Question

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 J = 0 the Maxwell equations come out to be: Y E = 0 7.B=0 7 x Ē = -1 1 x B = Poco 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 x c) = B(A.C) - (A•B) C 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) + B2 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 = fi(kx – wt) + F2(kx + wt) where fi and f2 are generic vector field functions. The following may be helpful with dealing with general functions such as fi and 52: af(s) OF(f() = F"(f(3) 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

(4) The Maxwell's equations in charge free space (p=0) Solution are 51 E = lu 7 3 DXĒ=-25 ЭЕ at JXB = M Eo at at a using at at Now taking curl of Ampere's law on both sides Ex 15 x B) = Moco 2lə xi XE) (9. B)-(7.5):- Mo Eo 3 3 (-3 Third equation 0-p?ő = MoEo 2²5 at [88- 10 2² wave equation at2 for magnetic field (B) for a one-dimensional propagation along equation 조 [using second equation 07 2 x-direction the wave 2 de B can be written as 22 Mo&o at² 2x² Now since 2. 2B dx2 = B = B, cos(kx-wt) + 3, cos(kx twt) dz [619 in (he-wt) - B, ke Sin (k24 wt) = -Bje?cos(kx-wt) - Bek?cos(kx+wt)] ]

B a x² similarly 28 - wb, sin (ke-wt) - wēsmlkx+ wt) at wa E, cos(kx-wt)-W2 B, cos(kx + wt) = ot² =-w2 B will be B Hence our wave equation -K² B = - ME w² w? - (MoE) => ek? =(46 EJErwack] 7 k² k² Hence e²= M, &o = 3X10° = 0 or os Moco [hit x107x8.86810-12 to Hence speed of the speed of this ware is equal light D²B = 1 a² e² at²