~ E = ˆ x ˆ y ˆ z ∂x ∂y ∂z E x E y E z Then - ∂B y ∂t = ˆ x ∂E z ∂y - ∂E y ∂z ! + ˆ y ∂E x ∂z - ∂E z ∂x ! + ˆ z ∂E y ∂x - ∂E x ∂y ! . Given ~ E = f ( x - ct x + g ( x - ct y + h ( x - ct z , 1
Background image of page 1

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

View Full Document Right Arrow Icon
then ∂E z ∂y = ∂E y ∂z = 0 , ∂E x ∂z = ∂E x ∂y = 0 , and ∂E y ∂x , ∂E z ∂x 6 = 0 .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}