Question

As you work through the parts of this question you are going to show that the Maxwell equations naturally contain electromagnetic waves. In a region of space that is void of all charges and currents, ρ = 0 and J~ = 0 the Maxwell equations come out to be:

Question 1: As you work through the parts of this question you are going to show that the Maxwell equations naturally contain electromagnetic waves. In a region of space that is void of all charges and currents, p= 0 and I = 0 the Maxwell equations come out to be: V. Ē= 0 ΗχB = μοεο2E 7. B=0 7 x Ē=- a) Using the same idea as I did in the lecture, derive the Wave Equation for magnetic fields. Start by taking the curl of both sides of the Ampere-Maxwell Law. You will find the following identity useful: Āx (B ) = B (A.C) - (A:B) b) Solutions to the Wave Equation always come in the form of a function that is traveling in the direction of the wave, and one that is traveling in the opposite that direction. To illustrate this, show that the following equation for the magnetic field is a solution to this equation: B = B, cos(kx - wt) + B, cos(kx + wt) where B and B2 are constant vectors. What is the speed of these waves? c) Extending this even further, show that the following equations are also solutions to the wave equation: f=fiku - wt) + F(kx + wt) where fi and 2 are generic vector field functions. The following may be helpful with dealing with general functions such as fi and f2: a af(s) -F(f(s)) = F'(f(s)) as as This implies that any function that comes with these kinds of variable dependence are solutions to the wave equations, and are thus waves. Does this make sense? What does that tell us about the variety of shapes that a "wave" can come in?

Answer 1

a) In the region of space that lacks charges and currents, $\rho$ =0 and J=0, the Maxwell's equations can be written as,

$\bigtriangledown$.E=0 (Gauss law in electrostatics) ---------(1)

$\bigtriangledown$.B=0 (Gauss law in magnetostatics) -----------(2)

$\bigtriangledown$$\times$E=- (Faradays law) ---------(3)

ОВ at

$\bigtriangledown$$\times$B=$\mu$_o$\epsilon$_o (Ampere-Maxwell law) ------------(4)

дЕ Әt

a) Taking curl of both sides of equation (4),

$\bigtriangledown$$\times$($\bigtriangledown$$\times$B)=$\mu$_o$\epsilon$_o$\bigtriangledown$$\times$

дЕ Әt

$\bigtriangledown$($\bigtriangledown$.B)-$\bigtriangledown$²B=-$\mu$_o$\epsilon$_{o
a2B 212
(from 3)}

Using (2),$\bigtriangledown$²B=$\mu$_o$\epsilon$_{o
a2B 212
--------(5)}

where c=(1/$\mu$_o$\epsilon$_o)^1/2

_{This is the wave equation for magnetic field.}

_{b) The wave equation can be written as a function that is travelling in the direction of the wave and the other travelling in the opposite direction.}

_{Let B=B1 cos(kx-$\omega$t)+B2 cos(kx+$\omega$t)}

where B1 and B2 are constant vectors

$\bigtriangledown$²B=-k² (B1 cos(kx-$\omega$t)+B2 cos(kx+$\omega$​​​​​​​t))

$\mu$_o$\epsilon$_{o
a2B 212}
=-$\mu$_o$\epsilon$_o$\omega$²(B1 cos(kx-$\omega$t)+B2 cos(kx+$\omega$​​​​​​​t)

Subsituting in (5),

k²=$\mu$_o$\epsilon$_o$\omega$² which is true, and hence it satisfies the wave equation

Speed of the wave c=(1/$\mu$_o$\epsilon$_o)^1/2

Thus, from the above equation c=$\omega$/k which is the speed of these waves

c) Given, f=f1(kx-$\omega$t)+f2(kx+$\omega$​​​​​​​t), where f1 and f2 are generic vector field functions

Substituting this into wave equation similar to equation (5)

$\bigtriangledown$²f=$\mu$_o$\epsilon$_{o
0 f Ꭷt2}

_{We get} k²=$\mu$_o$\epsilon$_o$\omega$² , which is a solution of wave equation.

thus, any function that comes with these kinds of variable dependence are solutions to the wave equations and are thus waves. Thus, for plane waves f1 and f2 are costants. For spherical waves, they depend inversely on r. Thus, the amplitudes of f1 and f2 give the shape of the wave.