31.5.2 Faraday’s Law

Gauss’s law tells us that an electromagnetic wave has to be a transverse wave. What does Faraday’s law have to say? Faraday’s law is concerned with the changing magnetic flux through a closed curve, so let’s apply Faraday’s law to the narrow rectangle in the xy-plane shown in FIGURE 31.21. We’ll assume that $Δ x$ is so small that $\vec{B}$ is essentially constant over the width of the rectangle.

The figure illustrates a narrow rectangle kept in the x y plane.

FIGURE 31.21 Full Alternative Text

The magnetic field $\vec{B}$ is perpendicular to the rectangle, so the magnetic flux is $Φ_{m} = B_{z} A_{rectangle} = B_{z} h Δ x$ . As the wave moves, the flux changes at the rate

\frac{d Φ_{m}}{d t} = \frac{d}{d t} (B_{z} h Δ x) = \frac{\partial B_{z}}{\partial t} h Δ x

(31.23)

The ordinary derivative $d B_{z} / d t$ , which is the full rate of change of B from all possible causes, becomes a partial derivative $\partial B_{z} / \partial t$ in this situation because the change in magnetic flux is due entirely to the change of $B_{z}$ with time and not at all to the spatial variation of $B_{z}$ .

According to our sign convention, we need to go around the rectangle in a counterclockwise direction to make the flux positive. Thus we must also use a counterclockwise direction to evaluate the line integral:

\oint^{​} \vec{E} \cdot d \vec{s} = {\int^{​}}_{right}^{} \vec{E} \cdot d \vec{s} + {\int^{​}}_{top}^{} \vec{E} \cdot d \vec{s} + \int_{left}^{} \vec{E} \cdot d \vec{s} + {\int^{​}}_{bottom}^{} \vec{E} \cdot d \vec{s}

(31.24)

The electric field $\vec{E}$ points in the y-direction; hence $\vec{E} \cdot d \vec{s} = 0$ at all points on the top and bottom edges and these two integrals are zero.

Along the left edge of the loop, at position x, $\vec{E}$ has the same value at every point. Figure 31.21 shows that the direction of $\vec{E}$ is opposite to $d \vec{s};$ thus $\vec{E} \cdot d \vec{s} = - E_{y} (x) d s$ . On the right edge of the loop, at position $x + Δ x$ , $\vec{E}$ is parallel to $d \vec{s}$ , and $\vec{E} \cdot d \vec{s} =^{} E_{y} (x + Δ x) d s$ . Thus the line integral of $\vec{E} \cdot d \vec{s}$ around the rectangle is

\oint^{​} \vec{E} \cdot d \vec{s} = - E_{y} (x) h + E_{y} (x + Δ x) h = [​ ​ ​ E_{y} (x + Δ x) - E_{y} (x) ​ ​ ​] h

(31.25)

NOTE $E_{y} (x)$ indicates that $E_{y}$ is a function of the position x. It is not $E_{y}$ multiplied by x.

You learned in calculus that the derivative of the function $f (x)$ is

\frac{d f}{d x} = lim_{Δ x \to 0} [\frac{f (x + Δ x) - f (x)}{Δ x}]

We’ve assumed that $Δ x$ is very small. If we now let the width of the rectangle go to zero, $Δ x \to 0$ , Equation 31.25 becomes

\oint \vec{E} \cdot d \vec{s} = \frac{\partial E_{y}}{\partial x} h Δ x

(31.26)

We’ve used a partial derivative because $E_{y}$ is a function of both position x and time t.

Now, using Equations 31.23 and 31.26, we can write Faraday’s law as

\oint \vec{E} \cdot d \vec{s} = \frac{\partial E_{y}}{\partial x} h Δ x = - \frac{d Φ_{m}}{d t} = - \frac{\partial B_{z}}{\partial t} h Δ x

The area $h Δ x$ of the rectangle cancels, and we’re left with

\frac{\partial E_{y}}{\partial x} = - \frac{\partial B_{z}}{\partial t}

(31.27)

Equation 31.27, which compares the rate at which $E_{y}$ varies with position to the rate at which $B_{z}$ varies with time, is a required condition that an electromagnetic wave must satisfy to be consistent with Maxwell’s equations.