Worked examples · Sheet 2

Harmonic waves and the relation between the fields

Theory used: Harmonic wave Relation between the electric field and the magnetic field Speed of electromagnetic waves Plane wave

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

A harmonic plane radio wave of frequency $f = 100 MHz$ propagates in vacuum, and the amplitude of its electric field is $E_{0} = 0.30 V/m$ . Compute the wavelength, the period, the angular frequency, the wave number and the amplitude of the magnetic field.

λ = \frac{c}{f}, T = \frac{1}{f}, ω = 2 π f, k = \frac{2 π}{λ}, B_{0} = \frac{E _{0}}{c}

λ T ω k B_{0} = \frac{3.00 \cdot 1 0 ^{8}}{1.00 \cdot 1 0 ^{8}} m = 3.00 m = \frac{1}{1.00 \cdot 1 0 ^{8}} s = 1.00 \cdot 1 0^{- 8} s = 2 π \cdot 1.00 \cdot 1 0^{8} rad/s \approx 6.28 \cdot 1 0^{8} rad/s = \frac{2 π}{3.00} rad/m \approx 2.09 rad/m = \frac{0.30}{3.00 \cdot 1 0 ^{8}} T = 1.0 \cdot 1 0^{- 9} T

In vacuum all electromagnetic waves propagate with speed $c$ , and for a harmonic wave $c = λ f$ , so the wavelength is the ratio of $c$ to the frequency. The period is the reciprocal of the frequency, the angular frequency is $2 π f$ and the wave number is $\frac{2 π}{λ}$ ; as a check, $\frac{ω}{k} = \frac{6.28 \cdot 1 0 ^{8}}{2.09} m/s \approx 3.00 \cdot 1 0^{8} m/s$ , which is exactly $c$ . By the relation between the electric field and the magnetic field of a plane wave, the amplitudes satisfy $E_{0} = c B_{0}$ , hence the magnetic field has an amplitude of only $1.0 nT$ .

Exercise 2

A harmonic plane wave of green light, with wavelength $λ = 500 nm$ , propagates in vacuum in the positive direction of the $x$ axis. Its electric field is directed along the $z$ axis and is

E (x, t) = E_{0} sin (k x - ω t) \hat{k}, E_{0} = 6.0 V/m .

Compute the frequency, angular frequency and wave number, and write the expression of the magnetic field $B (x, t)$ .

f = \frac{c}{λ}, ω = 2 π f, k = \frac{2 π}{λ}, B_{0} = \frac{E _{0}}{c}

\hat{k} \times (- \overset{}{^}) = \overset{}{^} ⟹ B (x, t) = - B_{0} sin (k x - ω t) \overset{}{^}

f ω k B_{0} = \frac{3.00 \cdot 1 0 ^{8}}{5.00 \cdot 1 0 ^{- 7}} Hz = 6.00 \cdot 1 0^{14} Hz = 2 π \cdot 6.00 \cdot 1 0^{14} rad/s \approx 3.77 \cdot 1 0^{15} rad/s = \frac{2 π}{5.00 \cdot 1 0 ^{- 7}} rad/m \approx 1.26 \cdot 1 0^{7} rad/m = \frac{6.0}{3.00 \cdot 1 0 ^{8}} T = 2.0 \cdot 1 0^{- 8} T

B (x, t) = - 2.0 \cdot 1 0^{- 8} sin (1.26 \cdot 1 0^{7} x - 3.77 \cdot 1 0^{15} t) \overset{}{^} T

Frequency, angular frequency and wave number follow from $c = λ f$ and from the definitions of the harmonic wave. For the magnetic field we use the relation between the electric field and the magnetic field: in a plane traveling wave $B$ oscillates in phase with $E$ , has magnitude $\frac{E}{c}$ , is perpendicular both to $E$ and to the direction of propagation, and $E \times B$ points in the direction of propagation. The electric field is along $\hat{k}$ and the wave travels toward $\overset{}{^}$ , so $B$ is along the $y$ axis. Since $\hat{k} \times \overset{}{^} = - \overset{}{^}$ and $\hat{k} \times (- \overset{}{^}) = \overset{}{^}$ , when $E_{z}$ is positive the magnetic field must point toward $- \overset{}{^}$ ; hence $B_{y} = - \frac{E _{z}}{c}$ and the magnetic field points along $- \overset{}{^}$ when $sin (k x - ω t) > 0$ .

Exercise 3

A harmonic plane wave propagates in vacuum in the positive direction of the $x$ axis, with the electric field directed along the $y$ axis. At the instant $t = 0$ the electric field is $E_{y} (x, 0) = E_{0} sin (k x)$ , with $E_{0} = 12 V/m$ , and two consecutive maxima are located at $x = 1.0 m$ and at $x = 5.0 m$ . Compute the wavelength and the frequency, the magnetic field vector at the point $x = 1.0 m$ at the instant $t = 0$ , and the first instant $t > 0$ at which the electric field at that point vanishes.

Electric field

E_{y}

(in

V/m

) as a function of

x

(in meters) at the instant

t = 0

: two consecutive maxima are one wavelength apart

λ = 5.0 m - 1.0 m = 4.0 m, f = \frac{c}{λ} = \frac{3.00 \cdot 1 0 ^{8}}{4.0} Hz = 7.5 \cdot 1 0^{7} Hz

E_{y} (x, t) = E_{0} sin (k x - ω t), B_{z} (x, t) = \frac{E _{0}}{c} sin (k x - ω t)

k \cdot 1.0 m = \frac{2 π}{4.0} \cdot 1.0 = \frac{π}{2} ⟹ B (1.0 m, 0) = \frac{E _{0}}{c} \hat{k} = \frac{12}{3.00 \cdot 1 0 ^{8}} \hat{k} T = 4.0 \cdot 1 0^{- 8} \hat{k} T

\frac{π}{2} - ω t = 0 ⟹ t = \frac{π}{2 ω} = \frac{T}{4} = \frac{1}{4 f} = \frac{1}{4 \cdot 7.5 \cdot 1 0 ^{7}} s \approx 3.3 \cdot 1 0^{- 9} s

At the instant $t = 0$ the field is periodic in space with period $λ$ , so the distance between two consecutive maxima is the wavelength, $4.0 m$ , and from $c = λ f$ it follows that $f = 75 MHz$ . The wave travels in the positive direction of the $x$ axis, hence its expression is $E_{0} sin (k x - ω t)$ , which for $t = 0$ gives back the given profile.

By the relation between the electric field and the magnetic field, with $E$ along $\overset{}{^}$ and propagation toward $\overset{}{^}$ , the magnetic field is $B_{z} = \frac{E _{y}}{c}$ along $\hat{k}$ . At the point $x = 1.0 m$ , at the instant $t = 0$ , the argument is $\frac{π}{2}$ and the electric field is maximum; since the two fields are in phase, the magnetic field is also maximum and equals $4.0 \cdot 1 0^{- 8} T$ in the direction of $\hat{k}$ .

At the same point the electric field vanishes for the first time when the argument goes from $\frac{π}{2}$ to $0$ , that is, after a quarter of a period, about 3.3 ns. The same result is obtained by observing that in that time the profile is translated by $c \frac{T}{4} = \frac{λ}{4} = 1.0 m$ , so the point of the profile that at $t = 0$ was at the origin, where the field is zero, arrives at $x = 1.0 m$ .