# A SIMPLE PLANE ELECTROMAGNETIC WAVE

A SIMPLE PLANE ELECTROMAGNETIC WAVE

Using an zyz-coordinate system (Fig. 33-2), we imagine that all space is divided into two regions by a plane perpendicular to the x-axis (parallel to the yz-plane). At every point to the left of this plane there are a uniform electric field E in the +y-direction and a uniform magnetic field jj in the +z-direction, as shown. Furthermore, we suppose that the boundary plane, which we call the wave front, moves to the right in the +x-direction
with a constant speed c, as yet unknown. Thus the E and jj fields travel to the right intopreviously field-free regions with a definite speed. The situation, in short, describes a rudimentary electromagnetic wave. A wave such as this, in which at any instant the fields are uniform over any plane perpendicular to the direction of propagation, is called a plane wave. In the case shown in Fig. 33-2, the fields are zero for planes to the right of
the wave front and have the same values on all planes to the left of the wave front; later we will consider more complex plane waves.

We won’t concern ourselves with the problem of actually producing such a field configuration. Instead, we simply ask whether it is consistent with the laws of electromagnetism, that is, with Maxwell’s equations. We’ll consider each of these four equations in turn. Let us first verify that our wave satisfies Maxwell’s first and second equations, that is, Gauss’s laws for electric and magnetic fields. To do this, we take as our Gaussian surface a rectangular box with sides parallel to the xy, zz, and yz coordinate planes (Fig. 33-3). The box encloses no electric charge, and you should be able to show that the total electric flux and magnetic flux through the box are both zero; this is true even if part of the box is in the region where E = B = O. This would not be the case·if E or jj had an x-component, parallel to the direction of propagation. We leave the proof as a problem (see Problem 33-30). Thus in order to satisfy Maxwell’s first and second equations, the electric and magnetic fields must be perpendicular to the direction of propagation; that is, the wave must be transverse.
The next of Maxwell’s equations to be considered is Faraday’s law:

To test whether our wave satisfies Faraday’s law, we apply this law to a rectangle efgh that is parallel to the xy-plane (Fig. 33-4a). As shown in Fig. 33-4b, a cross section in the xy-plane, this rectangle has height a and width ~x. At the time shown, the wave front
has progressed partway through the rectangle, and E is zero along the side ef In applying Faraday’s law we take the vector area dA of rectangle efgh to be in the +z-direction, With this choice the right-hand rule requires that we integrate E . dl counter clock wise.

Hence the left-hand side of Am..rere’s law, Eq. (33-5), is nonzero; ~e right-~and side must be nonzero as well. Thus E must have a y-component (perpendicular to B) so that the electric flux <l>Ethrough the rectangle and the time derivative d<l>idt can be nonzero. We come to the same conclusion that we inferred from Faraday’s law: In an electromagnetic wave, E and jj must be mutually perpendicular. In a time interval dt the electric flux <l>Ethrough the rectangle increases by d<l>E= E(ac dt). Since we-chose ciA to be in the +y-direction, this flux change is positive;